# 15: Equilibria

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## 15.1 Chemical Equilibrium

Textbook: Section 15.1

## 15.2 Equilibrium Constant and Reaction Quotient

Textbook: Section 15.2

Exercise $$\PageIndex{2.a}$$

Use the following reaction to answer

$$H_{2}(g)+Br_{2}(l)\rightleftharpoons 2HBr(g)$$

Which one of the following is the equilibrium constant expression for at 25°C?

1. $$k=\frac{\left [ HBr \right ]^{2}}{\left [ H_{2} \right ]\left [ Br_{2} \right ]}$$
2. $$k=\frac{\left [ HBr \right ]^{2}}{\left [ H_{2} \right ]}$$
3. $$k=\frac{\left [ HBr \right ]}{\left [ H_{2} \right ]}$$
4. none of the above

b. $k=\frac{\left [ HBr \right ]^{2}}{\left [ H_{2} \right ]} \nonumber$

You do not include the Br2(l) because the concentration of pure incompressible states of matter (liquids and solids) are always a constant value (related to their density). Therefore, they are absorbed into the equilibrium constant, (a constant divided or multiplied by another constant is still an constant).

Exercise $$\PageIndex{2.b}$$

Use the following reaction to answer

$$H_{2}(g)+Br_{2}(g)\rightleftharpoons 2HBr(g)$$

For a system at equilibrium, whatt is the equilibrium concentration of HBr if H2 and Br2 are both 0.800M, and K=8.0  at 70°C.

$k=\frac{\left [ HBr \right ]^{2}}{\left [ H_{2} \right ]\left [ Br_{2} \right ]} \nonumber$

$\left [ HBr \right ]=\sqrt{k\left [ H_{2} \right ]\left [ Br_{2} \right ]}=\sqrt{8.0*0.80^{2}}=2.26M \nonumber$

Exercise $$\PageIndex{2.c}$$

Why is Br2 a liquid in question 15.2.a and a gas in 15.2.b ?

The boiling point of Br2 is 58.8°C. In question 15.2.a the temperature is below the boiling point, and in 15.2.b it is above.

Exercise $$\PageIndex{2.d}$$

Use the following reaction to answer

$$H_{2}(g)+Br_{2}(l)\rightleftharpoons 2HBr(g)$$

Calculate Kp at 25° for a system at equilibrium if: PH2 = 2.50x10-2 atm, and PHBr = 1.500atm?

$K_{p}=\frac{\left ( P_{HBr} \right )^{2}}{P_{H_{2}}}=\frac{1.50^{2}}{2.50*10^{-2}}=90 \nonumber$

Exercise $$\PageIndex{2.e}$$

Use the following reaction to answer

$$H_{2}(g)+Br_{2}(l)\rightleftharpoons 2HBr(g)$$

What is Kc for the system in question 15.2d?

$K_{c}=\frac{\left ( P_{HBr}/RT \right )^{2}}{P_{H_{2}}/RT}=\frac{\left ( P_{HBr} \right )^{2}}{P_{H_{2}}}*\left ( RT \right )^{-1} \nonumber$

$K_{c}=90*\left ( 0.08206*298.2 \right )^{-1} \nonumber$

$K_{c}=3.68 \nonumber$

Exercise $$\PageIndex{2.f}$$

$2NO(g)\underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} N_{2}O_{2}(g),$

$N_{2}O_{2}(g)+O_{2}(g) \rightarrow 2NO_{2}(g)$

$2NO(g)+O_{2}(g) \rightarrow 2NO_{2}(g)$

$k_{1}=4.8*10^{-3}M^{-1}s^{-1},k_{-1}=3.6*10^{-6}s^{-1},K_{2}=6.4*10^{3}\nonumber$

a. What is the value of Kc for the first reaction above?

b.What is the value of Kc for the third reaction above?

c. Does reaction first reaction above favor reactants or products?

$K=\frac{k_{f}}{k_{r}}=\frac{k_{1}}{k_{-1}}=\frac{4.8*10^{-3}}{3.6*10^{-6}}=1.3333*10^{3}=1.3*10^{3}\nonumber$

$K_{3}=K_{1}*K_{2}=\left ( 1.3333*10^{3} \right )\left ( 6.4*10^{3} \right )=8.5*10^{6}\nonumber$

$K=\frac{k_{1}}{k_{-1}}=\frac{4.8*10^{-3}}{3.6*10^{-6}}=1.3*10^{3}\nonumber$

K>>1, the forward reaction is in favor, products

Exercise $$\PageIndex{2g}$$

$H_{2}(g)+Br_{2}(g)\rightleftharpoons 2HBr(g) \nonumber$

In what direction will the reaction at 440°C proceed for the following initial concentrations?

[HBr]= 1.0x10-2M, [H2]=5.0x10-3M, [Br2]=1.5x10-2M and Keq=50 at 440°C.

$Q=\frac{\left [ 1.0*10^{-2} \right ]^{2}}{\left [ 5.0*10^{-3} \right ]\left [ 1.5*10^{-2} \right ]}=1.3\nonumber$

Q < K, therefore, the reaction will proceed towards the product direction

Exercise $$\PageIndex{2h}$$

Which of the following is the relationship between the rate constants for the forward and reverse reactions and the equilibrium constant for a process?

1. $$K= k_{f} + k_{r}$$
2. $$K= k_{f}k_{r}$$
3. $$K=k_{f}-k_{r}$$
4. $$K= \frac{1}{\left ( k_{f}k_{r} \right )}$$
5. $$K=\frac{k_{f}}{k_{r}}$$

e. $$K=\frac{k_{f}}{k_{r}}$$

Exercise $$\PageIndex{2i}$$

What is the Kc for the following reaction?

$$CO+3H_{2}\rightleftharpoons CH_{4}+H_{2}O$$

$K_{c}=\frac{\left [ CH_{4} \right ]\left [ H_{2}O \right ]}{\left [ CO \right ]\left [ H_{2} \right ]^{3}}\nonumber$

Exercise $$\PageIndex{2j}$$

The value of Kc for the reaction below is 2.0*10-1.0 at 100°C.

$$COCl_{2}(g)\rightleftharpoons CO(g)+Cl_{2}(g)$$

What is the value of Kc for the reverse reaction at 100°C?

$$CO(g)+Cl_{2}(g)\rightleftharpoons COCl_{2}(g)$$

$K_{c1}=2.0*10^{-10} \nonumber$

$K_{c2}=\frac{1}{K_{c1}}=\frac{1}{2.0*10^{-10}}=5.0*10^{9}$

### Relating Kp and KC

Exercise $$\PageIndex{2.1A}$$

What is the relationship between Kp and Kc, where $\Delta n=\sum \text{Product Stoichiometric Coef }-\sum \text{Reactant Stoichiometric Coef}$

1. $$K_{p}=K_{c}\left ( RT \right )^{\Delta n}$$
2. $$K_{c}=K_{p}\left ( RT \right )^{\Delta n}$$
3.  $$K_{p}=K_{c}$$
4. None of the above

a. $$K_{p}=K_{c}\left ( RT \right )^{\Delta n}$$

Exercise $$\PageIndex{2.1b}$$

$$3A_{2}(g)+2B_{3}(s)\rightarrow 6A_{3}B_{2}(g)$$

What is the Δn for the equation above?

$3A_{2}(g)+2B_{3}(s)\rightarrow 6A_{3}B_{2}(g)\nonumber$

$\Delta n=6-3=3\nonumber$

Exercise $$\PageIndex{2.1c}$$

$$3A_{2}(g)+2B_{3}(s)\rightarrow 6A_{3}B_{2}(g)$$

Solve for Kp, given Kc = 2.3*204 at 30°C for the equation above?

$K_{p}=K_{c}\left ( RT \right )^{\Delta n}\nonumber$

$K_{p}=2.3*10^{4}\left (0.08206*303\right )^{\left(6-3 )\right)}\nonumber$

$K_{p}=3.5*10^{8}\nonumber$

Exercise $$\PageIndex{2.1d}$$

$$3A_{2}(g)+2B_{3}(s)\rightarrow 6A_{3}B_{2}(g)$$

Solve for Kc, given Kp = 2.3*104 at 30°C for the equation above?

$K_{p}=K_{c}\left ( RT \right )^{\Delta n}\nonumber$

$2.3*10^{4}=K_{c}\left (0.08206*303\right )^{\left (6-3\right )}\nonumber$

$K_{c}=\frac{2.3*10^{4}}{\left (0.08206*303\right )^{\left (6-3\right )}}\nonumber$

$K_{c}=1.5\nonumber$

Exercise $$\PageIndex{2.1e}$$

$$3A_{2}(g)+2B_{3}(s)\rightarrow 6A_{3}B_{2}(g)$$

Solve for Kc, given Kp = 3.1*104 at 30°C for the equation above?

$K_{p}=K_{c}\left ( RT \right )^{\Delta n}\nonumber$

$3.1*10^{4}=K_{c}\left (0.08206*298\right )^{\left (6-3\right )}\nonumber$

$K_{c}=\frac{3.1*10^{4}}{\left (0.08206*298\right )^{\left (6-3\right )}} \nonumber$

$K_{c}=2.1\nonumber$

Exercise $$\PageIndex{2.1f}$$

Solve for Kc, given Kp = 3.2*104 at 30°C for the following equation:

$3A_{2}(g)+4B_{3}(g)\rightarrow 2A_{3}B_{2}(g)\nonumber$

$K_{p}=K_{c}\left ( RT \right )^{\Delta n}\nonumber$

$3.2*10^{4}=K_{c}\left (0.08206*303\right )^{\left (2-3-4\right )}\nonumber$

$K_{c}=\frac{3.2*10^{4}}{\left (0.08206*303\right )^{\left (-5\right )}}\nonumber$

$K_{c}=3.0*10^{11}\nonumber$

Exercise $$\PageIndex{2.1g}$$

What is the temperature, given Kc = 3.5*108, and Kp = 5.8*105 for the following equation?

$A_{2}(g)+3B_{3}(g)\rightarrow 2AB_{3}(g)\nonumber$

$K_{p}=K_{c}\left ( RT \right )^{\Delta n}\nonumber$

$5.8*10^{5}=3.5*10^{8}\left (0.08206*T\right )\nonumber$

$T=\frac{\left ( \frac{5.8*10^{5}}{3.5*10^{8}} \right )^\frac{1}{\left ( 2-4 \right )}}{0.08206}\nonumber$

$T=299K\nonumber$

## 15.3 Determining an Equilibrium Constant

Textbook: Section 15.3

Exercise $$\PageIndex{3.a}$$

$H_{2}(g)+Br_{2}(g)\rightleftharpoons 2HBr(g) \nonumber$

What is the equilibrium concentration of Bromine at 440°C if given the following initial conditions?

[H2]= 0.60M, [HBr]=1.25M and [Br2]=0

 R H2 (g)            + Br2 (g) ⇌ 2HBr (g) I 0.60 M 0 1.25 M C x x -2x E 0.60+x x 1.25-2x

$K=\frac{\left [ 1.25-2x \right ]^{2}}{\left [ 0.60+x\right ]\left [ x \right ]}=50 \nonumber$

$\left ( 1.25-2x \right )^{2}=50\left ( 0.60+x^{2} \right ) \nonumber$

$46x^{2}+35x-1.5625=0 \nonumber$

$x=0.042M \nonumber$

Exercise $$\PageIndex{3.b}$$

$H_{2}(g)+Br_{2}(g)\rightleftharpoons 2HBr(g) \nonumber$

What is [HBr]eq at 440°C for the following initial conditions?

[H2]= 0.60M, [HBr]=1.25M and [Br2]=0

 R H2 (g)            + Br2 (g) ⇌ 2HBr (g) I 0.60 M 0 1.25 M C x x -2x E 0.60+x x 1.25-2x

$K=\frac{\left [ 1.25-2x \right ]^{2}}{\left [ 0.60+x\right ]\left [ x \right ]}=50 \nonumber$

$\left ( 1.25-2x \right )^{2}=50\left ( 0.60+x^{2} \right ) \nonumber$

$46x^{2}+35x-1.5625=0 \nonumber$

$x=0.042M \nonumber$

$1.25-\left( 2*0.042\right) =1.166M \nonumber$

Exercise $$\PageIndex{3.c}$$

$H_{2}(g)+Br_{2}(g)\rightleftharpoons 2HBr(g) \nonumber$

Determine Keq at a certain temperature if 0.80 mol HBr is produced after 0.50 mol H2 and Br2 react in a 2 L container.

 R H2 (g)            + Br2 (g) ⇌ 2HBr (g) I 0.25 M 0.25 M 0 C -x -x +2x E 0.25-x 0.25-x 2x

$0.8mol/2.0L=0.40M\nonumber$

$0.40M=2x \nonumber$

$x=0.2M \nonumber$

$K=\frac{[HBr]^2}{[H_2][Br_2]}=\frac{\left [ 0.4 \right ]^{2}}{\left [ 0.05 \right ]\left [ 0.05 \right ]}=64 \nonumber$

Exercise $$\PageIndex{3.d}$$

$H_{2}(g)+Br_{2}(g)\rightleftharpoons 2HBr(g) \nonumber$

At a certain temperature, Keq=6.30x102. What is the equilibrium concentration of H2 if 4.50mol of each of the three species were placed into a 3.00L flask and the system came to equilibrium?

 R H2 (g)            + Br2 (g) ⇌ 2HBr (g) I 1.5 M 1.5 M 1.5 M C -x -x +2x E 1.5-x 1.5-x 1.5+2x

$K=\frac{\left [ 1.5+2x \right ]^{2}}{\left [ 1.5-x \right ]\left [ 1.5-x \right ]}=6.3*10^{2} \nonumber$

$x=1.33 M \nonumber$

$\left [ H_{2} \right ]_{Equil}=1.5-1.33=0.17 M \nonumber$

Exercise $$\PageIndex{3.e}$$

$H_{2}(g)+Br_{2}(g)\rightleftharpoons 2HBr(g) \nonumber$

What is [HBr] after equilibrium is reached if 4.50mol of each of the three species were placed into a 3.00L flask?

 R H2 (g)            + Br2 (g) ⇌ 2HBr (g) I 1.5 M 1.5 M 1.5 M C -x -x +2x E 1.5-x 1.5-x 1.5+2x

$K=\frac{\left [ 1.5+2x \right ]^{2}}{\left [ 1.5-x \right ]\left [ 1.5-x \right ]}=6.3*10^{2} \nonumber$
$x=1.33 M \nonumber$
$1.5+\left( 2*1.33\right) = 4.16 M \nonumber$

Exercise $$\PageIndex{3.f}$$

Determine Keq for the following reaction if 0.0500M N2O4 is placed in a container and it decomposes to an equilibrium value of 0.0155M.

$$N_{2}O_{4}(g)\rightleftharpoons 2NO_{2}(g)$$

 R N2O4 (g) ⇌ 2NO2 (g) I 0.05 M 0 C -x +2x E 0.0155M 1.5+2x

$x=0.05-0.0155=0.0345M \nonumber$

$K=\frac{\left [ 2x \right ]^{2}}{\left [ 0.0155 \right ]}=\frac{\left [ 2*0.0345 \right ]^{2}}{\left [ 0.0155 \right ]}=0.307\nonumber$

## 15.4: Using Equilibrium Constants

Textbook: Section 15.4

Exercise $$\PageIndex{4a}$$

$H_{2}(g)+Br_{2}(g)\rightleftharpoons 2HBr(g) \nonumber$

What is the equilibrium concentration of Bromine at 440°C for the following initial conditions?

[H2]= 0.60M, [HBr]=1.25M and [Br2]=0  and Keq=50

 R H2 (g)            + Br2 (g) ⇌ 2HBr (g) I 0.60 M 0 1.25 M C x x -2x E 0.60+x x 1.25-2x

$K=\frac{\left [ 1.25-2x \right ]^{2}}{\left [ 0.60+x\right ]\left [ x \right ]}=50 \nonumber$

$\left ( 1.25-2x \right )^{2}=50\left ( 0.60+x^{2} \right ) \nonumber$

$46x^{2}+35x-1.5625=0 \nonumber$

$x=0.042M \nonumber$

Exercise $$\PageIndex{4b}$$

$H_{2}(g)+Br_{2}(g)\rightleftharpoons 2HBr(g) \nonumber$

What is [HBr]E at 440°C for the following initial concentrations?

[H2]= 0.60M, [HBr]=1.25M and [Br2]=0 and Keq=50

 R H2 (g)            + Br2 (g) ⇌ 2HBr (g) I 0.60 M 0 1.25 M C x x -2x E 0.60+x x 1.25-2x

$K=\frac{\left [ 1.25-2x \right ]^{2}}{\left [ 0.60+x\right ]\left [ x \right ]}=50 \nonumber$

$\left ( 1.25-2x \right )^{2}=50\left ( 0.60+x^{2} \right ) \nonumber$

$46x^{2}+35x-1.5625=0 \nonumber$

$x=0.042M \nonumber$

$1.25-\left( 2*0.042\right) =1.166M \nonumber$

Exercise $$\PageIndex{4c}$$

$H_{2}(g)+Br_{2}(g)\rightleftharpoons 2HBr(g) \nonumber$

At a certain temperature, Keq=6.30x102. What is the equilibrium concentration of H2 if 4.50mol of each of the three species were placed into a 3.00L flask?

 R H2 (g)            + Br2 (g) ⇌ 2HBr (g) I 1.5 M 1.5 M 1.5 M C -x -x +2x E 1.5-x 1.5-x 1.5+2x

$K=\frac{\left [ 1.5+2x \right ]^{2}}{\left [ 1.5-x \right ]\left [ 1.5-x \right ]}=6.3*10^{2} \nonumber$

$x=1.33 M \nonumber$

$\left [ H_{2} \right ]_{Equil}=1.5-1.33=0.17 M \nonumber$

Exercise $$\PageIndex{4d}$$

$H_{2}(g)+Br_{2}(g)\rightleftharpoons 2HBr(g) \nonumber$

At a certain temperature, Keq=6.30x102. What is the equilibrium concentration of HBr if 4.50mol of each of the three species were placed into a 3.00L flask?

 R H2 (g)            + Br2 (g) ⇌ 2HBr (g) I 1.5 M 1.5 M 1.5 M C -x -x +2x E 1.5-x 1.5-x 1.5+2x

$K=\frac{\left [ 1.5+2x \right ]^{2}}{\left [ 1.5-x \right ]\left [ 1.5-x \right ]}=6.3*10^{2} \nonumber$
$x=1.33 M \nonumber$
$1.5+\left( 2*1.33\right) = 4.16 M \nonumber$

### Complete the Square

Exercise $$\PageIndex{4.1a}$$

$$H_{2}(g)+Br_{2}(g)\rightleftharpoons 2HBr(g)$$

For the reaction, what is the equilibrium concentration of Br2 at 4400C if initially [H2]= 0.60M [HBr]=1.25M [Br2]=0.60 and k=50?

 R H2(g)            + Br2(g) ⇌ 2HBr I 0.60 M 0.60 M 1.25 M C -x -x +2x E 0.60-x 0.60-x 1.25+2x

$K=\frac{\left [ 1.25+2x \right ]^{2}}{\left [ 0.60-x \right ]\left [ 0.60-x \right ]}=50 \nonumber$

$\frac{\left [ 1.25+2x \right ]^{2}}{\left [ 0.60-x \right ]^{2}}=50\nonumber$

$\left (\frac{\left [ 1.25+2x \right ]}{\left [ 0.60-x \right ]}\right )^{2}=50\nonumber$

$\sqrt{\left (\frac{\left [ 1.25+2x \right ]}{\left [ 0.60-x \right ]}\right )^{2}}=\sqrt{50}\nonumber$

$\frac{\left [ 1.25+2x \right ]}{\left [ 0.60-x \right ]}=7.07\nonumber$

$x=0.33\,M\nonumber$

$0.60-0.33=0.27\,M\nonumber$

Exercise $$\PageIndex{4.1b}$$

$$H_{2}(g)+Br_{2}(g)\rightleftharpoons 2HBr(g)$$

For the reaction, what is the equilibrium concentration of HBr at 4400C if initially [H2]= 0.60M [HBr]=1.25M [Br2]=0.60 and k=50

 R H2(g)            + Br2(g) ⇌ 2HBr I 0.60 M 0.60 M 1.25 M C -x -x +2x E 0.60-x 0.60-x 1.25+2x

$K=\frac{\left [ 1.25+2x \right ]^{2}}{\left [ 0.60-x \right ]\left [ 0.60-x \right ]}=50 \nonumber$

$\frac{\left [ 1.25+2x \right ]^{2}}{\left [ 0.60-x \right ]^{2}}=50\nonumber$

$\left (\frac{\left [ 1.25+2x \right ]}{\left [ 0.60-x \right ]}\right )^{2}=50\nonumber$

$\sqrt{\left (\frac{\left [ 1.25+2x \right ]}{\left [ 0.60-x \right ]}\right )^{2}}=\sqrt{50}\nonumber$

$\frac{\left [ 1.25+2x \right ]}{\left [ 0.60-x \right ]}=7.07\nonumber$

$x=0.33\,M\nonumber$

$1.25+\left(2*0.33\right)=1.91\,M\nonumber$

Exercise $$\PageIndex{4.1}$$

For the reaction, What is [O2] at equilibrium if 0.050mol of N2O2, O2, and NO2 are mixed in a 1.0L container? k=25

$$N_{2}O_{2}(g)+O_{2}(g)\rightarrow 2NO_{2}(g)$$

 R N2O2(g)            + O2(g) ⇌ 2NO2(g) I 0.050 M 0.050 M 0.050 M C -x -x +2x E 0.050-x 0.050-x 0.050+x

$K=\frac{\left [ 0.05+2x \right ]^{2}}{\left [ 0.05-x \right ]\left [ 0.05-x \right ]}=25\nonumber$

$\frac{\left [ 0.05+2x \right ]^{2}}{\left [ 0.05-x \right ]^{2}}=25\nonumber$

$\sqrt{\frac{\left [ 0.05+2x \right ]^{2}}{\left [ 0.05-x \right ]^{2}}}=\sqrt{25}\nonumber$

$\frac{\left [ 0.05+2x \right ]}{\left [ 0.05-x \right ]}=25\nonumber$

$x=0.03 M\nonumber$

$0.05-0.03=0.02 M\nonumber$

### Complete the Cube

Exercise $$\PageIndex{4.2a}$$

$$Cl_{2}(g)+2HBr(g)\rightleftharpoons Br_{2}(g)+2HCl(g)$$

For the reaction, what is the equilibrium concentration of HCl if the initial [Cl2]=0.1M, [HBr]=0.20M, [Br2]=[HCl]=0 and k=15?

 R Cl2(g)            + 2HBr(g) ⇌ Br2(g) 2HCl(g) I 0.10M 0.20M 0 0 C -x -2x +x +2x E 0.10-x 0.20-2x x 2x

$K=\frac{\left [ Br_{2} \right ]\left [ HCl \right ]^{2}}{\left [ HBr \right ]^{2}\left [ Cl \right ]}=\frac{\left [ x \right ]\left [ 2x \right ]^{2}}{\left [ 0.2-2x \right ]^{2}\left [ 0.1-x \right ]}=\frac{\left [ x \right ]\left(4\left [ x \right ]^{2}\right)}{\left(2\left [ 0.1-x \right ]^{2}\right)\left [ 0.1-x \right ]}\nonumber$

$K=\frac{4\left [ x \right ]^{3}}{4\left [ 0.1-x \right ]^{3}}=\left (\frac{\left [ x \right ]}{\left [ 0.1-x \right ]}\right )^{3}\Rightarrow (K)^\frac{1}{3}=\frac{\left [ x \right ]}{\left [ 0.1-x \right ]}\nonumber$

$(K)^\frac{1}{3}\left [ 0.1-x \right ]=\left [ x \right ]\Rightarrow -\left [ x \right ]\left ( 1+K^{1/3} \right )=-0.1K^{1/3}\nonumber$

$\left [ x \right ]=\frac{0.1K^{1/3}}{K^{1/3}+1}=\frac{0.1*15^{1/3}}{15^{1/3}+1}=0.0712\nonumber$

$2x=0.14\nonumber$

Exercise $$\PageIndex{4.2b}$$

$$Cl_{2}(g)+2HBr(g)\rightleftharpoons Br_{2}(g)+2HCl(g)$$

For the reaction, what is the equilibrium concentration of HBr if the initial [Cl2]=0.1M, [HBr]=0.20M, [Br2]=[HCl]=0 and k=15?

 R Cl2(g)            + 2HBr(g) ⇌ Br2(g) 2HCl(g) I 0.10M 0.20M 0 0 C -x -2x +x +2x E 0.10-x 0.20-2x x 2x

$K=\frac{\left [ Br_{2} \right ]\left [ HCl \right ]^{2}}{\left [ HBr \right ]^{2}\left [ Cl \right ]}=\frac{\left [ x \right ]\left [ 2x \right ]^{2}}{\left [ 0.2-2x \right ]^{2}\left [ 0.1-x \right ]}=\frac{\left [ x \right ]\left(4\left [ x \right ]^{2}\right)}{\left(2\left [ 0.1-x \right ]^{2}\right)\left [ 0.1-x \right ]}\nonumber$

$K=\frac{4\left [ x \right ]^{3}}{4\left [ 0.1-x \right ]^{3}}=\left (\frac{\left [ x \right ]}{\left [ 0.1-x \right ]}\right )^{3}\Rightarrow (K)^\frac{1}{3}=\frac{\left [ x \right ]}{\left [ 0.1-x \right ]}\nonumber$

$(K)^\frac{1}{3}\left [ 0.1-x \right ]=\left [ x \right ]\Rightarrow -\left [ x \right ]\left ( 1+K^{1/3} \right )=-0.1K^{1/3}\nonumber$

$\left [ x \right ]=\frac{0.1K^{1/3}}{K^{1/3}+1}=\frac{0.1*15^{1/3}}{15^{1/3}+1}=0.0712\nonumber$

$0.20-2x=0.06M\nonumber$

## 15.5 Equilibrium Constants and Coupled Reactions

Textbook: Section 15.5

Exercise $$\PageIndex{5a}$$

The equilibrium constant for reaction (1) is K. What is the equilibrium contant for equation (2)?

(1)$$SO_{2}(g) + \frac{1}{2}O_{2} \rightleftharpoons SO_{3}(g)$$

(2)$$2SO_{3}(g) \rightleftharpoons 2SO_{2}(g) + O_{2}(g)$$

1. 1/2K
2. 1/K2
3. 2K
4. K2
5. -K2

b. 1/K2

Exercise $$\PageIndex{5b}$$

The equilibrium constant for reaction (1) is K. What is the equilibrium constant for equation (2)?

(1)$$\frac{1}{3}N_{2}(g) + H_{2}(g) \rightleftharpoons \frac{2}{3}NH_{3}(g)$$

(2)$$2NH_{3} \rightleftharpoons N_{2} + 3H_{2}$$

1. K3
2. 3K
3. K/3
4. 1/K3
5. -K3

d. 1/K3

Exercise $$\PageIndex{5c}$$

If the value of Kc for the following reaction is 0.25:

$$SO_{2}(g)+NO_{2}(g)\rightleftharpoons SO_{3}+NO(g)$$

What is the value of Kc for the reaction below?

$$2SO_{2}(g)+2NO_{2}(g)\rightleftharpoons 2SO_{3}+2NO(g)$$

$K_{c1}=\frac{\left [ SO_{3} \right ]\left [ NO \right ]}{\left [ SO_{2} \right ]\left [ NO_{2} \right ]}=0.25\nonumber$

$K_{c2}=\left ( K_{c1} \right )^{2}=\left (\frac{\left [ SO_{3} \right ]\left [ NO \right ]}{\left [ SO_{2} \right ]\left [ NO_{2} \right ]}\right )^{2}=0.25^{2}=0.0623 \nonumber$

## 15.6 Le Chatelier’s Principle

Textbook: Section 15.6

$$2SO_{2}(g)+O_{2}(g)\rightleftharpoons 2SO_{3} (g)\,\,\,\,\,\Delta H^{0}=-99\,kJ/mol$$

Use this reaction to answer questions 6.a-f

Exercise $$\PageIndex{6.a}$$

$$2SO_{2}(g)+O_{2}(g)\rightleftharpoons 2SO_{3} (g)\,\,\,\,\,\Delta H^{0}=-99\,kJ/mol$$

If the temperature increases while the pressure is constant, the reaction will proceed towards which direction?

Reactants, because of $$\Delta H<0$$

Exercise $$\PageIndex{6.b}$$

$$2SO_{2}(g)+O_{2}(g)\rightleftharpoons 2SO_{3} (g)\,\,\,\,\,\Delta H^{0}=-99\,kJ/mol$$

For the same reaction, if the temperature is held constant and the pressure is increased, which direction will the reaction proceed?

Products, increasing the pressure favors the side with fewer gas molecules

Exercise $$\PageIndex{6.c}$$

$$2SO_{2}(g)+O_{2}(g)\rightleftharpoons 2SO_{3} (g)\,\,\,\,\,\Delta H^{0}=-99\,kJ/mol$$

If more oxygen is added, which direction will the reaction proceed?

Products, by adding more reactant this favors the formation of more product

Exercise $$\PageIndex{6.d}$$

$$2SO_{2}(g)+O_{2}(g)\rightleftharpoons 2SO_{3} (g)\,\,\,\,\,\Delta H^{0}=-99\,kJ/mol$$

If the container in which the reaction occurs is enlarged, which direction will the reaction proceed?

Reactants, by enlarging the container the partial pressure decreases, therefore it favors the side with more gas molecules

Exercise $$\PageIndex{6.e}$$

$$2SO_{2}(g)+O_{2}(g)\rightleftharpoons 2SO_{3} (g)\,\,\,\,\,\Delta H^{0}=-99\,kJ/mol$$

The addition of catalyst will make the reaction shift towards which direction?

There will be no effect, catalysts only change the activation energy of a reaction

Exercise $$\PageIndex{6.f}$$

$$2SO_{2}(g)+O_{2}(g)\rightleftharpoons 2SO_{3} (g)\,\,\,\,\,\Delta H^{0}=-99\,kJ/mol$$

The addition of He gas will make the reaction shift towards which direction?

There will be no effect, the participation of an inert gas does not affect the reaction

## 15.7: General Questions

Exercise $$\PageIndex{7a}$$

$$C_{4}H_{10}(g)\rightleftharpoons C_{2}H_{6}(g)+C_{2}H_{4}(g)$$

What is the concentration of C4H10  if the equilibrium concentrations of C2H6 and C2H4 are both 0.014M? Kc = 0.07

$K_{c}=\frac{\left [ 0.014 \right ]^{2}}{\left [ C_{4}H_{10} \right ]}=0.070 \nonumber$

$\left [ C_{4}H_{10} \right ]=0.0028M \nonumber$

Exercise $$\PageIndex{7b}$$

$$C_{4}H_{10}(g)\rightleftharpoons C_{2}H_{6}(g)+C_{2}H_{4}(g)$$

For the same reaction, if the initial concentration of C4H10 is 0.035M, and there is no C2H6 C2H4 present initially. What is the equilibrium concentration of C4H10?

 R C4H10(g) ⇌ C2H6(g) + C2H4(g) I 0.035 M 0 0 C -x +x +x E 0.035-x x x

$K=\frac{\left [ x \right ]^{2}}{\left [ 0.035-x \right ]}=0.070 \nonumber$

$x=0.026 M \nonumber$

$0.035-0.026=0.009 M \nonumber$

Exercise $$\PageIndex{7c}$$

$$C_{4}H_{10}(g)\rightleftharpoons C_{2}H_{6}(g)+C_{2}H_{4}(g)$$

For the same reaction, if the initial concentration of C4H10 is 0.035M, and there is no C2H6 C2H4 present initially. What is the equilibrium concentration of C2H6?

 R C4H10(g) ⇌ C2H6(g)     + C2H4(g) I 0.035 M 0 0 C -x +x +x E 0.035-x x x
$K=\frac{\left [ x \right ]^{2}}{\left [ 0.035-x \right ]}=0.070 \nonumber$$x=0.026 M \nonumber$
(section 15.4.3.1.1)

Exercise $$\PageIndex{7d}$$

$$C_{4}H_{10}(g)\rightleftharpoons C_{2}H_{6}(g)+C_{2}H_{4}(g)$$

Calculate the value of Kc if the initial concentrations of C4H10 is 0.030M while both C2H6 and C2H4 are both 0.023MAt equilibrium. Once the reaction is over the concentration of C4H10 becomes 0.018M.

 R C4H10(g) ⇌ C2H6(g)     + C2H4(g) I 0.030 M 0.023 M 0.023 M C -x +x +x E 0.018 0.023+x 0.023+x

$x=0.030-0.018=0.012 \nonumber$

$K_{c}=\frac{\left [ 0.035 \right ]^{2}}{\left [ 0.018 \right ]}=0.070 \nonumber$

$K_{c}=0.068 \nonumber$

Exercise $$\PageIndex{7e}$$

$$C_{4}H_{10}(g)\rightleftharpoons C_{2}H_{6}(g)+C_{2}H_{4}(g)$$

For the same reaction, if the 2.0L container was evacuated, then pumped in with gases at the following pressure, C4H10 = 1.2atm, C2H6and C2H4 = 0.6 atm. What is the partial pressure (in atm) of C4H10 at equilibrium? Kp =0.64.

 R C4H10(g) ⇌ C2H6(g)     + C2H4(g) I 1.2 atm 0.6 atm 0.6 atm C -x +x +x E 1.2-x 0.6+x 0.6+x

$K_{p}=\frac{\left [ 0.6+x \right ]^{2}}{\left [ 1.2-x \right ]}=0.64 \nonumber$

$x=0.2 atm \nonumber$

$1.2-0.2=1.0atm \nonumber$

Exercise $$\PageIndex{7f}$$

If the equilibrium constant for reaction (1) is 4.22*10-3, what is the value of the equilibrium constant for the reaction (2) in the following mechanism?

3A + 2B ⇌ 2D + E    (1)

2D + E ⇌ 3A +2B     (2)

$k_{eq1}=\frac{\left [ D \right ]^{2}\left [ E \right ]}{\left [ A \right ]^{3}\left [ B \right ]^{2}}\nonumber$
$k_{eq2}=\frac{\left [ A \right ]^{3}\left [ B \right ]^{2}}{\left [ D \right ]^{2}\left [ E \right ]}\nonumber$
$k_{eq2}=\frac{\left [ A \right ]^{3}\left [ B \right ]^{2}}{\left [ D \right ]^{2}\left [ E \right ]}=\frac{1}{k_{eq1}}=\frac{1}{4.22*10^{-3}}=237\nonumber$

Exercise $$\PageIndex{7g}$$

The reaction

$$A+B \rightleftharpoons X+Y$$

has Kc = 1977 at 472K. At equilibrium _____.

1. only products exist
2. only reactants exist
3. products predominate
4. reactants predominate
5. roughly equal molar amounts of products and reactants are present

c. products predominate

Exercise $$\PageIndex{7h}$$

What is Kp for the following reaction at 25°C, Kc = 3.0*105?

$$2H_{2}S(g)+3O_{2}\rightleftharpoons 2H_{2}O(g)+2SO_{2}(g)$$

$T = 25+273.15 = 298.15\nonumber$

$\Delta n = 4-5 = -1\nonumber$

$K_{p}=K_{c}\left ( RT \right )^{\Delta n} \nonumber$

$K_{p}=\left ( 3.0*10^{5} \right )\left ( 0.0821*298.15 \right )^{-1}=1.2*10^{4}\nonumber$

Exercise $$\PageIndex{7i}$$

The value of Kc for the following reaction is 1.10 at 25.0°C. What is the value of Kp for this reaction?

$$4CuO(s)+CH_{4}(g) \rightleftharpoons CO_{2}(g) + 4Cu(s) + 2H_{2}O(g)$$

$T = 25+273.15 = 298.15\nonumber$

$\Delta n = 3-1 = 2 \nonumber$

$K_{p}=K_{c}\left ( RT \right )^{\Delta n} \nonumber$

$K_{p}=\left(1.10 \right )\left ( 0.0821*298.15 \right )^{2}=6.59*10^{2}\nonumber$

Exercise $$\PageIndex{7j}$$

What is the value of Kc for a flask at equilibrium that contains 0.0114 M HCl, 0.0931 M Cl2, and 0.0154 M H2 at a certain temperature?

$$2HCl(g) \rightleftharpoons Cl_{2}(g) + H_{2}(g)$$

$K_{c}=\frac{\left [ Cl_{2} \right ]\left [ H_{2} \right ]}{\left [ HCl \right ]^{2}}=\frac{\left [0.0931\right ]\left [0.0154\right ]}{\left [0.0114\right ]^{2}}=11.0\nonumber$

Exercise $$\PageIndex{7k}$$

Consider the gaseous equilibrium:$$2A\rightarrow 2B+C$$

Determine the value of the missing B concentration at equilibrium.

 Exp # [A] at equilibrium [B] at equilibrium [C] at equilibrium 1 0.10 M 0.10 M 0.20 M 2 0.20 M 0.50 M 0.032 M 3 0.35 M ? 0.15 M

Find k by solving the equilibrium constant expression using either experiment 1 or experiment 2 data

$k=\frac{\left [ B \right ]^{2}\left [ C \right ]}{\left [ A \right ]^{2}}=\frac{0.10^{2}*0.20}{0.10^{2}}=0.20\nonumber$

Solve the equilibrium constant expression for [B] then use k from above and solve with experiment 3 data

$\left [ B \right ]=\left ( \frac{k\left [ A \right ]^{2}}{\left [ C \right ]} \right )^{1/2}=\left ( \frac{0.20*\left [ 0.35 \right ]^{2}}{\left [ 0.15 \right ]} \right )^{1/2}=1.0 M\nonumber$

Exercise $$\PageIndex{7l}$$

Which of the following reactions at equilibrium has the following equilibrium constant expression?

$$\frac{\left [IBr\right ]^2}{\left [I_{2}\right ]\left [Br_{2}\right ]}$$

1. $$I_{2}(g)+Br_{2}(g)\rightleftharpoons IBr(g)$$
2. $$I_{2}(g)+Br_{2}(g)\rightleftharpoons 2 IBr(g)$$
3. $$2 IBr(g)\rightleftharpoons I_{2}(g)+Br_{2}(g)$$
4. $$2 I_{2}(g)+2 Br_{2}(g)\rightleftharpoons IBr(g)$$
5. $$IBr(g)\rightleftharpoons 2I_{2}(g)+2Br_{2}(g)$$

b. $$I_{2}(g)+Br_{2}(g)\rightleftharpoons 2 IBr(g)$$

## General Questions

Exercise $$\PageIndex{o}$$

The value of Kc for the following reaction at equilibrium is 54.0 at 427°C.

$$Eq \; 1: \;H_{2}(g)+I_{2}(g)\rightleftharpoons 2HI(g)$$

At this temperature, what is the value of Kc for:

$$Eq \; 2: \; HI(g)\rightleftharpoons \frac{1}{2}H_{2}(g)+\frac{1}{2}I_{2}(g)$$

If you write an equilibrium equation backwards the new constant is the reciprocal of the first, so

$2HI(g) \rightleftharpoons H_2(g)+I_2(g)\nonumber$ and

$K_{-1}=\frac{1}{K_1}\nonumber$

Since eq -1 is twice eq 2, it equilibrium constant is the square of eq 2, so eq2 is the square root of eq. 1.d.

$K_2 = \frac{1}{(K_{-1})^{1/2}}=\frac{1}{\sqrt{54}}=0.136$

Exercise $$\PageIndex{p}$$

What is the equilibrium constant expression for the following reaction?

$$Al_{2}\left (SO_{3}\right )_{3}(s)+6HCl(g)\rightleftharpoons 2AlCl_{3}(s)+3H_{2}O(l)+3SO_{2}(g)$$

$K_{c}=\frac{\left [SO_{2}\right ]^{3}}{\left [HCl\right ]^{6}}\nonumber$

Exercise $$\PageIndex{q}$$

What is the equilibrium constant expression for the following reaction?

$$3SO_{2}(g)\rightleftharpoons 2SO_{3}(g)+S(s)$$

$K_{c}=\frac{\left [SO_{3}\right ]^{2}}{\left [SO_{2}\right ]^{3}}\nonumber$

Exercise $$\PageIndex{r}$$

Consider the following chemical reaction:

$$H_{2}(g)+I_{2}(g)\rightleftharpoons 2HI(g)$$

At equilibrium, the concentrations of H2, I2, and HI were 0.15M, 0.033 M, and 0.55 M, respectively. What is the Kc for this reaction?

$K_{c}=\frac{\left [HI\right ]^{2}}{\left [H_{2}\right ]\left [I_{2}\right ]}=\frac{\left [0.55\right ]^{2}}{\left [0.15\right ]\left [0.033\right ]}=61\nonumber$

Exercise $$\PageIndex{s}$$

$$PCl_{5}(g)\rightleftharpoons PCl_{3}(g)+Cl_{2}(g)$$

Initially, 1.26 mol of PCl5(g) was placed in a 1.0 L flask. At equilibrium, 1.08 mol of PCl5(g) was present. What is the value of Kc for this reaction at this temperature?

 R PCl5 ⇌ PCl3     + Cl2 I 1.26 0 0 C -x +x +x E 1.08 x x

$1.08=1.26-x \nonumber$

$x=1.26-1.08=0.18 \nonumber$

$K_{c}=\frac{\left [ PCl_{3} \right ]\left [Cl_{2} \right ]}{\left [ PCl_{5} \right ]}=\frac{x^{2}}{1.26}=\frac{0.18^{2}}{1.26}=0.03\nonumber$

Exercise $$\PageIndex{t}$$

$$PCl_{5}(g)\rightleftharpoons PCl_{3}(g)+Cl_{2}(g)$$

What is the equilibrium partial pressure of PCl3? If in a 3.00 L vessel that was charged with 0.123 atm of PCl5 has a Kp of 0.0121?

 R PCl5 ⇌ PCl3     + Cl2 I 0.123 0 0 C -x +x +x E 0.123-x x x

$K_{p}=\frac{\left [PCl_{3}\right ]\left [Cl_{2}\right ]}{\left [PCl_{5}\right ]}=\frac{x^{2}}{0.123-x}=0.0121\nonumber$

$\frac{x^{2}}{0.123-x}=0.0121 \nonumber$

$x^{2}=\left (0.123-x\right )*0.0121 \nonumber$

$x^{2}=0.001488-0.0121x \nonumber$

$0=x^{2}+0.0121x-0.001488\nonumber$

Use the quadratic formual to solve for x

$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}=\frac{-0.0121\pm \sqrt{0.0121^{2}-\left ( 4*1*-0.001488 \right )}}{2*1}\nonumber$

$x=0.0330\nonumber$

The negative square is rejected because pressure cannot be negative.

Exercise $$\PageIndex{u}$$

The value of Kc for the following reaction is 0.070. What is the equilibrium concentration (M) of C4H10 if the equilibrium concentrations of C2H6 and C2H4 are both 0.035M?

$$C_{4}H_{10}(g)\rightleftharpoons C_{2}H_{6}(g)+C_{2}H_{4}(g)$$

 R C4H10 ⇌ C2H6     + C2H4 I x 0 0 C -y +y +y E x-y y y

$\left [C_{2}H_{6}\right ]=\left [C_{2}H_{4}\right ]=0.035\,M=y \nonumber$

$K_{c}=\frac{\left [C_{2}H_{6}\right ]\left [C_{2}H_{4}\right ]}{\left [C_{4}H_{10}\right ]}=\frac{0.035^{2}}{x-0.035}=0.070 \nonumber$

$0.070x-0.00245=0.001225 \nonumber$

$0.070x=0.003675 \nonumber$

$x=0.0525 \nonumber$

$\left [C_{4}H_{10}\right ]=x-y=0.0525-0.035=0.018M\nonumber$

Exercise $$\PageIndex{v}$$

Nitrosyl bromide decomposes according to the following equation:

$$2NOBr(g)\rightleftharpoons 2NO(g) + Br_{2}(g)$$

A sample of NOBr (0.64 mol) was placed in a 1.00 L flask containing no NO or Br2. At equilibrium, the flask contained 0.46 mol of NOBr. How many moles of NO and Br2 are in the flask at equilibrium?

 R 2NOBr ⇌ 2NO     + Br2 I 0.64 mol 0 0 C -2x +2x +x E 0.64-2x 2x x

$0.64-2x=0.46 \nonumber$

$-2x=-0.18 \nonumber$

$x=0.09 \nonumber$

$mol_{NO}=2x=2*0.09=0.18\,mol \nonumber$

$mol_{Br_{2}}=x=0.09\,mol \nonumber$

Exercise $$\PageIndex{w}$$

Which of the following will shift to the left in response to a decrease in volume?

1. $$H_{2}(g)+Cl_{2}(g)\rightleftharpoons 2HCl(g)$$
2. $$N_{2}(g) + 3H_{2}(g)\rightleftharpoons 2NH_{3}(g)$$
3. $$2SO_{3}(g)\rightleftharpoons 2SO_{2}(g)+O_{2}(g)$$
4. $$2HI(g) \rightleftharpoons H_{2}(g)+I_{2}(g)$$
5. $$4Fe(s)+3O_{2}(g)\rightleftharpoons 2Fe_{2}O_{3}(s)$$

c. $$2SO_{3}(g)\rightleftharpoons 2SO_{2}(g)+O_{2}(g)$$

Exercise $$\PageIndex{x}$$

For the endothermic reaction

$$CaCO_{3}(s)\rightleftharpoons CaO(s)+CO_{2}(g)$$

only _____ would favor shifting the equilibrium position to form more CO2 gas.

1. both decreasing the system temperature and increasing the system pressure
2. decreasing the system temperature
3. increasing both the system temperature and the system pressure
4. increasing the system pressure
5. increasing the system temperature

e. increasing the system temperature

Exercise $$\PageIndex{y}$$

Which of the following reactions would increase pressure at constant temperature not change the concentration of reactants and products?

1. $$2N_{2}(g)+O_{2}(g)\rightleftharpoons 2N_{2}O(g)$$
2. $$N_{2}(g)+2O_{2}(g)\rightleftharpoons 2NO_{2}(g)$$
3. $$N_{2}(g)+3H_{2}(g)\rightleftharpoons 2NH_{3}(g)$$
4. $$N_{2}O_{4}(g)\rightleftharpoons 2NO_{2}(g)$$
5. $$N_{2}(g)+O_{2}(g)\rightleftharpoons 2NO(g)$$

e. $$N_{2}(g)+O_{2}(g)\rightleftharpoons 2NO(g)$$

Exercise $$\PageIndex{z}$$

Consider the following reaction at equilibrium

$$2CO_{2}\rightleftharpoons 2CO(g)+O_{2}(g)$$

The yield of CO(g) in reaction can be maximized by carrying out the reaction _____.

1. at high temperature and high pressure
2. at high temperature and low pressure
3. at low temperature and high pressure
4. at low temperature and low pressure
5. in the presence of solid carbon

d. at low temperature and low pressure

Exercise $$\PageIndex{aa}$$

The effect of a catalyst on a chemical reaction is to _____.

1. accelerate the forward reaction only
2. increase the entropy change associated with a reaction
3. lower the energy of the transition state
4. make reactions more exothermic
5. react with product, effectively removing it and shifting the equilibrium to the right