## General Chemistry

This equilibrium practice problem set includes questions on writing the equilibrium constant of given chemical reactions, determining the value of equilibrium constant based on the concentrations and partial pressures of gases, deriving a new expression for an equilibrium constant from separate reactions, converting between Kc and Kp, calculating the quotient and determining the course of reaction,  calculations of concentrations based on the quotient and equilibrium constant, as well as working on equilibrium reactions based on the Le Châtelier’s principle.

#### Practice

1.

Write the equilibrium expression (K) for each of the following reactions.

A) N2(g) + O2(g) ⇆ 2NO(g)

B) 2NOCl(g) ⇆ 2NO(g) + Cl2(g)

C) CH3COOH(aq) ⇆ H+(aq) + CH3COO2(aq)

D) 2CuS(s) + 3O2(g) ⇆ 2CuO(s) + 2SO2(g)

E) CO2(g) + 3H2(g) ⇆ CH3OH(g) + H2O(g)

F) 2SO3(g) ⇆ 2SO2(g) + O2(g)

G) CaCO3(s) ⇆ CaO(s) + CO2(g)

H) Fe2O3(s) + 3CO(g) ⇆ 2Fe(s) + 3CO2(g)

I) 4HCl(g) + O2(g) ⇆ 2Cl2(g) + 2H2O(g)

J) CO32-(aq) + H2O(l) ⇆ HCO3(aq) + OH(aq)

K) P4(s) + 5O2(g) ⇆ P4O10(s)

A) The equilibrium constant is the ratio of the equilibrium concentrations of the products and reactants, each raised to a power equal to its stoichiometric coefficient in the balanced equation.

This is a gaseous reaction, and therefore, we can also express the equilibrium constant in terms of the partial pressures of the reactants and products.

N2(g) + O2(g) ⇆ 2NO(g)

$K = \frac{{{{[{\rm{NO}}]}^{\rm{2}}}}}{{[{{\rm{N}}_{\rm{2}}}][{{\rm{O}}_{\rm{2}}}]}}$

${K_p} = \frac{{{P^{\rm{2}}}_{{\rm{NO}}}}}{{{P_{{{\rm{N}}_{\rm{2}}}}}{P_{{{\rm{O}}_{\rm{2}}}}}}}$

B) All the reactants and products are in a gas state and therefore, both K (Kc) and Kp are applicable to express the equilibrium.

2NOCl(g) ⇆ 2NO(g) + Cl2(g)

$K = \frac{{{{[{\rm{NO}}]}^{\rm{2}}}[{\rm{C}}{{\rm{l}}_{\rm{2}}}]}}{{{{[{\rm{NOCl}}]}^{\rm{2}}}}}$

${K_p} = \frac{{{P^{\rm{2}}}_{{\rm{NO}}}{P_{{\rm{C}}{{\rm{l}}_{\rm{2}}}}}}}{{{P^2}_{{\rm{NOCl}}}}}$

C) Kp is not applicable for this reaction

CH3COOH(aq) ⇆ H+(aq) + CH3COO2(aq)

$K = \frac{{[{{\rm{H}}^ + }][{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{CO}}{{\rm{O}}_{\rm{2}}}^{\rm{–}}]}}{{{{[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{CO}}{{\rm{O}}_{\rm{2}}}{\rm{H}}]}^{}}}}$

D) All the reactants in a gas state are included in the expression for K (Kc) and Kp. Solids and liquids are omitted from the equilibrium expression:

2CuS(s) + 3O2(g) ⇆ 2CuO(s) + 2SO2(g)

$K = \frac{{{{[{\rm{S}}{{\rm{O}}_{\rm{2}}}]}^{\rm{2}}}}}{{{{[{{\rm{O}}_{\rm{2}}}]}^{\rm{3}}}}}$

${K_p} = \frac{{{P^{\rm{2}}}{{_{{\rm{SO}}}}_{_2}}}}{{{P^3}_{{{\rm{O}}_{\rm{3}}}}}}$

E) All the reactants and products are in a gas state and therefore, both K (Kc) and Kp are applicable to express the equilibrium.

CO2(g) + 3H2(g) ⇆ CH3OH(g) + H2O(g)

$K = \frac{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{OH][}}{{\rm{H}}_{\rm{2}}}{\rm{O]}}}}{{{\rm{[C}}{{\rm{O}}_{\rm{2}}}{\rm{][}}{{\rm{H}}_{\rm{2}}}{{\rm{]}}^{\rm{3}}}}}$

${K_p} = \frac{{{P_{{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{OH}}}}{P_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}}}{{{P_{{\rm{C}}{{\rm{O}}_{\rm{2}}}}}{P^3}_{{{\rm{H}}_{\rm{2}}}}}}$

F) All the reactants and products are in a gas state and therefore, both K (Kc) and Kp are applicable to express the equilibrium.

2SO3(g) ⇆ 2SO2(g) + O2(g)

$K = \frac{{{{[{\rm{S}}{{\rm{O}}_{\rm{2}}}]}^{\rm{2}}}[{{\rm{O}}_{\rm{2}}}]}}{{{{[{\rm{S}}{{\rm{O}}_{\rm{3}}}]}^{\rm{2}}}}}$

${K_p} = \frac{{{P^{\rm{2}}}{{_{{\rm{SO}}}}_{_2}}{P_{{{\rm{O}}_{\rm{2}}}}}}}{{{P^2}_{{\rm{S}}{{\rm{O}}_{\rm{3}}}}}}$

G) Solids and liquids are omitted from the equilibrium expression and therefore, only CO2 will be used for K and Kp.

CaCO3(s) ⇆ CaO(s) + CO2(g)

$K = [C{O_2}]$

${K_{\rm{p}}} = {P_{C{O_2}}}$

H) All the reactants in a gas state are included in the expression for K (Kc) and Kp. Solids and liquids are omitted from the equilibrium expression:

Fe2O3(s) + 3CO(g) ⇆ 2Fe(s) + 3CO2(g)

$K = \frac{{{{[{\rm{C}}{{\rm{O}}_{\rm{2}}}]}^3}}}{{{{[{\rm{CO}}]}^3}}}$

${K_p} = \frac{{{P^{\rm{3}}}{{_{{\rm{CO}}}}_{_2}}}}{{{P^3}_{{\rm{CO}}}}}$

I) All the reactants and products are in a gas state and therefore, both K (Kc) and Kp are applicable to express the equilibrium.

4HCl(g) + O2(g) ⇆ 2Cl2(g) + 2H2O(g)

$K = \frac{{{{{\rm{[C}}{{\rm{l}}_{\rm{2}}}{\rm{]}}}^{\rm{2}}}{{{\rm{[}}{{\rm{H}}_{\rm{2}}}{\rm{O]}}}^{\rm{2}}}}}{{{{{\rm{[HCl]}}}^{\rm{4}}}{\rm{[}}{{\rm{O}}_{\rm{2}}}{\rm{]}}}}$

${K_p} = \frac{{{P^2}_{{\rm{C}}{{\rm{l}}_{\rm{2}}}}{P^2}_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}}{{{P^4}_{{\rm{HCl}}}{P_{{{\rm{O}}_{\rm{2}}}}}}}$

J) There are no gases in this reaction, so only K (Kc) will be used to describe the equilibrium.  Water is not onluded since liquids are omitted from the equilibrium expression:

CO32-(aq) + H2O(l) ⇆ HCO3(aq) + OH(aq)

$K = \frac{{{\rm{[HC}}{{\rm{O}}_{\rm{3}}}^{\rm{–}}{\rm{][O}}{{\rm{H}}^{\rm{–}}}{\rm{]}}}}{{{\rm{[C}}{{\rm{O}}_{\rm{3}}}^{{\rm{2 – }}}{\rm{]}}}}$

K) Solids and liquids are omitted from the equilibrium expression and therefore, only O2 will be used for K and Kp.

P4(s) + 5O2(g) ⇆ P4O10(s)

$K = \frac{1}{{{{[{{\rm{O}}_2}]}^5}}}$

${K_{\rm{p}}} = \frac{1}{{{P^5}_{{\rm{C}}{{\rm{O}}_{\rm{2}}}}}}$

2.

Tree different reactions are represented in the diagram below. The reaction equations can be written as A + X ⇆ AX (X = B, C, or D).

(a) Which reaction has the largest equilibrium constant? (b) Which reaction has the smallest equilibrium constant?

(a) The reaction, A + D ⇆ AD has the largest equilibrium constant.

(b) The reaction, A + C ⇆ AC has the smallest equilibrium constant.

Solution

An important factor to notice here is that the reactions are similar since in all of them, two reactants are producing one product, and all the coefficients are one in the balanced equation.

(a) The reaction, A + D ⇆ AD has the largest equilibrium constant because it produces the most product at equilibrium.

(b) The reaction, A + C ⇆ AC has the smallest equilibrium constant because it produces the amount of product at equilibrium.

3.

The following graph represents an initial mixture of N2 and H2 at high temperature and pressure:

The gases react to form ammonia gas (NH3) as represented by the following concentration profile:

a) Label each plot on the graph as H2, N2, or NH3, and explain your answers.

b) What information do the relative shapes of the plots tell us?

c) At what time is equilibrium reached?

a) H2 – blue, H2 – green, NH3 – red

c) The equilibrium is established in about 4 hours.

Solution

a) Initially, hydrogen has the highest concentration judging from the number of spheres. Notice that blue spheres correspond to hydrogen since they are smaller atoms than those of nitrogen. Therefore, the blue line corresponds to the concentration of H2 which is decreasing with time.

The concentration of N2 is shown by the green line because it is decreasing as the reaction continues.

Ammonia, on the other hand, is not present in the beginning, and therefore, the red line represents its concentration which is increasing with time.

b) Let’s first write the balanced chemical equation for this reaction:

3H2(g) + N2(g) ⇆ 2NH3(g)

When a reaction starts, only H2 and N2 are present in the vessel and therefore, their concentrations decrease while the concentration of ammonia increases with time. Another observation is the difference in the concentration change of hydrogen and nitrogen.  Because the stoichiometry in the balanced equation dictates that for every 1 molecule of N2, 3 molecules of H2 are needed, the H2 concentration initially decreases more rapidly as compared to the initial decrease in N2 concentration.

Likewise, the NH3 plot initially increases faster than the N2 plot decreases since, here as well, the ratio of these two gases is 2 : 1 meaning there is 2 times more ammonia produced than nitrogen is consumed at any time of reaction.

c) The equilibrium is reached when the concentrations of reactants and products do not change anymore. This can be seen when the concentration lines hit a plateau at about 4-hour mark.

4.

At a particular temperature, it is determined for the reaction

2NO(g) + 2H2(g) ⇆ N2 (g) + 2H2O(g)

that at equilibrium, the concentrations are as follows: [NO(g)] = 3.2 x 10-3 M, [H2(g)] = 6.7 x 10-6 M, [N2(g)] = 4.8 x 10-2 M, and [H2O(g)] = 2.4 x 10-2 M. What is the value of the equilibrium constant K for the reaction at this temperature?

4.0 x 102

Solution

The equilibrium constant is the ratio of the equilibrium concentrations of the products and reactants, each raised to a power equal to its stoichiometric coefficient in the balanced equation.

$K = \frac{{{\rm{[}}{{\rm{N}}_{\rm{2}}}{\rm{][}}{{\rm{H}}_{\rm{2}}}{\rm{O}}{{\rm{]}}^{\rm{2}}}}}{{{{{\rm{[NO]}}}^{\rm{2}}}{\rm{[}}{{\rm{H}}_{\rm{2}}}{\rm{]}}}}$

$K = \frac{{{\rm{(}}4.8 \times {{10}^{ – 2}}{\rm{)(}}2.4 \times {{10}^{ – 2}}{{\rm{)}}^{\rm{2}}}}}{{{{(3.2 \times {{10}^{ – 3}})}^{\rm{2}}}(6.7 \times {{10}^{ – 6}})}} = 4.0 \times {10^2}$

5.

Given the equilibrium concentrations, calculate the value of the equilibrium constant K for the reaction between CO2 and H2 that produces methanol and water at high temperature.

CO2(g) + 3H2(g) ⇆ CH3OH(g) + H2O(g)

[CO2] = 0.061 M, [H2] = 0.079 M, [CH3OH] = 4.7 x 102 M, and [H2O] = 5.7 x 104 M Calculate the value of K for the reaction.

8.9 x 1011

Solution

$K = \frac{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{OH][}}{{\rm{H}}_{\rm{2}}}{\rm{O]}}}}{{{\rm{[C}}{{\rm{O}}_{\rm{2}}}{\rm{][}}{{\rm{H}}_{\rm{2}}}{{\rm{]}}^{\rm{3}}}}}$

$K = \frac{{{\rm{(}}4.7 \times {{10}^2}{\rm{)(}}5.7 \times {{10}^4}{\rm{)}}}}{{{\rm{(}}0.061{\rm{)(}}0.079{{\rm{)}}^{\rm{3}}}}} = 8.9 \times {10^{11}}$

6.

The reaction for converting methane to acetylene has an equilibrium constant of K = 0.154 at 2000 K.

2CH4(g) ⇆ C2H2(g) + 3H2(g)

Calculate the equilibrium constant for this process if the reaction is represented as follows:

a) CH4(g) ⇆ 1/2C2H2(g) + 3/2H2(g)

b) 4CH4(g) ⇆ 2C2H2(g) + 6H2(g)

c) 6CH4(g) ⇆ 3C2H2(g) + 9H2(g)

a) K = 0.392

b) K = 0.0237

c) K = 0.00365

Solution

Remember, when a reaction is reversed, then Knew = 1/Koriginal. When a reaction is multiplied by a value of n, then Knew = (Koriginal)n.

a) Comparing the equation CH4(g) ⇆ 1/2C2H2(g) + 3/2H2(g) to the original, we see that it is multiplied by 1/2. Therefore, Knew = (Koriginal)= (0.154)1/2= 0.392

b) Comparing the equation 4CH4(g) ⇆ 2C2H2(g) + 6H2(g) to the original, we see that it is multiplied by 2. Therefore, Knew = (Koriginal)= (0.154)2= 0.0237

c) Comparing the equation 6CH4(g) ⇆ 3C2H2(g) + 9H2(g) to the original, we see that it is multiplied by 3. Therefore, Knew = (Koriginal)= (0.154)3= 0.00365

7.

The following reaction has an equilibrium constant of Kp = 4.42 x 10-5 at 298 K:

CH3OH(g) ⇆ CO (g) + 2H2(g)

Calculate equilibrium constant for this process if the reaction is represented as follows:

a) CO(g) + 2H2(g) ⇆ CH3OH(g)

b) 2CH3OH(g) ⇆ 2CO(g) + 4H2(g)

c) 1/2CH3OH(g) ⇆ 1/2CO(g) + H2(g)

a) Knew = 2.26 x 104

b) Knew = 1.96 x 10-9

c) Knew = 6.65 x 10-3

Solution

Remember, when a reaction is reversed, then Knew = 1/Koriginal. When a reaction is multiplied by a value of n, then Knew = (Koriginal)n.

a) Comparing the equation CO (g) + 2H2(g) ⇆ CH3OH(g) to the original, we see that it is reversed. Therefore, Knew = 1/Koriginal= 1/4.42 x 10-5= 22,624.43 = 2.26 x 104

b) Comparing the equation 2CH3OH(g) ⇆ 2CO (g) + 4H2(g) to the original, we see that it is multiplied by 2. Therefore, Knew = (Koriginal)= (4.42 x 10-5)2= 1.96 x 10-9

c) Comparing the equation 1/2CH3OH(g) ⇆ 1/2CO (g) + H2(g) to the original, we see that it is multiplied by 1/2. Therefore, Knew = (Koriginal)= (4.42 x 10-5)1/2= 0.00665=6.65 x 10-3

8.

The equilibrium constant values for the reactions below were determined at a certain temperature:

S(s) + O2(g) ⇆ SO2(g)    Ka = 3.2 x 1045

2S(s) + 3O2(g) ⇆ 2SO3(g)    Kb = 3.2 x 10124

Using these data, determine the equilibrium constant Kc for the following reaction:

2SO2(g) + O2(g) ⇆ 2SO3(g)    Kc = ?

Kc = 3.1 x 1033

Solution

There are two SO2 reactant molecules in the target reaction, so reverse the first reaction and multiply by 2:

S(s) + O2(g) ⇆ SO2(g) ⇒ 2SO2(g) ⇆ 2S(s) + 2O2(g)

The equilibrium constant for this reaction is:

Kra = (1/ka)2 = (1/3.2 x 1045)= 9.77 x 10-92

Now, add the two reactions to make sure it matches the target equation:

2S(s) + 3O2(g) ⇆ 2SO3(g)

2SO2(g) ⇆ 2S(s) + 2O2(g)
__________________

2S(s) + 3O2(g) + 2SO2(g) ⇆ 2SO3(g) + 2S(s) + 2O2(g)

O2(g) + 2SO2(g) ⇆ 2SO3(g)

The equations add up, so the last part is to calculate the equilibrium constant.

Kc = Kra x K= 9.77 x 10-92 x 3.2 x 10124 = 3.1 x 1033

9.

The equilibrium constant values for the reactions below were determined at a certain temperature:

2NO(g) ⇆ N2(g) + O2(g) Ka = 3.8 x 1032

NO(g) + 1/2Cl2(g) ⇆ NOCl(g) Kb = 6.4

Using these data, determine the equilibrium constant Kc for the following reaction:

1/2N2(g) + 1/2O2(g) + 1/2Cl2(g) ⇆ NOCl(g) Kc = ?

Kc = 3.28 x 10-16

Solution

There are 1/2 moles of N2 reactant molecules in the target reaction, so reverse the first reaction and multiply by 1/2:

2NO(g) ⇆ N2(g) + O2(g) ⇒ 1/2N2(g) + 1/2O2(g) ⇆ NO(g)

The equilibrium constant for this reaction is:

Kra = (1/ka)1/2 = (1/3.8 x 1032)1/2 = 5.13 x 10-17

Now, add the two reactions to make sure it matches the target equation:

1/2N2(g) + 1/2O2(g) ⇆ NO(g)

NO(g) + 1/2Cl2(g) ⇆ NOCl(g)

_____________________

1/2N2(g) + 1/2O2(g) + NO(g) + 1/2Cl2(g) ⇆ NO(g) + NOCl(g)

1/2N2(g) + 1/2O2(g) + 1/2Cl2(g) ⇆ NOCl(g)

The equations add up, so the last part is to calculate the equilibrium constant.

Kc = Kra x K= 5.13 x 10-17 x 6.4 = 3.28 x 10-16

10.

The following equilibrium pressures were observed at a certain temperature for the Haber process

3H2(g) + N2(g) ⇆ 2NH3(g)

P(NH3) = 5.2 x 109 atm

P(N2) = 6.1 x 102 atm

P(H2) = 4.7 x 103 atm

Calculate the value for the equilibrium constant Kp at this temperature.

Kp = 4.3 x 105

Solution

${K_p} = \frac{{{P^2}_{{\rm{N}}{{\rm{H}}_{\rm{3}}}}}}{{{P^3}_{{{\rm{H}}_{\rm{2}}}}{P_{{{\rm{N}}_{\rm{2}}}}}}}$

${K_p} = \frac{{{{(5.2 \times {{10}^9})}^2}}}{{{{(4.7 \times {{10}^3})}^3}(6.1 \times {{10}^2})}} = 4.3 \times {10^5}$

11.

Consider the following reactions:

1) H2(g) + I2(g) ⇆ 2HI(g)

2) H2(g) + I2(s) ⇆ 2HI(g)

In which reaction are the K and Kp equal?

H2(g) + I2(g) ⇆ 2HI(g)

Solution

Kp = K(RT)Δn

Where Δn is the sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants. In the first reaction, Δn=0, therefore, Kp = K.

12.

At 300 K, the equilibrium concentrations for the following reaction are [CH3OH] = 0.240 M, [CO] = 0.350 M, and [H2] = 1.65 M for the reaction

CH3OH(g) ⇆ CO(g) + 2H2(g)

Calculate Kp at this temperature.

Kp = 2.41 x 103

Solution

Kp = Kc(RT)Δn

Where Δn is the sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants. For this reaction, Δn=1+2-1=2.

Next, we need to determine the Kc and use that value to calculate the Kp.

$K = \frac{{{\rm{[CO][}}{{\rm{H}}_{\rm{2}}}{{\rm{]}}^{\rm{2}}}}}{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{OH]}}}}$

$K = \frac{{(0.350){{({\rm{1}}{\rm{.65}})}^2}}}{{{\rm{(0}}{\rm{.240)}}}} = 3.97$

Kp = K(RT)2 = 3.97(0.08206 L atm/K•mol × 300. K)2 = 2,406 = 2.41 x 103

13.

Given the equilibrium constant, calculate Kp for each of the following reactions at 298 K.

a) N2O4(g) ⇆ 2NO2(g) Kc = 4.6 x 10-4

b) 3H2(g) + N2(g) ⇆ 2NH3(g) Kc = 6.7 x 109

c) H2(g) + B2(g) ⇆ 2HBr(g) Kc = 5.20 x 1018

a) Kp = 1.1 x 10-2

b) Kp = 1.6 x 1011

c) Kp = 5.20 x 1018

Solution

Kp = K(RT)Δn

Where Δn is the sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants.

a) Δn = 2-1 = 1

Kp = K(RT) = 4.6 x 10-4(0.08206 L atm/K•mol × 298 K) = 1.1 x 10-2

b) Δn = 2- (3+ 1) = -2

Kp = K(RT) = 6.7 x 109(0.08206 L atm/K•mol × 298 K)-2 = 1.6 x 1011

c) Δn = 2- (1+ 1) = 0

Kp = K(RT)= K = 5.20 x 1018

14.

Calculate Kc for each reaction.

a) CO(g) + Cl2(g) ⇆ COCl2(g) Kp = 5.3 x 106

b) CH4(g) + H2O(g) ⇆ CO(g) + 3H2(g) Kp = 7.7 x 108

c) 2SO2(g) + O2(g) ⇆ 2SO3(g) Kp = 6.26 x 105

a) Kc = 1.3 x 105

b) Kc =3.1 x 104

c) Kc = 1.5 x 104

Solution

Kp = Kc(RT)Δn , therefore, Kc = Kp/(RT)Δn

a) Δn = 1-(1+1) = -1

Kc = 5.3 x 106/(0.08206 L atm/K•mol × 298 K)-1 = 129605 = 1.3 x 105

b) Δn = 1+3-(1+1) = 2

Kc = 7.7 x 108/(0.08206 L atm/K•mol × 298 K)2 = 31487 =3.1 x 104

c) Δn = 2-(1+2) = -1

Kc = 6.26 x 105/(0.08206 L atm/K•mol × 298 K)-1 = 15308 = 1.5 x 104

15.

Consider the reaction between nitrogen and hydrogen gases:

3H2(g) + N2(g) ⇆ 2NH3(g)

Using the data given in the table, complete the missing numbers assuming that all concentrations are measured at equilibrium.

K = 8.80 x 102

[NH3] = 3.9 x 10-2

[H2] = 3.86

Solution

1) Determine the equilibrium constant using the concentration of the reactants and products in the table (row 1).

$K = \frac{{{{{\rm{[N}}{{\rm{H}}_{\rm{3}}}{\rm{]}}}^{\rm{2}}}}}{{{\rm{[}}{{\rm{N}}_{\rm{2}}}{\rm{][}}{{\rm{H}}_{\rm{2}}}{{\rm{]}}^{\rm{3}}}}}$

$K = \frac{{{{{\rm{[0}}{\rm{.518]}}}^{\rm{2}}}}}{{{\rm{[0}}{\rm{.142][0}}{\rm{.129}}{{\rm{]}}^{\rm{3}}}}} = 8,80 \times {10^2}$

2) Assign the concentration of ammonia as x, and set up an equation based on the equilibrium constant.

$K = \frac{{{{{\rm{(X)}}}^{\rm{2}}}}}{{{\rm{(0}}{\rm{.125)(0}}{\rm{.116}}{{\rm{)}}^{\rm{3}}}}} = 7.8$

X2 = 0.001521874, X = 0.0390 = 3.9 x 10-2

3) Assign the concentration of hydrogen as x, and set up an equation based on the equilibrium constant.

$K = \frac{{{{{\rm{(0}}{\rm{.712)}}}^{\rm{2}}}}}{{{\rm{(0}}{\rm{.136)(X}}{{\rm{)}}^{\rm{3}}}}} = 0.0647$

0.0087992 X3 = 0.506944

X = 3.86

16.

Consider the following reaction at 298 K:

2NO(g) + Cl2(g) ⇆ 2NOCl(g) Kp = 31.6

In a reaction mixture at equilibrium, the partial pressure of NO is 128 torr and that of Cl2 is 176 torr. Calculate the partial pressure of NOCl at equilibrium.

P(NOCl) = 0.455 atm, or 346 torr

Solution

Remember, when converting Kc and Kp, the value for R is most often used as 0.08206 L atm/ mol K, and therefore, the units for pressure should also be in atm.

${P_{({\rm{NO}})}}{\rm{ = }}\;\;{\rm{128torr}}\;\;{\rm{x}}\;\frac{{{\rm{1}}\;{\rm{atm}}}}{{{\rm{760}}\;{\rm{torr}}}}\;{\rm{ = }}\;\;{\rm{0}}.{\rm{168}}\;{\rm{atm}}$

${P_{({\rm{C}}{{\rm{l}}_{\rm{2}}})}}\; = \;{\rm{176}}\;{\rm{torr\;}}\; \times \;{\rm{\;}}\frac{{{\rm{1}}\;{\rm{atm}}}}{{{\rm{760}}\;{\rm{tor}}}}\,{\rm{\; = \;}}\,{\rm{0}}.{\rm{232}}\,{\rm{atm}}$

Next, use the Kp expression to assign the numbers and an unknown for the P(NOCl):

${K_p} = \frac{{{P^2}_{{\rm{NOCl}}}}}{{{P^2}_{{\rm{NO}}}{P_{{\rm{C}}{{\rm{l}}_{\rm{2}}}}}}}$

${K_p} = \frac{{{{\rm{X}}^{\rm{2}}}}}{{{{\left( {{\rm{0}}{\rm{.168}}} \right)}^{\rm{2}}}{\rm{0}}{\rm{.232}}}}\;{\rm{ = }}\;{\rm{31}}{\rm{.6}}$

X = 0.455 atm = 346 torr

Check:

${K_p} = \frac{{{{\rm{0.455}}^{\rm{2}}}}}{{{{\left( {{\rm{0}}{\rm{.168}}} \right)}^{\rm{2}}}{\rm{0}}{\rm{.232}}}}\;{\rm{ = }}\;{\rm{31}}{\rm{.6}}$

17.

The equilibrium constant for the following reaction at 600 oC is determined to be Kc = 0.495:

H2O(g) + CO(g) ⇆ H2(g) + CO2(g)

Calculate the number of H2 moles that are present at equilibrium if a mixture of 0.400 mole CO and 0.500 mole H2O is heated to 600°C in a 10.0-L container.

0.184 mol H2

Solution

The concentrations of the gases are going to be moles over the container volume (10.0 L). So, [CO] = 0.0400 mol/L and [H2O] = 0.0500 mol/L.

Let’s set up an ICE table assigning x mol/L as the depletion in the concentration of either CO or H2O. The increase of H2 and COconcentration is going to be x which is their equilibrium concentration. The equilibrium concentration of CO and H2O is going to be 0.0400-x and 0.0500-x respectively.

H2O(g) + CO(g) ⇆ H2(g) + CO2(g)

The equilibrium constat is:

${K_{\rm{c}}}\;{\rm{ = }}\;\frac{{\left[ {{{\rm{H}}_{\rm{2}}}} \right]\left[ {{\rm{C}}{{\rm{O}}_{\rm{2}}}} \right]}}{{\left[ {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right]\left[ {{\rm{CO}}} \right]}}\; = \,0.495$

${K_{\rm{c}}}\;{\rm{ = }}\;\frac{{\left( {\rm{x}} \right)\left( {\rm{x}} \right)}}{{\left( {{\rm{0}}{\rm{.0500 – x}}} \right)\left( {{\rm{0}}{\rm{.0400 – x}}} \right)}}\;{\rm{ = }}\,{\rm{0}}{\rm{.495}}$

${K_{\rm{c}}}\;{\rm{ = }}\;\frac{{{{\rm{x}}^{\rm{2}}}}}{{\left( {{\rm{0}}.{\rm{0500 – x}}} \right)\left( {{\rm{0}}.{\rm{0400 – x}}} \right)}}\;{\rm{ = }}{\mkern 1mu} {\rm{0}}.{\rm{495}}$

${K_{\rm{c}}}\;{\rm{ = }}\;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{{\rm{x}}^{\rm{2}}}\; – \;0.09x\; + \;0.002}}\;{\rm{ = }}{\mkern 1mu} {\rm{0}}.{\rm{495}}$

0.505x2 + 0.04455x – 0.00099 = 0

${\rm{x}}\;{\rm{ = }}\;\frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}$

a = 0.505, b = 0.04455, c = -0.00099

x = 0.0184 mol/L

The last step is to convert the concentration to the number of moles by multiplying by 10 since the reaction is carried out in a 10.0 L container.

n (H2) = 0.0184 mol/L x 10.0 L = 0.184 mol

18.

The equilibrium constant Kc for the following reaction at 800°C is 3.74 x 105

H2(g) + I2(g) ⇆ 2HI(g)

If 6.25 moles of HI were initially added to a 15.0-L empty vessel, what would the concentrations of H2, I2, and HI be at equilibrium.

[H2] = [I2] = 6.8 x 10-4 mol/L

[HI] = 0.416 mol/L

Solution

First, determine the initial concentration of HI.

M (HI) = 6.25 mol ÷ 15.0 L = 0.417 mol/L

Then, set up the ICE table, assuming x mol/L of HI decomposed to form H2 and I2. From the stoichiometric ratio, we know that 0.5x mol/l of H2 and Iwill be formed:

H2(g) + I2(g) ⇆ 2HI(g)

The equilibrium constant is:

${K_{\rm{c}}}\;{\rm{ = }}\;\frac{{{{\left[ {{\rm{HI}}} \right]}^2}}}{{\left[ {{{\rm{H}}_{\rm{2}}}} \right]\left[ {{{\rm{I}}_{\rm{2}}}} \right]}}\; = \,3.74{\rm{ }}x{\rm{ }}{10^5}$

${K_{\rm{c}}}\;{\rm{ = }}\;\frac{{{{\left( {{\rm{0}}{\rm{.417 – x}}} \right)}^2}}}{{\left( {{\rm{0}}{\rm{.5x}}} \right)\left( {{\rm{0}}{\rm{.5x}}} \right)}}\; = \,3.74{\rm{ }} \times {\rm{ }}{10^5}$

${K_{\rm{c}}}\;{\rm{ = }}\;\frac{{{{\left( {{\rm{0}}{\rm{.417 – x}}} \right)}^{\rm{2}}}}}{{{\rm{0}}{\rm{.25}}{{\rm{x}}^{\rm{2}}}}}\;{\rm{ = }}\,{\rm{3}}{\rm{.74 \times 1}}{{\rm{0}}^{\rm{5}}}$

Taking the square root of both sides we obtain:

${K_{\rm{c}}}\;{\rm{ = }}\;\frac{{\left( {{\rm{0}}{\rm{.417 – x}}} \right)}}{{{\rm{0}}{\rm{.5x}}}}\;{\rm{ = }}\,{\rm{612}}$

x = 0.00136

Therefore, the equilibrium concentrations are:

[H2] = [I2] = 0.5 x 0.00136 = 0.00068 = 6.8 x 10-4 mol/L

[HI] = 0.417 – 0.00136 = 0.416 mol/L

Check:

${K_{\rm{c}}}\;{\rm{ = }}\;\frac{{{{0.416}^2}}}{{{{0.00068}^2}}}\; = \,3.74{\rm{ }} \times {\rm{ }}{10^5}$

19.

The equilibrium constant for the following reaction at 700 K is Kp = 6.7 x 10-3

CO(g) + 2H2(g) ⇆ CH3OH(g)

A reaction mixture contains 0.248 atm of H2, 0.085 atm of CO, and 0.598 atm of CH3OH. Is the reaction mixture at equilibrium? If not, in what direction will the reaction proceed?

The reaction is not at equilibrium and will proceed to the left.

Solution

The equilibrium constant for this reaction is:

${K_p}\; = \;\frac{{{P_{{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{OH}}}}}}{{{P^2}_{{{\rm{H}}_{\rm{2}}}}{P_{{\rm{CO}}}}}}\; = \;6.7{\rm{ }} \times {\rm{ }}{10^{ – 3}}$

Now, if the mixture is at equilibrium, then plugging the concentration values in this equation, we should get the value for Kp.

${K_p}\; = \;\frac{{0.598}}{{{{0.248}^2}0.085}}\; = \;114\; > \;6.7{\rm{ }}x{\rm{ }}{10^{ – 3}}$

This is greater than Kp (6.7 x 10-3), and therefore, the reaction will proceed to the left forming more CO and H2. This can also be predicted when we compare the partial pressures of the gases. Despite a small equilibrium constant, CH3OH has a higher partial pressure than CO and H2. The system, cannot stay in this state, and thus some of the CH3OH will break into CO and H2.

20.

For the reaction shown below, Kc = 0.654 at 600 K.

N2O4(g) ⇆ 2NO2(g)

If initially, 0.0600 M of N2O4 are present in the reaction vessel, what are the equilibrium concentrations of the gases at 600 K?

[N2O4] = 0.0133 mol/L

[NO2] = 0.0934 mol/L

Solution

To set up the ICE table, we need to determine which gas will increase and which one will decrease. Because only N2Ois initially present in the system, then its concentration is going to decrease, while the concentration of NOwill increase.

Let’s suppose x mol/L of N2Odecomposes (reacts) to form 2x mol/L NO2. This is based on the stoichiometric ratio of the gases – every one mole of N2O4 produces 2 mol NO2.

N2O4(g) ⇆ 2NO2(g)

The equilibrium constant for this reaction is:

${K_{\rm{c}}}\;{\rm{ = }}\;\frac{{{{\left[ {{\rm{N}}{{\rm{O}}_{\rm{2}}}} \right]}^2}}}{{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_4}} \right]}}\; = \,0.654$

${K_{\rm{c}}}\;{\rm{ = }}\;\frac{{{{{\rm{(2x)}}}^2}}}{{0.0600 – x}}\; = \,0.654$

${K_{\rm{c}}}\;{\rm{ = }}\;\frac{{{{{\rm{4x}}}^2}}}{{0.0600 – x}}\; = \,0.654$

4x2 + 0.654x – 0.003924 = 0

${\rm{x}}\;{\rm{ = }}\;\frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}$

a = 4, b = 0.654, c = -0.003924

x=0.0467 or x=−0.210 (doesn’t work)

Therefore, the equilibrium concentrations are:

[N2O4] = 0.0600 – 0.0467 = 0.0133 mol/L

[NO2] = 2 x 0.0467 = 0.0934 mol/L

21.

For the reaction shown below, Kc = 255 at 800 K.

PCl3(g) + Cl2(g) ⇆ PCl5(g)

If a reaction mixture initially contains 0.3500 M PCl3 and 0.375 M Cl2 at 800 K, what are the equilibrium concentrations of all the species in the mixture?

[PCl3] = 0.0253 mol/L

[Cl2] = 0.0503 mol/L

[PCl5] = 0.3247  mol/L

Solution

The equilibrium constant for this reaction is:

${K_{\rm{c}}}\;{\rm{ = }}\;\frac{{\left[ {{\rm{PC}}{{\rm{l}}_{\rm{5}}}} \right]}}{{\left[ {{\rm{PC}}{{\rm{l}}_{\rm{3}}}} \right]\left[ {{\rm{C}}{{\rm{l}}_{\rm{2}}}} \right]}}\;{\rm{ = }}\,{\rm{255}}$

To set up the ICE table, we need to determine which gas will increase and which one will decrease. Because no PClis present in the system, the reaction will proceed to the right, thus decreasing the concentration of PCl3 and Cl2.

Let’s suppose x mol/L of PCl3 and Cl2 reacts to form x mol/L PCl5. This is based on the stoichiometric ratio of the gases – every one mole of PCl3 and Cl2 produces one mol PCl5.

PCl3(g) + Cl2(g) ⇆ PCl5(g)

Plugging the numbers, we get an equation based on the equilibrium constant:

${K_{\rm{c}}}\;{\rm{ = }}\;\frac{{\rm{x}}}{{\left( {{\rm{0}}{\rm{.3500 – x}}} \right)\left( {{\rm{0}}{\rm{.375 – x}}} \right)}}\;{\rm{ = }}\,{\rm{255}}$

255x2 – 185.875x + 33.46875 = 0

${\rm{x}}\;{\rm{ = }}\;\frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}$

a = -255, b = 185.9, c = -33.47

x=0.324691 or x=0.404231

x=0.404231 doesn’t work becasue 0.3500-0.404231 gives a negative value for the concentration.

x=0.3247

Therefore, the equilibrium concentrations are:

[PCl3] = 0.3500 – 0.3247 = 0.0253 mol/L

[Cl2] = 0.375 – 0.3247 = 0.0503 mol/L

[PCl5] = x=0.3247  mol/L

22.

Consider the following reaction characterized with Kp = 2.34 x 10-4 at 250 K:

I2(g) + Cl2(g) ⇆ 2ICl(g)

A reaction mixture initially contains I2 with partial pressure of 655 torr and Cl2 with partial pressure of 864 torr at 250 K. Calculate the equilibrium partial pressure of ICl.

0.0141 atm = 11.4 torr

Solution

Remember, when converting Kc and Kp, the value for R is most often used as 0.08206 L atm/ mol K, and therefore, the units for pressure should also be in atm.

${P_{({{\rm{I}}_{\rm{2}}})}}\; = \;{\rm{655}}\;{\rm{torr\;}} \times {\rm{\;}}\frac{{{\rm{1}}\;{\rm{atm}}}}{{{\rm{760}}\;{\rm{torr}}}}{\rm{\; = }}\,{\rm{0}}{\rm{.862}}\;{\rm{atm}}$

${P_{({\rm{C}}{{\rm{l}}_{\rm{2}}})}}\; = \;{\rm{864}}\;{\rm{torr\;}} \times {\rm{\;}}\frac{{{\rm{1}}\;{\rm{atm}}}}{{{\rm{760}}\;{\rm{torr}}}}{\rm{\; = }}\,{\rm{1}}{\rm{.14}}\;{\rm{atm}}$

The equilibrium constant is:

${K_p}\; = \;\frac{{{P^2}_{{\rm{ICl}}}}}{{{P_{{{\rm{I}}_{\rm{2}}}}}{P_{{\rm{C}}{{\rm{l}}_{\rm{2}}}}}}}\; = \;2.34{\rm{ }} \times {\rm{ }}{10^{ – 4}}$

However, remember that these are the partial pressures at equilibrium so, we cannot use the numbers above.

First, we need to set up an ICE table to determine the partial pressures at equilibrium. Because the system does not contain any ICl at the beginning, its pressure (concentration) is going to increase, while the pressure of Cl2 and I2 is going to decrease. Let’s assume the pressure of Cl2 and I2 drops by x atm which means the pressure of ICl will increase by 2x atm:

Plug in the numbers and assign x for the partial pressure of ICl:

${K_p}\; = \;\frac{{{{{\rm{(2x)}}}^{\rm{2}}}}}{{\left( {{\rm{0}}{\rm{.862 – x}}} \right)\left( {{\rm{1}}{\rm{.14 – x}}} \right)}}\;{\rm{ = }}\;{\rm{2}}{\rm{.34 \times 1}}{{\rm{0}}^{{\rm{ – 4}}}}$

${K_p}\; = \;\frac{{{\rm{4}}{{\rm{x}}^{\rm{2}}}}}{{{{\rm{x}}^{\rm{2}}}\;{\rm{ – }}\;{\rm{2x}}\;{\rm{ + }}\;{\rm{098268}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.34 \times 1}}{{\rm{0}}^{{\rm{ – 4}}}}$

Solving this quadratic equation, we get x=0.00753

Therefore, the partial pressure of ICl at equilibrium is 2 x 0.00753 atm = 0.0141 atm = 11.4 torr

A good approach to avoid solving long quadratic equations is to make an approximation/assumption that x << 0.862, meaning that there is very little pressure drop (reaction).

In this case, we can simplify the quadratic equation as:

${K_p}\; = \;\frac{{{{{\rm{(2x)}}}^{\rm{2}}}}}{{\left( {{\rm{0}}{\rm{.862}}} \right)\left( {{\rm{1}}{\rm{.14}}} \right)}}\;{\rm{ = }}\;{\rm{2}}{\rm{.34 \times 1}}{{\rm{0}}^{{\rm{ – 4}}}}$

Solving for x, we find that x = 0.007582 which is very close to 0.00753 (error < 1%).

Let’s assume though that we did not know the actual value for x and only solve for it by the approximation. To test whether the assumption was applicable to this reaction, plug the values in the equilibrium expression and compare the result to the actual Kc given in the problem.

${K_p}\; = \;\frac{{{{{\rm{(2}} \times 0.007582{\rm{)}}}^{\rm{2}}}}}{{\left( {{\rm{0}}{\rm{.862 – }}0.007582} \right)\left( {{\rm{1}}{\rm{.14 – }}0.007582} \right)}}\;{\rm{ = }}\;{\rm{2}}{\rm{.38 \times 1}}{{\rm{0}}^{{\rm{ – 4}}}}$

And this confirms that the approximation was suitable for this problem. If it wasn’t, and the difference between the Kc values was more than 5%, then the equation must be solved without the approximation.

23.

Consider the following reversible reaction:

POCl3(g) ⇌ POCl(g) + Cl2(g)     Kc = 0.650

The following initial amounts of reactants and products were mixed: [POCl3] = 0.650 M, [POCl] = 0.450 M, and [Cl2] = 0.250 M.

Calculate the reaction quotient, Qc, and determine the equilibrium concentration of POCl?

0.590 mol/L

Solution

Since the system initially contains both reactants and products, we need to first determine the reaction quotient:

$Q\;{\rm{ = }}\;\frac{{\left[ {{\rm{POCl}}} \right]\left[ {{\rm{C}}{{\rm{l}}_{\rm{2}}}} \right]}}{{\left[ {{\rm{POC}}{{\rm{l}}_3}} \right]}}\;$

$Q\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.450 }} \times \;{\rm{0}}{\rm{.250}}}}{{0.650}}\; = 0.173$

Q < K, so, the reaction will proceed to the right. This means, the concentrations of POCl and Clare going to increase, while the concentration of POCl3 will decrease.

Let’s set up an ICE table and assign x as the depletion in the concentration of POCl3:

POCl3(g) ⇌ POCl(g) + Cl2(g)

The equilibrium constant is:

$K\;{\rm{ = }}\;\frac{{\left[ {{\rm{POCl}}} \right]\left[ {{\rm{C}}{{\rm{l}}_{\rm{2}}}} \right]}}{{\left[ {{\rm{POC}}{{\rm{l}}_3}} \right]}}\;$

$K\;{\rm{ = }}\;\frac{{\left( {{\rm{0}}{\rm{.450 + x}}} \right)\left( {{\rm{0}}{\rm{.250 + x}}} \right)}}{{0.650 – x}}\; = 0.650$

x=0.13958

Therefore, the [POCl] at equilibrium is 0.450 + 0.140 = 0.590 mol/L

24.

Consider the following equilibrium:

2NO(g) + 2H2(g) ⇆ N2(g) + 2H2O(g)

Initially, there are 0.15 moles of NO and 0.25 moles of H2, in a 10.0-L container. If there are 0.056 moles of NO at equilibrium, how many moles of N2 are present at equilibrium?

0.047 mol

Solution

The reaction is going to proceed to the right because there are only reactants present in the mixture initially (Q = 0).

Even though we don’t need an ICE table here, let’s put one to visualize the changes. Here is the date we have:

2NO(g) + 2H2(g) ⇆ N2(g) + 2H2O(g)

Next is to determine how many moles of NO have reacted to establish the equilibrium. For this, subtract the final (equilibrium) and initial concentrations of NO:

Change (NO) = 0.15-0.056=0.094 mol

Add this to the ICE table and determine the moles of NO based on the stoichiometric ratio. Every 2 moles of NO produce one mole of N2, and therefore,

n (N2) = 0.094/2 = 0.047 mol

2NO(g) + 2H2(g) ⇆ N2(g) + 2H2O(g)

And since there is no N2 present in the beginning, this is also the number of moles that are present at equilibrium.

25.

Consider the following equilibrium:

2NOCl(g) ⇆ 2NO(g) + Cl2(g)

2.00 mole of pure NOCl and 1.65 mole of pure Cl2 are placed in a 1.00-L container. Calculate the equilibrium concentration of NO(g) considering that with K = 2.4 x 10–6.

0.00242 mol/L

Solution

The first here is to determine the direction of the reaction. Now, there are both reactants and products, however, there is no NO, and remember, when one of the reactants or products is missing from the system, the equilibrium is going to be established by producing some of that compound. Therefore, in this case, the reaction is going to proceed to the right to produce some NO.

So, let’s set up an ICE table assigning 2x as the depletion in NOCl concentration:

2NOCl(g) ⇆ 2NO(g) + Cl2(g)

The equilibrium constant is:

$K = \frac{{{{[{\rm{NO}}]}^{\rm{2}}}[{\rm{C}}{{\rm{l}}_{\rm{2}}}]}}{{{{[{\rm{NOCl}}]}^{\rm{2}}}}}$

$K = \;\frac{{{{\left( {{\rm{2x}}} \right)}^{\rm{2}}}\left( {{\rm{1}}{\rm{.65 + x}}} \right)}}{{{{\left( {{\rm{2}}{\rm{.00 – 2x}}} \right)}^{\rm{2}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.4 \times 1}}{{\rm{0}}^{{\rm{–6}}}}$

This is a cubic equiation, and we can approach it by an approximation that 2x<<2.00 or x<<1.00. This allows to simplify the equation as follows:

$K = \;\frac{{{{\left( {{\rm{2x}}} \right)}^{\rm{2}}}\left( {{\rm{1}}{\rm{.65}}} \right)}}{{{{\left( {{\rm{2}}{\rm{.00}}} \right)}^{\rm{2}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.4 \times 1}}{{\rm{0}}^{{\rm{–6}}}}$

$K = \;\frac{{{\rm{4}}{{\rm{x}}^{\rm{2}}}\left( {{\rm{1}}{\rm{.65}}} \right)}}{{\rm{4}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.4 \times 1}}{{\rm{0}}^{{\rm{–6}}}}$

6.6x2 = 9.6×10-6

x=0.00121

The equilibrium concentratio of NO is 2x, so that is 0.00242 mol/L.

Now, we need to plug the numbers and see if the value for the equilibrium constant is close enough to the given number (K = 2.4 x 10–6).

$K = \;\frac{{{{\left( {{\rm{2x}}} \right)}^{\rm{2}}}\left( {{\rm{1}}{\rm{.65 + x}}} \right)}}{{{{\left( {{\rm{2}}{\rm{.00 – 2x}}} \right)}^{\rm{2}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.4 \times 1}}{{\rm{0}}^{{\rm{–6}}}}$

$K = \;\frac{{{{\left( {{\rm{2}} \times {\rm{0}}{\rm{.00121}}} \right)}^{\rm{2}}}\left( {{\rm{1}}{\rm{.65 + 0}}{\rm{.00121}}} \right)}}{{{{\left( {{\rm{2}}{\rm{.00 – 2}} \times {\rm{0}}{\rm{.00121}}} \right)}^{\rm{2}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.42 \times 1}}{{\rm{0}}^{{\rm{–6}}}}$

And that is very close to the value of K, and therefore, out approximation was correct and the equilibrium concentratio of NO is 0.00242 mol/L.

26.

Consider the following equilibrium process

2SO3(g) ⇆ 2SO2(g) + O2(g)   ΔH° = 2198 kJ/mol

How will the concentrations of SO2, O2, and SO3 be affected in each scenario?

(a) the temperature is increased

(b) the pressure is decreased by increasing the volume of the container

(c) the concentration of O2 is increased

(e) an inert gas is added at constant volume

(a) The reaction will shift to the right.

(b) The reaction will shift to the right.

(c) The reaction will shift left.

(d) It has no effect on the equilibrium position.

(e) It has no effect on the equilibrium composition.

Solution

(a) the temperature is increased: We need to first realize that this is an endothermic reaction (ΔHo positive). This means the reaction consumes heat, or, in other words, heat is a reactant. According to Le Châtelier’s principle, if we increase the temperature, the reaction will go in the direction that absorbs heat. Therefore, the reaction will shift right.

(b) the pressure on the gases is decreased: The number of gas molecules is different, so the reaction will respond in a way that brings the pressure up again. The products contain more gas molecules, therefore, the reaction will shift to the right.

(c) the concentration of O2 is increased: The reaction will shift left to use up some of the added oxygen gas and decrease its concentration.

(d) a is added catalyst: The catalyst is used to speed up the reaction so it reaches equilibrium more quickly. However, it has no effect on the equilibrium position.

(e) an inert gas is added at constant volume: Remember, changing the pressure of a gaseous reaction only has an effect on the equilibrium position if it results in changes of partial pressures of gases in the chemical equation. And although adding an inert gas to the system, increases the total pressure, the partial pressures of gases do not change. Therefore, it has no effect on the equilibrium composition.

27.

Suppose you need to increase the amount of C3H6Cl2 produced in the following exothermic reaction:

C3H6(g) + Cl2(g) ⇆ C3H6Cl2(g)

Which of the following strategies will work once the reaction mixture reaches equilibrium?

a) decreasing the reaction volume

b) removing C3H6Cl2 from the reaction mixture as it forms

b) removing C3H6Cl2 from the reaction mixture as it forms

Solution

a) decreasing the reaction volume – Has no effect since the number of gas molecules is equal on both sides.

b) removing C3H6Cl2 from the reaction mixture as it forms – Will increase the yield of the reaction since the equilibrium will shift to the right to reinstate the amount of product removed from the system.

c) adding a catalyst – Has no effect on the equilibrium position

d) adding Cl– Will shift the reaction right and thus, increase the amount of product.

28.

Predict the shift in the equilibrium position that will occur for each of the following reactions at equilibrium when the volume of the reaction container is increased.

1) PCl3(g) + Cl2(g) ⇆ PCl5(g)

2) 2NBr3(g) ⇆ N2(g) + 3Br2(g)

3) CO(g) + Cl2(g) ⇆ COCl2(g)

4) H2(g) + B2(g) ⇆ 2HBr(g)

5) MgCO3(s) ⇆ MgO(s) + CO2(g)

1) The reaction will shift left

2) The reaction will shift right

3) The equilibrium will not change

4) The equilibrium will not change

5) The reaction will shift right.

Solution

If we increase the volume of a reaction container, the reaction itself will shift to the side that contains more gas molecules since this increases its own volume. If there are an equal number of gas molecules on both sides of the reaction, then the reaction will remain at equilibrium regardless of how we change the volume of the container.

1) The reaction will shift left producing more reactants because there are 2 molecules of gas reactants compared with 1 molecule of gas on the product side.

2) The reaction will shift right producing more products because there are 4 molecules of gas products compared with 2 molecules of gas on the reactants side.

3) The equilibrium will not change because both sides contain an equal number of molecules in gas state.

4) The equilibrium will not change because both sides contain an equal number of molecules in gas state.

5) Only the products contain a gas molecule, therefore, the reaction will shift right. Remember, we ignore the solids and liquids and only concentrate on the gases because of their extremely large volume compared with solids and liquids.

29.

Consider the following equilibrium process for the commercial production of hydrogen:

CO(g) + H2O(g) ⇆ CO2(g) + H2(g)    ΔH° = +42 kJ/mol

Predict the direction of the shift in equilibrium when

(a) the temperature is raised.

(b) more CO gas is added to the reaction mixture.

(c) some CO2 is removed from the mixture.

(d) the pressure on the gases is increased by changing the volume of the container.

(e) a catalyst is added to the reaction mixture

(a) The reaction will shift right.

(b) The reaction will shift right.

(c) The reaction will shift right.

(d) No effect on the equilibrium.

(e) It has no effect on the equilibrium position.

Solution

(a) the temperature is raised: We need to first realize that this is an endothermic reaction (ΔHo positive). This means the reaction consumes heat, or, in other words, heat is a reactant. According to Le Châtelier’s principle, if we increase the temperature, the reaction will go in the direction that absorbs heat. Therefore, the reaction will shift right.

(b) more CO gas is added to the reaction mixture: The reaction will shift right to consume some of the added CO gas and decrease its concentration.

(c) some CO2 is removed from the mixture: The reaction will shift right to reinstate the original concentration of CO2.

(d) the pressure on the gases is increased: If the number of gas molecules was different, the reaction would respond in a way that brings the pressure back down. However, the number of gas molecules is equal here, therefore, there will be no effect on the equilibrium.

(e) a catalyst is added to the reaction mixture: The catalyst is used to speed up the reaction so it reaches equilibrium more quickly. However, it has no effect on the equilibrium position.

30.

Consider this reaction at equilibrium:

N2 + O2(g) ⇆ 2NO(g)

Predict whether the reaction will shift left, shift right, or remain unchanged after each disturbance.

a) NO is added to the reaction mixture.

b) N2 is added to the reaction mixture.

c) NO is removed from the reaction mixture.

a) The reaction will shift left.

b) The reaction will shift right.

c) The reaction will shift right.

Solution

According to the Le Châtelier’s principle, when a system in chemical equilibrium is disturbed by a change of temperature, pressure, or concentration, the system shifts in the direction that counteracts this change.

a) NO is added: The reaction will shift left to use up some of the added nitrogen monoxide and decrease its concentration.

b) N2 is added: The reaction will shift right to consume some of the added nitrogen gas and decrease its concentration.

c) NO is removed: The reaction will shift right to reinstate the original concentration of NO.