If a Sample of a(G) at 1.80 Atm Is Heated to 500 K, What Is the Pressure of B(G) at Equilibrium?

Chapter 13. Fundamental Equilibrium Concepts

thirteen.four Equilibrium Calculations

Learning Objectives

By the end of this section, you lot will be able to:

• Write equations representing changes in concentration and pressure for chemic species in equilibrium systems
• Utilize algebra to perform various types of equilibrium calculations

We know that at equilibrium, the value of the reaction quotient of whatever reaction is equal to its equilibrium abiding. Thus, nosotros can utilise the mathematical expression for Q to determine a number of quantities associated with a reaction at equilibrium or approaching equilibrium. While we accept learned to identify in which direction a reaction will shift to reach equilibrium, we want to extend that understanding to quantitative calculations. We practise so past evaluating the ways that the concentrations of products and reactants alter as a reaction approaches equilibrium, keeping in mind the stoichiometric ratios of the reaction. This algebraic arroyo to equilibrium calculations will be explored in this section.

Changes in concentrations or pressures of reactants and products occur as a reaction organisation approaches equilibrium. In this department we will see that nosotros tin can relate these changes to each other using the coefficients in the counterbalanced chemic equation describing the organisation. We apply the decomposition of ammonia every bit an example.

On heating, ammonia reversibly decomposes into nitrogen and hydrogen co-ordinate to this equation:

[latex]two\text{NH}_3(one thousand)\;{\rightleftharpoons}\;\text{North}_2(g)\;+\;3\text{H}_2(g)[/latex]

If a sample of ammonia decomposes in a closed system and the concentration of Due north₂ increases by 0.11 G, the modify in the N₂ concentration, Δ[N₂], the final concentration minus the initial concentration, is 0.11 M. The change is positive considering the concentration of N₂ increases.

The change in the H₂ concentration, Δ[H₂], is besides positive—the concentration of H₂ increases as ammonia decomposes. The chemical equation tells us that the change in the concentration of H₂ is three times the change in the concentration of N₂ considering for each mole of N_ii produced, 3 moles of H₂ are produced.

[latex]{\Delta}[\text{H}_2] = 3\;\times\;{\Delta}[\text{N}_2][/latex]

[latex]= 3\;\times\;(0.xi\;K) = 0.33\;K[/latex]

The change in concentration of NH_iii, Δ[NH₃], is twice that of Δ[N₂]; the equation indicates that two moles of NH_iii must decompose for each mole of N_two formed. Notwithstanding, the change in the NH₃ concentration is negative because the concentration of ammonia decreases as it decomposes.

[latex]{\Delta}[\text{NH}_3] = -2\;\times\;{\Delta}[\text{N}_2] = -ii\;\times\;(0.11\;M) = -0.22\;G[/latex]

We can relate these relationships direct to the coefficients in the equation

[latex]\begin{assortment}{ccccc} 2\text{NH}_3(thousand) & {\rightleftharpoons} & \text{Due north}_2(g) & + & three\text{H}_2(g) \\[0.5em] {\Delta}[\text{NH}_3] = -2\;\times\;{\Delta}[\text{N}_2] & & {\Delta}[\text{N}_2] = 0.11\;M & & {\Delta}[\text{H}_2] = 3\;\times\;{\Delta}[\text{N}_2] \cease{assortment}[/latex]

Note that all the changes on i side of the arrows are of the same sign and that all the changes on the other side of the arrows are of the opposite sign.

If we did not know the magnitude of the change in the concentration of North₂, we could represent it by the symbol x.

[latex]{\Delta}[\text{North}_2] = x[/latex]

The changes in the other concentrations would then be represented as:

[latex]{\Delta}[\text{H}_2] = 3\;\times\;{\Delta}[\text{N}_2] = 3x[/latex]

[latex]{\Delta}[\text{NH}_3] = -ii\;\times\;{\Delta}[\text{Northward}_2] = -2x[/latex]

The coefficients in the Δ terms are identical to those in the balanced equation for the reaction.

[latex]\begin{assortment}{ccccc} 2\text{NH}_3(yard) & {\rightleftharpoons} & \text{Due north}_2(k) & + & 3\text{H}_2(g) \\[0.5em] -2x & & x & & 3x \cease{array}[/latex]

The simplest mode for us to find the coefficients for the concentration changes in any reaction is to use the coefficients in the counterbalanced chemic equation. The sign of the coefficient is positive when the concentration increases; it is negative when the concentration decreases.

Case i

Determining Relative Changes in Concentration
Consummate the changes in concentrations for each of the following reactions.

(a) [latex]\begin{array}{lcccc} \text{C}_2\text{H}_2(g) & + & 2\text{Br}_2(g) & {\rightleftharpoons} & \text{C}_2\text{H}_2\text{Br}_4(1000) \\[0.5em] x & & \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} \end{array}[/latex]

(b) [latex]\begin{array}{lcccc} \text{I}_2(aq) & + & \text{I}^{-}(aq) & {\rightleftharpoons} & \text{I}_3^{\;\;-}(aq) \\[0.5em] \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} & & ten \end{array}[/latex]

(c) [latex]\begin{array}{lcccccc} \text{C}_3\text{H}_8(one thousand) & + & 5\text{O}_2(thou) & {\rightleftharpoons} & iii\text{CO}_2(g) & + & 4\text{H}_2\text{O}(g) \\[0.5em] x & & \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} \terminate{array}[/latex]

Solution
(a) [latex]\begin{assortment}{lcccc} \text{C}_2\text{H}_2(one thousand) & + & 2\text{Br}_2(g) & {\rightleftharpoons} & \text{C}_2\text{H}_2\text{Br}_4(1000) \\[0.5em] x & & 2x & & -x \end{array}[/latex]

(b) [latex]\begin{array}{lcccc} \text{I}_2(aq) & + & \text{I}^{-}(aq) & {\rightleftharpoons} & \text{I}_3^{\;\;-}(aq) \\[0.5em] -ten & & -x & & x \end{array}[/latex]

(c) [latex]\begin{array}{lcccccc} \text{C}_3\text{H}_8(g) & + & 5\text{O}_2(g) & {\rightleftharpoons} & 3\text{CO}_2(thousand) & + & 4\text{H}_2\text{O}(g) \\[0.5em] x & & 5x & & -3x & & -4x \end{assortment}[/latex]

Cheque Your Learning
Complete the changes in concentrations for each of the following reactions:

(a) [latex]\begin{assortment}{lcccc} ii\text{SO}_2(g) & + & \text{O}_2(g) & {\rightleftharpoons} & ii\text{And so}_3(g) \\[0.5em] \dominion[0ex]{ii.5em}{0.1ex} & & x & & \dominion[0ex]{2.5em}{0.1ex} \end{array}[/latex]

(b) [latex]\begin{array}{lcc} \text{C}_4\text{H}_8(1000) & {\rightleftharpoons} & 2\text{C}_2\text{H}_4(g) \\[0.5em] \rule[0ex]{2.5em}{0.1ex} & & -2x \end{array}[/latex]

(c) [latex]\brainstorm{array}{lcccccc} 4\text{NH}_3(chiliad) & + & 7\text{H}_2\text{O}(g) & {\rightleftharpoons} & 4\text{NO}_2(g) & + & vi\text{H}_2\text{O}(g) \\[0.5em] \rule[0ex]{two.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} & & \dominion[0ex]{ii.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} \end{array}[/latex]

Answer:

(a) twox, 10, −ii10; (b) x, −210; (c) ivten, vii10, −4x, −6ten or −fourten, −viix, 4ten, 6ten

Calculations Involving Equilibrium Concentrations

Because the value of the reaction caliber of whatever reaction at equilibrium is equal to its equilibrium constant, we tin use the mathematical expression for Q_c (i.e., the law of mass action) to determine a number of quantities associated with a reaction at equilibrium. It may aid if we proceed in mind that Q_c = K_c (at equilibrium) in all of these situations and that at that place are only three bones types of calculations:

1. Calculation of an equilibrium abiding. If concentrations of reactants and products at equilibrium are known, the value of the equilibrium constant for the reaction can be calculated.
2. Calculation of missing equilibrium concentrations. If the value of the equilibrium constant and all of the equilibrium concentrations, except one, are known, the remaining concentration tin can be calculated.
3. Calculation of equilibrium concentrations from initial concentrations. If the value of the equilibrium constant and a set of concentrations of reactants and products that are not at equilibrium are known, the concentrations at equilibrium can be calculated.

A like list could exist generated using Q_P , Grand_P , and partial pressure level. Nosotros will expect at solving each of these cases in sequence.

Calculation of an Equilibrium Constant

Since the police force of mass action is the merely equation nosotros have to depict the relationship between K_c and the concentrations of reactants and products, any problem that requires united states to solve for K_c must provide plenty information to make up one's mind the reactant and product concentrations at equilibrium. Armed with the concentrations, nosotros can solve the equation for Grand_c , equally information technology volition be the only unknown.

Example 2 showed us how to determine the equilibrium constant of a reaction if we know the concentrations of reactants and products at equilibrium. The following example shows how to use the stoichiometry of the reaction and a combination of initial concentrations and equilibrium concentrations to determine an equilibrium abiding. This technique, ordinarily called an ICE nautical chart—for Initial, Change, and Eastwardquilibrium–volition exist helpful in solving many equilibrium problems. A nautical chart is generated beginning with the equilibrium reaction in question. Underneath the reaction the initial concentrations of the reactants and products are listed—these conditions are usually provided in the problem and nosotros consider no shift toward equilibrium to have happened. The next row of information is the modify that occurs as the system shifts toward equilibrium—do not forget to consider the reaction stoichiometry as described in a previous section of this chapter. The concluding row contains the concentrations once equilibrium has been reached.

Example 2

Calculation of an Equilibrium Constant
Iodine molecules react reversibly with iodide ions to produce triiodide ions.

[latex]\text{I}_2(aq)\;+\;\text{I}^{-}(aq)\;{\rightleftharpoons}\;\text{I}_3^{\;\;-}(aq)[/latex]

If a solution with the concentrations of I₂ and I⁻ both equal to 1.000 × 10^−three Grand before reaction gives an equilibrium concentration of I₂ of 6.61 × ten⁻⁴ M, what is the equilibrium abiding for the reaction?

Solution
We volition brainstorm this problem by calculating the changes in concentration every bit the system goes to equilibrium. And so we determine the equilibrium concentrations and, finally, the equilibrium constant. First, we set a table with the initial concentrations, the changes in concentrations, and the equilibrium concentrations using −10 equally the change in concentration of I_two.

Since the equilibrium concentration of I_ii is given, we tin can solve for ten. At equilibrium the concentration of I_two is 6.61 × x^−iv K and then that

[latex]1.000\;\times\;10^{-three}\;-\;ten = vi.61\;\times\;x^{-4}[/latex]

[latex]x = ane.000\;\times\;x^{-3}\;-\;6.61\;\times\;ten^{-4}[/latex]

[latex]= iii.39\;\times\;10^{-4}\;M[/latex]

Now nosotros can fill in the tabular array with the concentrations at equilibrium.

Nosotros now summate the value of the equilibrium constant.

[latex]K_c = Q_c = \frac{[\text{I}_3^{\;\;-}]}{[\text{I}_2][\text{I}^{-}]}[/latex]

[latex]= \frac{3.39\;\times\;10^{-iv}\;M}{(6.61\;\times\;10^{-4}\;M)(6.61\;\times\;ten^{-4}\;M)} = 776[/latex]

Cheque Your Learning
Ethanol and acetic acid react and form h2o and ethyl acetate, the solvent responsible for the odor of some smash polish removers.

[latex]\text{C}_2\text{H}_5\text{OH}\;+\;\text{CH}_3\text{CO}_2\text{H}\;{\rightleftharpoons}\;\text{CH}_3\text{CO}_2\text{C}_2\text{H}_5\;+\;\text{H}_2\text{O}[/latex]

When i mol each of C_twoH₅OH and CH₃CO₂H are allowed to react in 1 L of the solvent dioxane, equilibrium is established when 1313 mol of each of the reactants remains. Summate the equilibrium constant for the reaction. (Note: Water is not a solvent in this reaction.)

Calculation of a Missing Equilibrium Concentration

If we know the equilibrium constant for a reaction and know the concentrations at equilibrium of all reactants and products except 1, nosotros tin calculate the missing concentration.

Case 3

Calculation of a Missing Equilibrium Concentration
Nitrogen oxides are air pollutants produced by the reaction of nitrogen and oxygen at loftier temperatures. At 2000 °C, the value of the equilibrium constant for the reaction, [latex]\text{Due north}_2(g)\;+\;\text{O}_2(k)\;{\rightleftharpoons}\;2\text{NO}(g)[/latex], is 4.1 × 10⁻⁴. Detect the concentration of NO(yard) in an equilibrium mixture with air at 1 atm pressure at this temperature. In air, [N_ii] = 0.036 mol/L and [O_two] 0.0089 mol/L.

Solution
Nosotros are given all of the equilibrium concentrations except that of NO. Thus, we can solve for the missing equilibrium concentration by rearranging the equation for the equilibrium constant.

[latex]K_c = Q_c = \frac{[\text{NO}]^two}{[\text{Due north}_2][\text{O}_2]}[/latex]

[latex][\text{NO}]^2 = K_c[\text{N}_2][\text{O}_2][/latex]

[latex][\text{NO}] = \sqrt{K_c[\text{N}_2][\text{O}_2]}[/latex]

[latex]= \sqrt{(4.1\;\times\;10^{-four})(0.036)(0.0089)}[/latex]

[latex]= \sqrt{i.31\;\times\;10^{-seven}}[/latex]

[latex]= 3.6\;\times\;10^{-four}[/latex]

Thus [NO] is 3.6 × 10⁻⁴ mol/L at equilibrium nether these conditions.

We tin can check our answer by substituting all equilibrium concentrations into the expression for the reaction quotient to see whether it is equal to the equilibrium constant.

[latex]Q_c = \frac{[\text{NO}]^2}{[\text{Due north}_2][\text{O}_2]}[/latex]

[latex]= \frac{(3.half dozen\;\times\;10^{-iv})^2}{(0.036)(0.0089)}[/latex]

[latex]Q_c = 4.0\;\times\;10^{-four} = K_c[/latex]

The reply checks; our calculated value gives the equilibrium abiding inside the mistake associated with the significant figures in the problem.

Check Your Learning
The equilibrium constant for the reaction of nitrogen and hydrogen to produce ammonia at a certain temperature is half dozen.00 × 10⁻². Calculate the equilibrium concentration of ammonia if the equilibrium concentrations of nitrogen and hydrogen are 4.26 M and 2.09 M, respectively.

Calculation of Changes in Concentration

If we know the equilibrium constant for a reaction and a set of concentrations of reactants and products that are not at equilibrium, we can summate the changes in concentrations as the system comes to equilibrium, as well every bit the new concentrations at equilibrium. The typical procedure tin can be summarized in four steps.

1. Decide the direction the reaction proceeds to come to equilibrium.
   1. Write a balanced chemic equation for the reaction.
   2. If the direction in which the reaction must proceed to accomplish equilibrium is not obvious, calculate Q_c from the initial concentrations and compare to K_c to make up one's mind the direction of change.
2. Make up one's mind the relative changes needed to reach equilibrium, and then write the equilibrium concentrations in terms of these changes.
   1. Define the changes in the initial concentrations that are needed for the reaction to accomplish equilibrium. Mostly, we correspond the smallest change with the symbol x and limited the other changes in terms of the smallest alter.
   2. Ascertain missing equilibrium concentrations in terms of the initial concentrations and the changes in concentration adamant in (a).
3. Solve for the alter and the equilibrium concentrations.
   1. Substitute the equilibrium concentrations into the expression for the equilibrium constant, solve for x, and cheque whatsoever assumptions used to detect x.
   2. Calculate the equilibrium concentrations.
4. Check the arithmetics.
   1. Bank check the calculated equilibrium concentrations by substituting them into the equilibrium expression and determining whether they give the equilibrium constant.

      Sometimes a detail step may differ from trouble to problem—it may be more circuitous in some problems and less complex in others. However, every calculation of equilibrium concentrations from a set of initial concentrations volition involve these steps.

      In solving equilibrium issues that involve changes in concentration, sometimes it is convenient to set an Water ice table, equally described in the previous department.

Example 4

Calculation of Concentration Changes as a Reaction Goes to Equilibrium
Nether certain conditions, the equilibrium constant for the decomposition of PCl₅(m) into PCl_iii(g) and Cl₂(g) is 0.0211. What are the equilibrium concentrations of PCl₅, PCl₃, and Cl_two if the initial concentration of PCl_five was ane.00 M?

Solution
Apply the stepwise process described before.

1. Determine the direction the reaction proceeds.

   The balanced equation for the decomposition of PCl_five is

   [latex]\text{PCl}_5(thousand)\;{\rightleftharpoons}\;\text{PCl}_3(g)\;+\;\text{Cl}_2(g)[/latex]

   Because we have no products initially, Q_c = 0 and the reaction volition proceed to the right.

2. Determine the relative changes needed to reach equilibrium, so write the equilibrium concentrations in terms of these changes.

   Allow us represent the increase in concentration of PCl_iii by the symbol 10. The other changes may be written in terms of x by because the coefficients in the chemical equation.

   [latex]\begin{array}{lcccc} \text{PCl}_5(m) & {\rightleftharpoons} & \text{PCl}_3(g) & + & \text{Cl}_2(k) \\[0.5em] -x & & x & & x \end{assortment}[/latex]

   The changes in concentration and the expressions for the equilibrium concentrations are:
   

3. Solve for the change and the equilibrium concentrations.

   Substituting the equilibrium concentrations into the equilibrium abiding equation gives

   [latex]K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} = 0.0211[/latex]

   [latex]= \frac{(ten)(x)}{(1.00\;-\;x)}[/latex]

   This equation contains but ane variable, x, the change in concentration. We can write the equation every bit a quadratic equation and solve for 10 using the quadratic formula.

   [latex]0.0211 = \frac{(ten)(10)}{(1.00\;-\;10)}[/latex]

   [latex]0.0211(1.00\;-\;ten) = ten^two[/latex]

   [latex]x^2\;+\;0.0211x\;-\;0.0211 = 0[/latex]

   Appendix B shows us an equation of the grade ax ² + bx + c = 0 tin can be rearranged to solve for x:

   [latex]x = \frac{-b\;{\pm}\;\sqrt{b^2\;-\;4ac}}{2a}[/latex]

   In this case, a = 1, b = 0.0211, and c = −0.0211. Substituting the advisable values for a, b, and c yields:

   [latex]x = \frac{-0.0211\;{\pm}\;\sqrt{(0.0211)^2\;-\;four(one)(-0.0211)}}{2(1)}[/latex]

   [latex]= \frac{-0.0211\;{\pm}\;\sqrt{(four.45\;\times\;x^{-four})\;+\;(viii.44\;\times\;10^{-two})}}{2}[/latex]

   [latex]= \frac{-0.0211\;{\pm}\;0.291}{two}[/latex]

   Hence

   [latex]x = \frac{-0.0211\;+\;0.291}{two} = 0.135[/latex]

   or

   [latex]10 = \frac{-0.0211\;-\;0.291}{two} = -0.156[/latex]

   Quadratic equations frequently have two different solutions, 1 that is physically possible and one that is physically impossible (an inapplicable root). In this case, the 2nd solution (−0.156) is physically impossible considering we know the change must exist a positive number (otherwise we would cease up with negative values for concentrations of the products). Thus, x = 0.135 M.

   The equilibrium concentrations are

   [latex][\text{PCl}_5] = 1.00\;-\;0.135 = 0.87\;One thousand[/latex]

   [latex][\text{PCl}_3] = x = 0.135\;Grand[/latex]

   [latex][\text{Cl}_2] = 10 = 0.135\;K[/latex]

4. Cheque the arithmetic.

   Exchange into the expression for Thou_c (to cheque the calculation) gives

   [latex]K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} = \frac{(0.135)(0.135)}{0.87} = 0.021[/latex]

   The equilibrium constant calculated from the equilibrium concentrations is equal to the value of K_c given in the problem (when rounded to the proper number of pregnant figures). Thus, the calculated equilibrium concentrations cheque.

Check Your Learning
Acetic acid, CH_iiiCO₂H, reacts with ethanol, C₂H_fiveOH, to grade water and ethyl acetate, CH_threeCO₂C_iiH₅.

[latex]\text{CH}_3\text{CO}_2\text{H}\;+\;\text{C}_2\text{H}_5\text{OH}\;{\leftrightharpoons}\;\text{CH}_3\text{CO}_2\text{C}_2\text{H}_5\;+\;\text{H}_2\text{O}[/latex]

The equilibrium abiding for this reaction with dioxane as a solvent is 4.0. What are the equilibrium concentrations when a mixture that is 0.15 M in CH_iiiCO₂H, 0.fifteen Yard in C_twoH₅OH, 0.twoscore Thousand in CH₃CO₂C₂H_five, and 0.40 M in H₂O are mixed in enough dioxane to make 1.0 L of solution?

Respond:

[CH_threeCO_iiH] = 0.36 M, [C₂H₅OH] = 0.36 M, [CH_threeCO₂C₂H₅] = 0.17 G, [H₂O] = 0.17 K

Check Your Learning
A i.00-L flask is filled with 1.00 moles of H_ii and 2.00 moles of I₂. The value of the equilibrium constant for the reaction of hydrogen and iodine reacting to form hydrogen iodide is 50.5 under the given conditions. What are the equilibrium concentrations of H₂, I₂, and How-do-you-do in moles/L?

[latex]\text{H}_2(g)\;+\;\text{I}_2(1000)\;{\leftrightharpoons}\;two\text{HI}(g)[/latex]

Answer:

[H₂] = 0.06 G, [I₂] = 1.06 Thou, [Hi] = one.88 G

Sometimes it is possible to use chemical insight to find solutions to equilibrium problems without really solving a quadratic (or more than complicated) equation. Beginning, however, information technology is useful to verify that equilibrium can be obtained starting from two extremes: all (or mostly) reactants and all (or more often than not) products (similar to what was shown in Figure two in Chapter 13.two Equilibrium Constants).

Consider the ionization of 0.150 M HA, a weak acrid.

[latex]\text{HA}(aq)\;{\rightleftharpoons}\;\text{H}^{+}(aq)\;+\;\text{A}^{-}(aq)\;\;\;\;\;\;\;K_c = 6.fourscore\;\times\;10^{-four}[/latex]

The most obvious mode to determine the equilibrium concentrations would be to start with only reactants. This could exist chosen the "all reactant" starting indicate. Using 10 for the amount of acrid ionized at equilibrium, this is the Ice tabular array and solution.

Setting upwardly and solving the quadratic equation gives

[latex]K_c = \frac{[\text{H}^{+}][\text{A}^{-}]}{[\text{HA}]} = \frac{(x)(x)}{(0.150\;-\;ten)} = 6.eighty\;\times\;10^{-4}[/latex]

[latex]x^2\;+\;(6.80\;\times\;x^{-4}x)\;-\;(1.02\;\times\;10^{-four}) = 0[/latex]

[latex]x = \frac{-vi.eighty\;\times\;10^{-4}\;{\pm}\;\sqrt{(6.80\;\times\;10^{-4})^2\;-\;(4)(ane)(-1.02\;\times\;10^{-4})}}{(ii)(1)}[/latex]

[latex]x = 0.00977\;M\;\text{or}\;-0.0104\;M[/latex]

Using the positive (physical) root, the equilibrium concentrations are

[latex][\text{HA}] = 0.150\;-\;10 = 0.140\;M[/latex]

[latex][\text{H}^{+}] = [\text{A}^{-}] = x = 0.00977\;M[/latex]

A less obvious way to solve the problem would be to assume all the HA ionizes first, then the system comes to equilibrium. This could exist chosen the "all product" starting point. Bold all of the HA ionizes gives

[latex][\text{HA}] = 0.150\;-\;0.150 = 0\;M[/latex]

[latex][\text{H}^{+}] = 0\;+\;0.150 = 0.150\;K[/latex]

[latex][\text{A}^{-}] = 0\;+\;0.150 = 0.150\;M[/latex]

Using these as initial concentrations and "y" to represent the concentration of HA at equilibrium, this is the ICE table for this starting point.

Setting up and solving the quadratic equation gives

[latex]K_c = \frac{[\text{H}^{+}][\text{A}^{-}]}{[\text{HA}]} = \frac{(0.150\;-\;y)(0.150\;-\;y)}{(y)} = six.80\;\times\;x^{-4}[/latex]

[latex]6.80\;\times\;10^{-4}y = 0.0225\;-\;0.300y\;+\;y^2[/latex]

Retain a few extra significant figures to minimize rounding problems.

[latex]y^ii\;-\;0.30068y\;+\;0.022500 = 0[/latex]

[latex]y = \frac{0.30068\;{\pm}\;\sqrt{(0.30068)^2\;-\;(4)(1)(0.022500)}}{(ii)(1)}[/latex]

[latex]y = \frac{0.30068\;{\pm}\;0.020210}{2}[/latex]

Rounding each solution to three significant figures gives

[latex]y = 0.160\;G\;\;\;\;\;\;\;\text{or}\;\;\;\;\;\;\;y = 0.140\;M[/latex]

Using the physically significant root (0.140 Yard) gives the equilibrium concentrations equally

[latex][\text{HA}] = y = 0.140\;Yard[/latex]

[latex][\text{H}^{+}] = 0.150\;-\;y = 0.010\;M[/latex]

[latex][\text{A}^{-}] = 0.150\;-\;y = 0.010\;M[/latex]

Thus, the 2 approaches give the aforementioned results (to 3 decimal places), and show that both starting points lead to the same equilibrium atmospheric condition. The "all reactant" starting bespeak resulted in a relatively small change (x) considering the system was shut to equilibrium, while the "all product" starting point had a relatively large change (y) that was about the size of the initial concentrations. Information technology can be said that a system that starts "close" to equilibrium volition require only a "small" change in conditions (x) to achieve equilibrium.

Think that a modest K_c means that very picayune of the reactants course products and a large Yard_c means that most of the reactants form products. If the organisation can be arranged and so it starts "shut" to equilibrium, and then if the change (10) is pocket-sized compared to whatsoever initial concentrations, it can be neglected. Small-scale is commonly defined as resulting in an fault of less than 5%. The following two examples demonstrate this.

Instance 5

Judge Solution Starting Close to Equilibrium
What are the concentrations at equilibrium of a 0.15 M solution of HCN?

[latex]\text{HCN}(aq)\;{\rightleftharpoons}\;\text{H}^{+}(aq)\;+\;\text{CN}^{-}(aq)\;\;\;\;\;\;\;K_c = 4.9\;\times\;ten^{-10}[/latex]

Solution
Using "10" to represent the concentration of each product at equilibrium gives this Ice table.

The exact solution may be obtained using the quadratic formula with

[latex]K_c = \frac{(10)(x)}{0.15\;-\;x}[/latex]

solving

[latex]x^ii\;+\;four.9\;\times\;ten^{-10}\;-\;7.35\;\times\;10^{-11} = 0[/latex]

[latex]x = 8.56\;\times\;10^{-6}\;M\;(3\;\text{sig.\;figs.}) = viii.6\;\times\;10^{-half dozen}\;M\;(2\;\text{sig.\;figs.})[/latex]

Thus [H⁺] = [CN^–] = ten = viii.6 × 10^{–half dozen} M and [HCN] = 0.15 – ten = 0.xv Yard.

In this case, chemical intuition can provide a simpler solution. From the equilibrium abiding and the initial conditions, x must be pocket-size compared to 0.15 K. More formally, if [latex]x\;{\ll}\;0.15[/latex], then 0.15 – x ≈ 0.15. If this assumption is true, then it simplifies obtaining x

[latex]K_c = \frac{(10)(x)}{0.15\;-\;x}\;{\approx}\;\frac{x^ii}{0.15}[/latex]

[latex]4.9\;\times\;10^{-10} = \frac{ten^2}{0.fifteen}[/latex]

[latex]x^2 = (0.15)(4.9\;\times\;10^{-ten}) = 7.4\;\times\;ten^{-11}[/latex]

[latex]10 = \sqrt{7.4\;\times\;ten^{-xi}} = eight.6\;\times\;ten^{-6}\;M[/latex]

In this example, solving the exact (quadratic) equation and using approximations gave the same result to ii significant figures. While virtually of the time the approximation is a flake dissimilar from the exact solution, as long as the error is less than 5%, the judge solution is considered valid. In this trouble, the 5% applies to IF (0.xv – 10) ≈ 0.15 1000, so if

[latex]\frac{10}{0.fifteen}\;\times\;100\% = \frac{8.6\;\times\;x^{-6}}{0.fifteen}\;\times\;100\% = 0.006\%[/latex]

is less than 5%, as it is in this instance, the supposition is valid. The approximate solution is thus a valid solution.

Bank check Your Learning
What are the equilibrium concentrations in a 0.25 1000 NH_three solution?

[latex]\text{NH}_3(aq)\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{NH}_4^{\;\;+}(aq)\;+\;\text{OH}^{-}(aq)\;\;\;\;\;\;\;K_c = one.8\;\times\;x^{-five}[/latex]

Assume that 10 is much less than 0.25 K and calculate the error in your assumption.

Answer:

[latex][\text{OH}^{-}] = [\text{NH}_4^{\;\;+}] = 0.0021\;G[/latex]; [NH₃] = 0.25 Thousand, fault = 0.84%

The second case requires that the original information be processed a bit, just it still tin exist solved using a small 10 approximation.

Example 6

Approximate Solution After Shifting Starting Concentration
Copper(II) ions form a complex ion in the presence of ammonia

[latex]\text{Cu}^{2+}(aq)\;+\;4\text{NH}_3(aq)\;{\rightleftharpoons}\;\text{Cu(NH}_3)_4^{\;\;2+}(aq)\;\;\;\;\;\;\;K_c = five.0\;\times\;10^{thirteen} = \frac{[\text{Cu(NH}_3)_4^{\;\;ii+}]}{[\text{Cu}^{ii+}(aq)][\text{NH}_3]^4}[/latex]

If 0.010 mol Cu^two+ is added to 1.00 L of a solution that is 1.00 M NH_three what are the concentrations when the organization comes to equilibrium?

Solution
The initial concentration of copper(Ii) is 0.010 M. The equilibrium constant is very large and so it would exist meliorate to beginning with equally much product as possible because "all products" is much closer to equilibrium than "all reactants." Note that Cu²⁺ is the limiting reactant; if all 0.010 M of it reacts to form product the concentrations would be

[latex][\text{Cu}^{2+}] = 0.010\;-\;0.010 = 0\;M[/latex]

[latex][\text{Cu(NH}_3)_4^{\;\;2+}] = 0.010\;M[/latex]

[latex][\text{NH}_3] = 1.00\;-\;4\;\times\;0.010 = 0.96\;M[/latex]

Using these "shifted" values as initial concentrations with x as the gratis copper(2) ion concentration at equilibrium gives this Ice tabular array.

Since we are starting close to equilibrium, 10 should be small so that

[latex]0.96\;+\;4x\;{\approx}\;0.96\;M[/latex]

[latex]0.010\;-\;10\;{\approx}\;0.010\;M[/latex]

[latex]K_c = \frac{(0.010\;-\;x)}{x(0.96\;-\;4x)^4}\;{\approx}\;\frac{(0.010)}{ten(0.96)^4} = v.0\;\times\;10^{13}[/latex]

[latex]x = \frac{(0.010)}{K_c(0.96)^4} = two.4\;\times\;ten^{-16}\;M[/latex]

Select the smallest concentration for the five% rule.

[latex]\frac{2.four\;\times\;ten^{-sixteen}}{0.010}\;\times\;100\% = ii\;\times\;x^{-12}\%[/latex]

This is much less than 5%, so the assumptions are valid. The concentrations at equilibrium are

[latex][\text{Cu}^{ii+}] = x = 2.4\;\times\;10^{-sixteen}\;1000[/latex]

[latex][\text{NH}_3] = 0.96\;-\;4x = 0.96\;M[/latex]

[latex][\text{Cu(NH}_3)_4^{\;\;2+}] = 0.010\;-\;x = 0.010\;M[/latex]

By starting with the maximum amount of product, this system was nearly equilibrium and the modify (10) was very small-scale. With only a small change required to get to equilibrium, the equation for ten was profoundly simplified and gave a valid effect well within the v% error maximum.

Bank check Your Learning
What are the equilibrium concentrations when 0.25 mol Ni²⁺ is added to i.00 L of 2.00 Thousand NH₃ solution?

[latex]\text{Ni}^{2+}(aq)\;+\;6\text{NH}_3(aq)\;{\rightleftharpoons}\;\text{Ni(NH}_3)_6^{\;\;two+}(aq)\;\;\;\;\;\;\;K_c = five.5\;\times\;10^eight[/latex]

With such a big equilibrium abiding, outset form as much product as possible, then assume that simply a pocket-sized amount (x) of the product shifts left. Calculate the error in your assumption.

Answer:

[latex][\text{Ni(NH}_3)_6^{\;\;2+}] = 0.25\;K[/latex], [NH_iii] = 0.l Thou, [Ni²⁺] = 2.9 × 10^–8 M, error = 1.two × 10^–five%

Key Concepts and Summary

The ratios of the rate of change in concentrations of a reaction are equal to the ratios of the coefficients in the balanced chemical equation. The sign of the coefficient of X is positive when the concentration increases and negative when it decreases. We learned to approach 3 basic types of equilibrium problems. When given the concentrations of the reactants and products at equilibrium, we can solve for the equilibrium constant; when given the equilibrium constant and some of the concentrations involved, we can solve for the missing concentrations; and when given the equilibrium constant and the initial concentrations, we can solve for the concentrations at equilibrium.

Chemistry End of Affiliate Exercises

1. A reaction is represented by this equation: [latex]\text{A}(aq)\;+\;ii\text{B}(aq)\;{\rightleftharpoons}\;2\text{C}(aq)\;\;\;\;\;\;\;K_c = one\;\times\;x^three[/latex]

   (a) Write the mathematical expression for the equilibrium constant.

   (b) Using concentrations ≤1 K, make up two sets of concentrations that describe a mixture of A, B, and C at equilibrium.

2. A reaction is represented by this equation: [latex]two\text{W}(aq)\;{\rightleftharpoons}\;\text{X}(aq)\;+\;two\text{Y}(aq)\;\;\;\;\;\;\;K_c = 5\;\times\;10^{-four}[/latex]

   (a) Write the mathematical expression for the equilibrium constant.

   (b) Using concentrations of ≤i M, make upwards two sets of concentrations that describe a mixture of Westward, 10, and Y at equilibrium.

3. What is the value of the equilibrium constant at 500 °C for the formation of NH₃ co-ordinate to the following equation?

   [latex]\text{North}_2(one thousand)\;+\;3\text{H}_2(yard)\;{\rightleftharpoons}\;2\text{NH}_3(g)[/latex]

   An equilibrium mixture of NH₃(g), H₂(m), and Northward₂(thousand) at 500 °C was found to contain 1.35 M H₂, 1.fifteen Thousand N_two, and 4.12 × 10^−ane G NH_three.

4. Hydrogen is prepared commercially by the reaction of methyl hydride and water vapor at elevated temperatures.

   [latex]\text{CH}_4(g)\;+\;\text{H}_2\text{O}(g)\;{\rightleftharpoons}\;3\text{H}_2(yard)\;+\;\text{CO}(thou)[/latex]

   What is the equilibrium constant for the reaction if a mixture at equilibrium contains gases with the following concentrations: CH_iv, 0.126 M; H₂O, 0.242 K; CO, 0.126 M; H₂ 1.15 M, at a temperature of 760 °C?

5. A 0.72-mol sample of PCl₅ is put into a 1.00-Fifty vessel and heated. At equilibrium, the vessel contains 0.xl mol of PCl₃(yard) and 0.40 mol of Cl_ii(g). Calculate the value of the equilibrium abiding for the decomposition of PCl₅ to PCl₃ and Cl_two at this temperature.
6. At 1 atm and 25 °C, NO_ii with an initial concentration of i.00 K is 3.3 × 10⁻³% decomposed into NO and O₂. Calculate the value of the equilibrium constant for the reaction.

   [latex]2\text{NO}_2(g)\;{\rightleftharpoons}\;ii\text{NO}(g)\;+\;\text{O}_2(yard)[/latex]

7. Summate the value of the equilibrium constant K_P for the reaction [latex]2\text{NO}(g)\;+\;\text{Cl}_2(thousand)\;{\rightleftharpoons}\;2\text{NOCl}(g)[/latex] from these equilibrium pressures: NO, 0.050 atm; Cl₂, 0.xxx atm; NOCl, ane.2 atm.
8. When heated, iodine vapor dissociates according to this equation:

   [latex]\text{I}_2(g)\;{\rightleftharpoons}\;2\text{I}(thou)[/latex]

   At 1274 K, a sample exhibits a partial pressure level of I_two of 0.1122 atm and a partial pressure due to I atoms of 0.1378 atm. Make up one's mind the value of the equilibrium constant, K_P , for the decomposition at 1274 Yard.

9. A sample of ammonium chloride was heated in a closed container.

   [latex]\text{NH}_4\text{Cl}(s)\;{\rightleftharpoons}\;\text{NH}_3(chiliad)\;+\;\text{HCl}(thou)[/latex]

   At equilibrium, the pressure level of NH₃(g) was found to exist 1.75 atm. What is the value of the equilibrium constant K_P for the decomposition at this temperature?

10. At a temperature of 60 °C, the vapor pressure level of water is 0.196 atm. What is the value of the equilibrium constant Chiliad_P for the transformation at 60 °C?

    [latex]\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{H}_2\text{O}(g)[/latex]

11. Complete the changes in concentrations (or pressure, if requested) for each of the following reactions.

    (a)

    [latex]\begin{assortment}{lcccc} two\text{SO}_3(g) & {\rightleftharpoons} & 2\text{SO}_2(g) & + & \text{O}_2(thou) \\[0.5em] \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} & & +ten \\[0.5em] \dominion[0ex]{2.5em}{0.1ex} & & \rule[0ex]{ii.5em}{0.1ex} & & 0.125\;M \finish{assortment}[/latex]

    (b)

    [latex]\brainstorm{array}{lcccccc} 4\text{NH}_3(one thousand) & + & 3\text{O}_2(g) & {\rightleftharpoons} & 2\text{Due north}_2(g) & + & 6\text{H}_2\text{O}(g) \\[0.5em] \rule[0ex]{2.5em}{0.1ex} & & 3x & & \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} \\[0.5em] \rule[0ex]{two.5em}{0.1ex} & & 0.24\;1000 & & \rule[0ex]{ii.5em}{0.1ex} & & \rule[0ex]{ii.5em}{0.1ex} \finish{array}[/latex]

    (c) Alter in force per unit area:

    [latex]\begin{array}{lcccc} two\text{CH}_4(g) & {\rightleftharpoons} & \text{C}_2\text{H}_2(g) & + & 3\text{H}_2(g) \\[0.5em] \dominion[0ex]{2.5em}{0.1ex} & & x & & \rule[0ex]{two.5em}{0.1ex} \\[0.5em] \rule[0ex]{2.5em}{0.1ex} & & 25\;\text{torr} & & \rule[0ex]{ii.5em}{0.1ex} \end{array}[/latex]

    (d) Alter in force per unit area:

    [latex]\begin{array}{lcccccc} \text{CH}_4(g) & + & \text{H}_2\text{O}(g) & {\rightleftharpoons} & \text{CO}(thou) & + & 3\text{H}_2(g) \\[0.5em] \dominion[0ex]{two.5em}{0.1ex} & & x & & \dominion[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} \\[0.5em] \rule[0ex]{two.5em}{0.1ex} & & five\;\text{atm} & & \dominion[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} \cease{assortment}[/latex]

    (e)

    [latex]\begin{array}{lcccc} \text{NH}_4\text{Cl}(s) & {\rightleftharpoons} & \text{NH}_3(chiliad) & + & \text{HCl}(g) \\[0.5em] & & x & & \rule[0ex]{2.5em}{0.1ex} \\[0.5em] & & 1.03\;\times\;10^{-four}\;M & & \rule[0ex]{two.5em}{0.1ex} \finish{array}[/latex]

    (f) change in pressure:

    [latex]\begin{assortment}{lcccc} \text{Ni}(s) & + & 4\text{CO}(g) & {\leftrightharpoons} & \text{Ni(CO)}_4(g) \\[0.5em] & & 4x & & \rule[0ex]{ii.5em}{0.1ex} \\[0.5em] & & 0.forty\;\text{atm} & & \dominion[0ex]{2.5em}{0.1ex} \end{array}[/latex]

12. Complete the changes in concentrations (or pressure, if requested) for each of the following reactions.

    (a)

    [latex]\begin{array}{lcccc} 2\text{H}_2(thou) & + & \text{O}_2(g) & {\rightleftharpoons} & two\text{H}_2\text{O}(k) \\[0.5em] \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} & &+2x \\[0.5em] \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} & & 1.50\;M \end{assortment}[/latex]

    (b)

    [latex]\begin{array}{lcccccc} \text{CS}_2(g) & + & 4\text{H}_2(yard) & {\rightleftharpoons} & \text{CH}_4(g) & + & 2\text{H}_2\text{S}(grand) \\[0.5em] x & & \dominion[0ex]{2.5em}{0.1ex} & & \rule[0ex]{ii.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} \\[0.5em] 0.020\;M & & \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} \end{array}[/latex]

    (c) Alter in pressure:

    [latex]\begin{assortment}{lcccc} \text{H}_2(thou) & + & \text{Cl}_2(chiliad) & {\rightleftharpoons} & 2\text{HCl}(g) \\[0.5em] x & & \dominion[0ex]{ii.5em}{0.1ex} & & \dominion[0ex]{2.5em}{0.1ex} \\[0.5em] 1.50\;\text{atm} & & \rule[0ex]{2.5em}{0.1ex} & & \dominion[0ex]{2.5em}{0.1ex} \terminate{array}[/latex]

    (d) Change in pressure:

    [latex]\begin{array}{lcccccc} ii\text{NH}_3(yard) & + & 2\text{O}_2(g) & {\rightleftharpoons} & \text{North}_2\text{O}(g) & + & 3\text{H}_2\text{O}(g) \\[0.5em] \rule[0ex]{2.5em}{0.1ex} & & \dominion[0ex]{two.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} & & ten \\[0.5em] \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} & & 60.vi\;\text{torr} \end{assortment}[/latex]

    (east)

    [latex]\brainstorm{array}{lcccc} \text{NH}_4\text{HS}(s) & {\leftrightharpoons} & \text{NH}_3(g) & + & \text{H}_2\text{S}(g) \\[0.5em] & & x & & \rule[0ex]{2.5em}{0.1ex} \\[0.5em] & & 9.8\;\times\;10^{-6}\;1000 & & \dominion[0ex]{2.5em}{0.1ex} \end{array}[/latex]

    (f) Modify in force per unit area:

    [latex]\begin{array}{lcccc} \text{Fe}(s) & + & five\text{CO}(1000) & {\leftrightharpoons} & \text{Atomic number 26(CO)}_4(g) \\[0.5em] & & \rule[0ex]{2.5em}{0.1ex} & & ten \\[0.5em] & & \rule[0ex]{2.5em}{0.1ex} & & 0.012\;\text{atm} \end{array}[/latex]

13. Why are at that place no changes specified for Ni in Practice xi, office (f)? What belongings of Ni does modify?
14. Why are in that location no changes specified for NH₄HS in Exercise 12, office (eastward)? What holding of NH₄HS does alter?
15. Analysis of the gases in a sealed reaction vessel containing NH_iii, N_two, and H₂ at equilibrium at 400 °C established the concentration of N₂ to exist one.2 Thou and the concentration of H_ii to be 0.24 M.

    [latex]\text{N}_2(thousand)\;+\;3\text{H}_2(1000)\;{\rightleftharpoons}\;2\text{NH}_3(1000)\;\;\;\;\;\;\;K_c = 0.50\;\text{at}\;400\;^{\circ}\text{C}[/latex]

    Calculate the equilibrium molar concentration of NH₃.

16. Summate the number of moles of Howdy that are at equilibrium with i.25 mol of H₂ and 1.25 mol of I₂ in a 5.00−L flask at 448 °C.

    [latex]\text{H}_2\;+\;\text{I}_2\;{\rightleftharpoons}\;two\text{Hello}\;\;\;\;\;\;\;K_c = 50.2\;\text{at}\;448\;^{\circ}\text{C}[/latex]

17. What is the pressure of BrCl in an equilibrium mixture of Cl_ii, Br₂, and BrCl if the pressure of Cl₂ in the mixture is 0.115 atm and the pressure of Br₂ in the mixture is 0.450 atm?

    [latex]\text{Cl}_2(thousand)\;+\;\text{Br}_2(g)\;{\rightleftharpoons}\;2\text{BrCl}(grand)\;\;\;\;\;\;\;K_P = four.seven\;\times\;ten^{-2}[/latex]

18. What is the force per unit area of CO_ii in a mixture at equilibrium that contains 0.50 atm H_ii, two.0 atm of H₂O, and 1.0 atm of CO at 990 °C?

    [latex]\text{H}_2(chiliad)\;+\;\text{CO}_2(g)\;{\rightleftharpoons}\;\text{H}_2\text{O}(thou)\;+\;\text{CO}(g)\;\;\;\;\;\;\;K_P = 1.six\;\text{at}\;990\;^{\circ}\text{C}[/latex]

19. Cobalt metal tin can be prepared by reducing cobalt(Ii) oxide with carbon monoxide.

    [latex]\text{CoO}(s)\;+\;\text{CO}(chiliad)\;{\rightleftharpoons}\;\text{Co}(s)\;+\;\text{CO}_2(chiliad)\;\;\;\;\;\;\;K_c = 4.90\;\times\;10^two\;\text{at}\;550\;^{\circ}\text{C}[/latex]

    What concentration of CO remains in an equilibrium mixture with [CO_ii] = 0.100 G?

20. Carbon reacts with water vapor at elevated temperatures.

    [latex]\text{C}(s)\;+\;\text{H}_2\text{O}(one thousand)\;{\rightleftharpoons}\;\text{CO}(thousand)\;+\;\text{H}_2(m)\;\;\;\;\;\;\;K_c = 0.ii\;\text{at}\;grand\;^{\circ}\text{C}[/latex]

    What is the concentration of CO in an equilibrium mixture with [H₂O] = 0.500 Thou at 1000 °C?

21. Sodium sulfate 10−hydrate, Na₂SO₄·10H₂O, dehydrates according to the equation

    [latex]\text{Na}_2\text{And then}_4{\cdot}x\text{H}_2\text{O}(due south)\;{\rightleftharpoons}\;\text{Na}_2\text{SO}_4(s)\;+\;10\text{H}_2\text{O}(g)\;\;\;\;\;\;\;K_P = 4.08\;\times\;ten^{-25}\;\text{at}\;25\;^{\circ}\text{C}[/latex]

    What is the pressure of water vapor at equilibrium with a mixture of Na₂So₄·10H₂O and NaSO₄?

22. Calcium chloride six−hydrate, CaCl₂·6H_iiO, dehydrates according to the equation

    [latex]\text{CaCl}_2{\cdot}six\text{H}_2\text{O}(south)\;{\rightleftharpoons}\;\text{CaCl}_2(s)\;+\;half-dozen\text{H}_2\text{O}(yard)\;\;\;\;\;\;\;K_P = 5.09\;\times\;10^{-44}\;\text{at}\;25\;^{\circ}\text{C}[/latex]

    What is the pressure level of h2o vapor at equilibrium with a mixture of CaCl₂·6H₂O and CaCl₂?

23. A student solved the post-obit problem and plant the equilibrium concentrations to exist [So_ii] = 0.590 M, [O_two] = 0.0450 Yard, and [SO₃] = 0.260 M. How could this student check the piece of work without reworking the problem? The problem was: For the following reaction at 600 °C:

    [latex]2\text{And then}_2(g)\;+\;\text{O}_2(thou)\;{\rightleftharpoons}\;2\text{SO}_3(m)\;\;\;\;\;\;\;K_c = 4.32[/latex]

    What are the equilibrium concentrations of all species in a mixture that was prepared with [SO₃] = 0.500 M, [And so₂] = 0 M, and [O₂] = 0.350 Thou?

24. A student solved the following problem and establish [N₂O_iv] = 0.16 M at equilibrium. How could this educatee recognize that the respond was wrong without reworking the problem? The trouble was: What is the equilibrium concentration of N₂O₄ in a mixture formed from a sample of NO_two with a concentration of 0.10 M?

    [latex]2\text{NO}_2(g)\;{\rightleftharpoons}\;\text{N}_2\text{O}_4(thousand)\;\;\;\;\;\;\;K_c = 160[/latex]

25. Assume that the change in concentration of N₂O₄ is minor enough to exist neglected in the following problem.
    (a) Calculate the equilibrium concentration of both species in i.00 L of a solution prepared from 0.129 mol of N₂O_iv with chloroform as the solvent.

    [latex]\text{Northward}_2\text{O}_4(m)\;{\leftrightharpoons}\;2\text{NO}_2(thou)\;\;\;\;\;\;\;K_c = 1.07\;\times\;10^{-v}[/latex] in chloroform

    (b) Prove that the change is minor plenty to be neglected.

26. Presume that the change in concentration of COCl_two is modest enough to exist neglected in the following problem.

    (a) Summate the equilibrium concentration of all species in an equilibrium mixture that results from the decomposition of COCl_ii with an initial concentration of 0.3166 Thousand.

    [latex]\text{COCl}_2(g)\;{\rightleftharpoons}\;\text{CO}(g)\;+\;\text{Cl}_2(g)\;\;\;\;\;\;\;K_c = 2.2\;\times\;10^{-ten}[/latex]

    (b) Show that the change is pocket-size enough to be neglected.

27. Assume that the change in pressure of H₂South is pocket-sized enough to be neglected in the post-obit trouble.

    (a) Calculate the equilibrium pressures of all species in an equilibrium mixture that results from the decomposition of H_iiSouth with an initial pressure level of 0.824 atm.

    [latex]2\text{H}_2\text{South}(m)\;{\rightleftharpoons}\;2\text{H}_2(g)\;+\;\text{S}_2(g)\;\;\;\;\;\;\;K_P = two.ii\;\times\;10^{-6}[/latex]

    (b) Prove that the change is small enough to be neglected.

28. What are all concentrations after a mixture that contains [H₂O] = 1.00 M and [Cl₂O] = ane.00 M comes to equilibrium at 25 °C?

    [latex]\text{H}_2\text{O}(g)\;+\;\text{Cl}_2\text{O}(g)\;{\rightleftharpoons}\;2\text{HOCl}(g)\;\;\;\;\;\;\;K_c = 0.0900[/latex]

29. What are the concentrations of PCl₅, PCl_iii, and Cl₂ in an equilibrium mixture produced by the decomposition of a sample of pure PCl₅ with [PCl₅] = two.00 Thousand?

    [latex]\text{PCl}_5(m)\;{\rightleftharpoons}\;\text{PCl}_3(thou)\;+\;\text{Cl}_2(g)\;\;\;\;\;\;\;K_c = 0.0211[/latex]

30. Calculate the pressures of all species at equilibrium in a mixture of NOCl, NO, and Cl₂ produced when a sample of NOCl with a pressure level of 10.0 atm comes to equilibrium according to this reaction:
    [latex]2\text{NOCl}(chiliad)\;{\rightleftharpoons}\;2\text{NO}(g)\;+\;\text{Cl}_2(g)\;\;\;\;\;\;\;K_P = 4.0\;\times\;10^{-iv}[/latex]
31. Calculate the equilibrium concentrations of NO, O₂, and NO₂ in a mixture at 250 °C that results from the reaction of 0.twenty M NO and 0.x One thousand O_two. (Hint: K is big; assume the reaction goes to completion then comes back to equilibrium.)

    [latex]2\text{NO}(g)\;+\;\text{O}_2(chiliad)\;{\rightleftharpoons}\;two\text{NO}_2(yard)\;\;\;\;\;\;\;K_c = 2.iii\;\times\;10^5\;\text{at}\;250\;^{\circ}\text{C}[/latex]

32. Calculate the equilibrium concentrations that consequence when 0.25 K O₂ and 1.0 M HCl react and come to equilibrium.

    [latex]4\text{HCl}(g)\;+\;\text{O}_2(m)\;{\rightleftharpoons}\;2\text{Cl}_2(g)\;+\;ii\text{H}_2\text{O}(g)\;\;\;\;\;\;\;K_c = 3.1\;\times\;x^{13}[/latex]

33. One of the important reactions in the formation of smog is represented by the equation

    [latex]\text{O}_3(g)\;+\;\text{NO}(m)\;{\rightleftharpoons}\;\text{NO}_2(g)\;+\;\text{O}_2(g)\;\;\;\;\;\;\;K_P = vi.0\;\times\;ten^{34}[/latex]

    What is the pressure of O_iii remaining after a mixture of O₃ with a pressure of 1.2 × 10⁻⁸ atm and NO with a force per unit area of ane.2 × ten⁻⁸ atm comes to equilibrium? (Hint: K_P is large; assume the reaction goes to completion and so comes back to equilibrium.)

34. Summate the pressures of NO, Cl₂, and NOCl in an equilibrium mixture produced by the reaction of a starting mixture with iv.0 atm NO and 2.0 atm Cl_ii. (Hint: Thousand_P is pocket-sized; presume the reverse reaction goes to completion then comes back to equilibrium.)

    [latex]ii\text{NO}(g)\;+\;\text{Cl}_2(k)\;{\rightleftharpoons}\;two\text{NOCl}(g)\;\;\;\;\;\;\;K_P = 2.5\;\times\;10^3[/latex]

35. Calculate the number of grams of HI that are at equilibrium with i.25 mol of H₂ and 63.v k of iodine at 448 °C.

    [latex]\text{H}_2\;+\;\text{I}_2\;{\rightleftharpoons}\;two\text{Hello}\;\;\;\;\;\;\;K_c = 50.2\;\text{at}\;448\;^{\circ}\text{C}[/latex]

36. Butane exists as 2 isomers, due north−butane and isobutane.
    

    One thousand_P = ii.v at 25 °C

    What is the pressure of isobutane in a container of the two isomers at equilibrium with a total force per unit area of 1.22 atm?

37. What is the minimum mass of CaCO₃ required to establish equilibrium at a certain temperature in a half dozen.50-Fifty container if the equilibrium constant (1000_c ) is 0.050 for the decomposition reaction of CaCO_iii at that temperature?

    [latex]\text{CaCO}_3(s)\;{\rightleftharpoons}\;\text{CaO}(due south)\;+\;\text{CO}_2(g)[/latex]

38. The equilibrium constant (K_c ) for this reaction is 1.lx at 990 °C:

    [latex]\text{H}_2(grand)\;+\;\text{CO}_2(k)\;{\rightleftharpoons}\;\text{H}_2\text{O}(thousand)\;+\;\text{CO}(1000)[/latex]

    Summate the number of moles of each component in the final equilibrium mixture obtained from adding 1.00 mol of H₂, ii.00 mol of CO_ii, 0.750 mol of H_iiO, and 1.00 mol of CO to a 5.00-L container at 990 °C.

39. At 25 °C and at 1 atm, the partial pressures in an equilibrium mixture of Due north₂O₄ and NO₂ are [latex]\text{P}_{\text{N}_2\text{O}_4} = 0.seventy\;\text{atm}[/latex] and [latex]\text{P}_{\text{NO}_2} = 0.30\;\text{atm}[/latex].

    (a) Predict how the pressures of NO_ii and N₂O₄ will change if the total pressure increases to nine.0 atm. Volition they increase, subtract, or remain the same?

    (b) Calculate the partial pressures of NO₂ and North_iiO₄ when they are at equilibrium at 9.0 atm and 25 °C.

40. In a 3.0-L vessel, the following equilibrium partial pressures are measured: North₂, 190 torr; H₂, 317 torr; NH_iii, 1.00 × 10³ torr.

    [latex]\text{N}_2(g)\;+\;3\text{H}_2(k)\;{\rightleftharpoons}\;2\text{NH}_3(yard)[/latex]

    (a) How volition the partial pressures of H_ii, N_two, and NH₃ change if H₂ is removed from the arrangement? Volition they increase, decrease, or remain the aforementioned?

    (b) Hydrogen is removed from the vessel until the partial force per unit area of nitrogen, at equilibrium, is 250 torr. Calculate the partial pressures of the other substances under the new conditions.

41. The equilibrium constant (K_c ) for this reaction is 5.0 at a given temperature.

    [latex]\text{CO}(thousand)\;+\;\text{H}_2\text{O}(k)\;{\rightleftharpoons}\;\text{CO}_2(g)\;+\;\text{H}_2(k)[/latex]

    (a) On analysis, an equilibrium mixture of the substances present at the given temperature was found to contain 0.20 mol of CO, 0.30 mol of water vapor, and 0.ninety mol of H₂ in a liter. How many moles of CO_two were in that location in the equilibrium mixture?

    (b) Maintaining the aforementioned temperature, additional H_two was added to the system, and some h2o vapor was removed past drying. A new equilibrium mixture was thereby established containing 0.40 mol of CO, 0.thirty mol of water vapor, and 1.two mol of H₂ in a liter. How many moles of CO₂ were in the new equilibrium mixture? Compare this with the quantity in part (a), and discuss whether the second value is reasonable. Explain how it is possible for the water vapor concentration to exist the same in the two equilibrium solutions even though some vapor was removed before the second equilibrium was established.

42. Antimony pentachloride decomposes according to this equation:

    [latex]\text{SbCl}_5(g)\;{\rightleftharpoons}\;\text{SbCl}_3(g)\;+\;\text{Cl}_2(1000)[/latex]

    An equilibrium mixture in a 5.00-50 flask at 448 °C contains 3.85 g of SbCl_five, ix.14 g of SbCl₃, and 2.84 g of Cl₂. How many grams of each will be establish if the mixture is transferred into a 2.00-L flask at the same temperature?

43. Consider the reaction between H_two and O_two at 1000 Thou
    [latex]2\text{H}_2(g)\;+\;\text{O}_2(g)\;{\rightleftharpoons}\;2\text{H}_2\text{O}(g)\;\;\;\;\;\;\;K_P = \frac{(P_{\text{H}_2\text{O}})^2}{(P_{\text{O}_2})(P_{\text{H}_2})^3} = ane.33\;\times\;10^{twenty}[/latex]

    If 0.500 atm of H₂ and 0.500 atm of O_two are allowed to come to equilibrium at this temperature, what are the partial pressures of the components?

44. An equilibrium is established according to the following equation

    [latex]\text{Hg}_2^{\;\;2+}(aq)\;+\;\text{NO}_3^{\;\;-}(aq)\;+\;three\text{H}^{+}(aq)\;{\rightleftharpoons}\;2\text{Hg}^{2+}(aq)\;+\;\text{HNO}_2(aq)\;+\;\text{H}_2\text{O}(l)\;\;\;\;\;\;\;K_c = iv.half dozen[/latex]

    What will happen in a solution that is 0.20 M each in [latex]\text{Hg}_2^{\;\;2+}[/latex], [latex]\text{NO}_3^{\;\;-}[/latex], H⁺, Hg²⁺, and HNO₂?

    (a) [latex]\text{Hg}_2^{\;\;2+}[/latex] volition be oxidized and [latex]\text{NO}_3^{\;\;-}[/latex] reduced.

    (b) [latex]\text{Hg}_2^{\;\;2+}[/latex] will exist reduced and [latex]\text{NO}_3^{\;\;-}[/latex] oxidized.

    (c) Hg²⁺ will be oxidized and HNO₂ reduced.

    (d) Hg²⁺ will be reduced and HNO₂ oxidized.

    (eastward) At that place will be no alter because all reactants and products take an activity of 1.

45. Consider the equilibrium

    [latex]4\text{NO}_2(thousand)\;+\;vi\text{H}_2\text{O}(thou)\;{\rightleftharpoons}\;4\text{NH}_3(yard)\;+\;seven\text{O}_2(thou)[/latex]

    (a) What is the expression for the equilibrium constant (Yard_c ) of the reaction?

    (b) How must the concentration of NH_iii change to reach equilibrium if the reaction caliber is less than the equilibrium constant?

    (c) If the reaction were at equilibrium, how would a decrease in pressure level (from an increment in the volume of the reaction vessel) affect the pressure of NO₂?

    (d) If the change in the pressure of NO₂ is 28 torr as a mixture of the four gases reaches equilibrium, how much will the pressure of O₂ modify?

46. The binding of oxygen by hemoglobin (Hb), giving oxyhemoglobin (HbO₂), is partially regulated by the concentration of H_threeO⁺ and dissolved CO₂ in the blood. Although the equilibrium is complicated, it can be summarized as

    [latex]\text{HbO}_2(aq)\;+\;\text{H}_3\text{O}^{+}(aq)\;+\;\text{CO}_2(g)\;{\rightleftharpoons}\;\text{CO}_2\;-\;\text{Hb}\;-\;\text{H}^{+}\;+\;\text{O}_2(yard)\;+\;\text{H}_2\text{O}(l)[/latex]

    (a) Write the equilibrium abiding expression for this reaction.

    (b) Explicate why the production of lactic acid and CO₂ in a muscle during exertion stimulates release of O₂ from the oxyhemoglobin in the claret passing through the muscle.

47. The hydrolysis of the saccharide sucrose to the sugars glucose and fructose follows a first-order rate equation for the disappearance of sucrose.

    [latex]\text{C}_{12}\text{H}_{22}\text{O}_{11}(aq)\;+\;\text{H}_2\text{O}(l)\;{\longrightarrow}\;\text{C}_6\text{H}_{12}\text{O}_6(aq)\;+\;\text{C}_6\text{H}_{12}\text{O}_6(aq)[/latex]

    Rate = k[C₁₂H₂₂O_xi]

    In neutral solution, thou = 2.ane × 10⁻¹¹/s at 27 °C. (Every bit indicated by the rate constant, this is a very tiresome reaction. In the human trunk, the charge per unit of this reaction is sped up past a type of goad called an enzyme.) (Note: That is not a fault in the equation—the products of the reaction, glucose and fructose, accept the same molecular formulas, C_viH₁₂O₆, just differ in the organisation of the atoms in their molecules). The equilibrium constant for the reaction is 1.36 × 10⁵ at 27 °C. What are the concentrations of glucose, fructose, and sucrose afterward a 0.150 Thousand aqueous solution of sucrose has reached equilibrium? Call back that the activity of a solvent (the constructive concentration) is 1.

48. The density of trifluoroacetic acid vapor was adamant at 118.1 °C and 468.five torr, and found to be 2.784 k/50. Calculate Chiliad_c for the association of the acid.
    
49. Liquid N_iiO₃ is dark blueish at depression temperatures, but the color fades and becomes greenish at higher temperatures as the compound decomposes to NO and NO₂. At 25 °C, a value of One thousand_P = 1.91 has been established for this decomposition. If 0.236 moles of Due north₂O_three are placed in a 1.52-50 vessel at 25 °C, summate the equilibrium partial pressures of North₂O₃(g), NO₂(g), and NO(g).
50. A ane.00-50 vessel at 400 °C contains the following equilibrium concentrations: N₂, 1.00 M; H₂, 0.50 M; and NH₃, 0.25 M. How many moles of hydrogen must exist removed from the vessel to increment the concentration of nitrogen to 1.1 M?
51. A 0.010 M solution of the weak acid HA has an osmotic pressure (see chapter on solutions and colloids) of 0.293 atm at 25 °C. A 0.010 M solution of the weak acrid HB has an osmotic pressure of 0.345 atm nether the same conditions.

    (a) Which acid has the larger equilibrium constant for ionization

    HA [latex][\text{HA}(aq)\;{\rightleftharpoons}\;\text{A}^{-}(aq)\;+\;\text{H}^{+}(aq)][/latex] or HB [latex][\text{HB}(aq)\;{\rightleftharpoons}\;\text{H}^{+}(aq)\;+\;\text{B}^{-}(aq)][/latex]?

    (b) What are the equilibrium constants for the ionization of these acids?

    (Hint: Remember that each solution contains three dissolved species: the weak acrid (HA or HB), the cohabit base (A⁻ or B⁻), and the hydrogen ion (H⁺). Call up that osmotic pressure level (like all colligative properties) is related to the total number of solute particles. Specifically for osmotic pressure, those concentrations are described by molarities.)

Solutions

Answers to Chemistry End of Chapter Exercises

1. [latex]K_c = \frac{[\text{C}]^two}{[\text{A}][\text{B}]^ii}[/latex]. [A] = 0.1 Chiliad, [B] = 0.1 Yard, [C] = 1 M; and [A] = 0.01, [B] = 0.250, [C] = 0.791.

three. K_c = 6.00 × 10^−two

5. 1000_c = 0.50

7. The equilibrium equation is
K_P = one.9 × 10ⁱⁱⁱ

9. Chiliad_P = 3.06

11. (a) −210, 2x, −0.250 M, 0.250 One thousand; (b) 4ten, −2ten, −610, 0.32 M, −0.16 Chiliad, −0.48 Thou; (c) −2x, iiix, −50 torr, 75 torr; (d) x, − 10, −3x, 5 atm, −5 atm, −15 atm; (due east) 10, ane.03 × x⁻⁴ Yard; (f) x, 0.i atm.

thirteen. Activities of pure crystalline solids equal one and are constant; however, the mass of Ni does change.

15. [NH_three] = nine.i × 10⁻ⁱⁱ M

17. P _BrCl = iv.9 × 10^−two atm

19. [CO] = two.0 × ten⁻⁴ M

21. [latex]P_{\text{H}_2\text{O}} = iii.64\;\times\;ten^{-3}\;\text{atm}[/latex]

23. Calculate Q based on the calculated concentrations and run across if it is equal to K_c . Because Q does equal 4.32, the system must exist at equilibrium.

25. (a) [NO₂] = 1.17 × 10⁻³ M
[N_twoO₄] = 0.128 Thousand
(b) Percentage error [latex]= \frac{5.87\;\times\;10^{-4}}{0.129}\;\times\;100\% = 0.455\%[/latex]. The change in concentration of N₂O_iv is far less than the five% maximum immune.

27. (a) [H₂S] = 0.810 atm
[H₂] = 0.014 atm
[S₂] = 0.0072 atm
(b) The 2x is dropped from the equilibrium adding because 0.014 is negligible when subtracted from 0.824. The percent error associated with ignoring twox is [latex]\frac{0.014}{0.824}\;\times\;100\% = 1.7\%[/latex], which is less than allowed past the "5% exam." The error is, indeed, negligible.

29. [PCl₃] = 1.80 M; [PC₃] = 0.195 M; [PCl₃] = 0.195 M.

31. [NO₂] = 0.19 M
[NO] = 0.0070 1000
[O₂] = 0.0035 G

33. [latex]P_{\text{O}_3} = 4.ix\;\times\;10^{-26}\;\text{atm}[/latex]

35. 507 g

37. 330 thou

39. (a) Both gases must increment in pressure.
(b)[latex]P_{\text{N}_2\text{O}_4} = 8.0\;\text{atm\;and}\;P_{\text{NO}_2} = 1.0\;\text{atm}[/latex]

41. (a) 0.33 mol.
(b) [CO]ⁱⁱ = 0.50 Yard Added H₂ forms some water to compensate for the removal of water vapor and every bit a consequence of a shift to the left after H₂ is added.

43. [latex]P_{\text{H}_2} = viii.64\;\times\;ten^{-11}\;\text{atm}[/latex]
[latex]P_{\text{O}_2} = 0.250\;\text{atm}[/latex]
[latex]P_{\text{H}_2\text{O}} = 0.500\;\text{atm}[/latex]

45. (a) [latex]K_c = \frac{[\text{NH}_3]^4[\text{O}_2]^7}{[\text{NO}_2]^4[\text{H}_2\text{O}]^half-dozen}[/latex]. (b) [NH_iii] must increment for Q_c to reach Yard_c . (c) That decrease in pressure would decrease [NO_ii]. (d) [latex]P_{\text{O}_2} = 49\;\text{torr}[/latex]

47. [fructose] = 0.15 M

49. [latex]P_{\text{Northward}_2\text{O}_3} = ane.90\;\text{atm\;and}\;P_{\text{NO}} = P_{\text{NO}_2} = ane.90\;\text{atm}[/latex]

51. (a) HB ionizes to a greater degree and has the larger One thousand_c .
(b) K_c (HA) = v × 10⁻⁴
One thousand_c (HB) = 3 × 10⁻³

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Source: https://opentextbc.ca/chemistry/chapter/13-4-equilibrium-calculations/

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