# 15: Equilibria

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### Equilibrium Constant

Exercise \(\PageIndex{a}\)

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?

- \(k=\frac{\left [ HBr \right ]^{2}}{\left [ H_{2} \right ]\left [ Br_{2} \right ]} \)
- \(k=\frac{\left [ HBr \right ]^{2}}{\left [ H_{2} \right ]} \)
- \(k=\frac{\left [ HBr \right ]}{\left [ H_{2} \right ]}\)
- none of the above

**Answer**-
b. \[k=\frac{\left [ HBr \right ]^{2}}{\left [ H_{2} \right ]} \]

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 \(\PageIndex{b}\)

Use the following reaction to answer

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

What is the equilibrium concentration of HBr if H_{2} and Br_{2} are both 0.800M, and K=8.0 at 60°C.

**Answer**-
\[k=\frac{\left [ HBr \right ]^{2}}{\left [ H_{2} \right ]\left [ Br_{2} \right ]}\]

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

Exercise \(\PageIndex{c}\)

Why is Br_{2} a liquid in question 15.1 and a gas in 15.2 when the pressure is constant?

**Answer**-
The boiling point of Br

_{2}is 58.8°C. In question 15.1 the temperature is below the boiling point, and in 15.2 it is above.

Exercise \(\PageIndex{d}\)

Use the following reaction to answer

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

Calculate K_{p} at 25° for a system at equilibrium if: P_{H2 }= 2.50x10^{-2 }atm, and P_{HBr} = 1.500atm?

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

Exercise \(\PageIndex{e}\)

Use the following reaction to answer

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

What is K_{c} for the system in question 15.3?

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

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

\[K_{c}=3.68\]

Exercise \(\PageIndex{f}\)

Use the reaction from Figure 15.1 to answer:

What is the value of K_{c} for reaction (1) in Figure 15.1?

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

Exercise \(\PageIndex{g}\)

Use the reaction from Figure 15.1 to answer:

What is the value of K_{c} for step (3) of question 15.6?

**Answer**-
\[K_{3}=K_{1}*K_{2}=\left ( 1.3333*10^{3} \right )\left ( 6.4*10^{3} \right )=8.5*10^{6}\]

Exercise \(\PageIndex{h}\)

Use the reaction from Figure 15.2 to answer:

Does step 1 in question 15.6 favor reactants or products?

**Answer**-
\[K=\frac{k_{1}}{k_{-1}}=\frac{4.8*10^{-3}}{3.6*10^{-6}}=1.3*10^{3}\]

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

### Calculations of Equilibrium Constant

Exercise \(\PageIndex{i}\)

Use the reaction from Figure 15.2 to answer:

In what direction will the reaction at 440°C proceed if:

[HBr]= 1.0x10^{-2}M, [H_{2}]=5.0x10^{-3}M, [Br_{2}]=1.5x10^{-2}M and K_{eq}=50 at 440°C.

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

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

Exercise \(\PageIndex{j}\)

Use the reaction from Figure 15.2 to answer:

What is the equilibrium concentration of Bromine at 440°C if:

[H_{2}]= 0.60M, [HBr]=1.25M and [Br_{2}]=0

**Answer**-
R H _{2}(g) +Br _{2}(g)⇌ 2HBr (g) I 0.60 M 0 1.25 M C x x -2x E 0.60+x x 1.25-2x \[K=\frac{\left [ 1.25-2x \right ]^{2}}{\left [ 0.60+x\right ]\left [ x \right ]}=50\]

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

\[46x^{2}+35x-1.5625=0\]

\[x=0.042M\]

Exercise \(\PageIndex{k}\)

Use the reaction from Figure 15.2 to answer:

What is [HBr] after equilibrium is reached for the problem described in question 15.9?

**Answer**-
R H _{2}(g) +Br _{2}(g)⇌ 2HBr (g) I 0.60 M 0 1.25 M C x x -2x E 0.60+x x 1.25-2x - \[\left ( 1.25-2x \right )^{2}=50\left ( 0.60+x^{2} \right )\]
- \[46x^{2}+35x-1.5625=0\]
- \[x=0.042M\]
- \[1.25-\left( 2*0.042\right) =1.166M\]

Exercise \(\PageIndex{l}\)

Use the reaction from Figure 15.2 to answer:

Determine K_{eq} at a certain temperature if 0.80 mol HBr is produced after 0.50 mol H_{2} and Br_{2} react in a 2 L container.

**Answer**-
R H _{2}(g) +Br _{2}(g)⇌ 2HBr (g) I 0.25 M 0.25 M 0 C -x -x +2x E 0.25-x 0.25-x 2x - \[0.40M=2x\]
- \[x=0.2M\]
- \[K=\frac{\left [ 0.4 \right ]^{2}}{\left [ 0.05 \right ]\left [ 0.05 \right ]}=64\]

Exercise \(\PageIndex{m}\)

Use the reaction from Figure 15.32 to answer:

At a certain temperature, K_{eq}=6.30x10^{2}. What is the equilibrium concentration of H_{2} if 4.50mol of each of the three species were placed into a 3.00L flask?

**Answer**-
R H _{2}(g) +Br _{2}(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 - \[x=1.33 M\]
- \[\left [ H_{2} \right ]_{Equil}=1.5-1.33=0.17 M\]

Exercise \(\PageIndex{n}\)

Use the reaction from Figure 15.2 to answer:

What is [HBr] after equilibrium is reached for the problem described in question 15.12?

**Answer**-
R H _{2}(g) +Br _{2}(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 - \[x=1.33 M\]
- \[1.5+\left( 2*1.33\right) = 4.16 M\]

Exercise \(\PageIndex{o}\)

Determine K_{eq} for the following reaction if 0.0500M N_{2}O_{4} is placed in a container and it decomposes to an equilibrium value of 0.0155M.

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

**Answer**-
R N _{2}O_{4}(g)⇌ 2NO _{2}(g)I 0.05 M 0 C -x +2x E 0.0155M 1.5+2x - \[K=\frac{\left [ 2x \right ]^{2}}{\left [ 0.0155 \right ]}=\frac{\left [ 2*0.0345 \right ]^{2}}{\left [ 0.0155 \right ]}=0.307\]

### Le Chatelier’s Principle

Exercise \(\PageIndex{p}\)

Use the reaction from Figure 15.3 to answer:

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

**Answer**-
Reactants, because of \(\Delta H<0\)

Exercise \(\PageIndex{q}\)

Use the reaction from Figure 15.3 to answer:

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

**Answer**-
Products, increasing the pressure favors the side with fewer gas molecules

Exercise \(\PageIndex{r}\)

Use the reaction from Figure 15.3 to answer:

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

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

Exercise \(\PageIndex{s}\)

Use the reaction from Figure 15.4 to answer:

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

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

Exercise \(\PageIndex{t}\)

Use the reaction from Figure 15.3 to answer:

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

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

Exercise \(\PageIndex{u}\)

Use the reaction from Figure 15.3 to answer:

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

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

### Equilibrium Constant

Exercise \(\PageIndex{v}\)

Use the reaction from Figure 15.4 to answer:

What is the concentration of C_{4}H_{10} at equilibrium if the concentrations of C_{2}H_{6} and C_{2}H_{4} are both 0.014M? K_{c} = 0.07

**Answer**-
\[K_{c}=\frac{\left [ 0.014 \right ]^{2}}{\left [ C_{4}H_{10} \right ]}=0.070 \]

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

Exercise \(\PageIndex{w}\)

Use the reaction from Figure 15.4 to answer:

For the same reaction, if the initial concentration of C_{4}H_{10} is 0.035M, and there is no C_{2}H_{6} C_{2}H_{4} present initially. What is the equilibrium concentration of C_{4}H_{10}?

**Answer**-
R C _{4}H_{10}(g)⇌ C _{2}H_{6}(g) +C _{2}H_{4}(g)I 0.035 M 0 0 C -x +x +x E 0.035-x x x \[K=\frac{\left [ x \right ]^{2}}{\left [ 0.035-x \right ]}=0.070 \]

\[x=0.026 M \]

\[0.035-0.026=0.009 M\]

Exercise \(\PageIndex{x}\)

Use the reaction from Figure 15.4 to answer:

Following Q 15.23, what is the equilibrium concentration of C_{2}H_{6}?

**Answer**-
R C4H10(g) ⇌ C2H6(g) + C2H4(g) I 0.035 M 0 0 C -x +x +x E 0.035-x x x - \[x=0.026 M\]

Exercise \(\PageIndex{y}\)

Use the reaction from Figure 15.4 to answer:

If the initial concentration of C_{4}H_{10} is 0.030M, and the ones of C_{2}H_{6} and C_{2}H_{4} are both 0.023M. At equilibrium, the concentration of C_{4}H_{10} becomes 0.018M, what is the value of K_{c} of the reaction.

**Answer**-
R C _{4}H_{10}(g)⇌ C _{2}H_{6}(g) +C _{2}H_{4}(g)I 0.030 M 0.023 M 0.023 M C -x +x +x E 0.018 0.023+x 0.023+x - \[x=0.030-0.018=0.012\]
- \[K_{c}=\frac{\left [ 0.035 \right ]^{2}}{\left [ 0.018 \right ]}=0.070 \]
- \[K_{c}=0.068\]

Exercise \(\PageIndex{z}\)

Use the reaction from Figure 15.4 to answer:

For the same reaction, if the 2.0L container was evacuated, then pumped in with gases at the following pressure, C_{4}H_{10} = 1.2atm, C_{2}H_{6}and C_{2}H_{4} = 0.6 atm. What is the partial pressure (in atm) of C_{4}H_{10} at equilibrium? K_{p} =0.64.

**Answer**-
R C _{4}H_{10}(g)⇌ C _{2}H_{6}(g) +C _{2}H_{4}(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 - \[K_{p}=\frac{\left [ 0.6+x \right ]^{2}}{\left [ 1.2-x \right ]}=0.64 \]
- \[x=0.2 atm \]
- \[1.2-0.2=1.0atm\]

### Equilibria Calculations Through "Completing the Power"

Exercise \(\PageIndex{aa}\)

Use the reaction from Figure 15.5 to answer:

For the reaction, what is the equilibrium concentration of Br_{2} at 440^{0}C if initially [H_{2}]= 0.60M [HBr]=1.25M [Br_{2}]=0.6 and k=50?

**Answer**-
R H _{2}(g) +Br _{2}(g)⇌ 2HBr I 0.60 M 0.06 M 1.25 M C -x -x +2x E 0.60-x 0.60-x 1.25+2x \[K=\frac{\left [ 1.25+2x \right ]^{2}}{\left [ 0.60-x \right ]\left [ 0.60-x \right ]}=50 \]

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

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

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

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

\[x=0.33\,M\]

\[0.60-0.33=0.27\,M\]

Exercise \(\PageIndex{ab}\)

Use the reaction from Figure 15.5 to answer:

What is the equilibrium concentration of HBr?

**Answer**-
R H _{2}(g) +Br _{2}(g)⇌ 2HBr I 0.60 M 0.06 M 1.25 M C -x -x +2x E 0.60-x 0.60-x 1.25+2x \[K=\frac{\left [ 1.25+2x \right ]^{2}}{\left [ 0.60-x \right ]\left [ 0.60-x \right ]}=50 \]

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

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

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

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

\[x=0.33\,M\]

\[1.25+\left(2*0.33\right)=1.91\,M\]

Exercise \(\PageIndex{ac}\)

Use the reaction from Figure 15.6 to answer:

For the reaction, what is the equilibrium concentration of HCl if the initial [Cl_{2}]=0.1M, [HBr]=0.20M, [Br_{2}]=[HCl]=0 and k=15?

**Answer**-
R _{Cl2}(g) +2HBr(g) ⇌ Br _{2}(g)2HCl(g) I 0.10M 0.20M 0 0 C -x -2x +x +2x E 0.10-x 0.20-2x x 2x \[K=\frac{\left [ Br_{2} \right ]\left [ HCl \right ]^{2}}{\left [ HBr \right ]^{2}\left [ Cl \right ]}=\frac{\left [ x \right ]\left [ 2x \right ]^{2}}{\left [ 0.2-2x \right ]^{2}\left [ 0.1-x \right ]}=\frac{\left [ x \right ]\left(4\left [ x \right ]^{2}\right)}{\left(2\left [ 0.1-x \right ]^{2}\right)\left [ 0.1-x \right ]}\]

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

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

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

\[2x=0.14\]

Exercise \(\PageIndex{ad}\)

Use the reaction from Figure 15.6 to answer:

What is the equilibrium concentration of HBr in Q15.29?

**Answer**-
R _{Cl2}(g) +2HBr(g) ⇌ Br _{2}(g)2HCl(g) I 0.10M 0.20M 0 0 C -x -2x +x +2x E 0.10-x 0.20-2x x 2x \[K=\frac{\left [ Br_{2} \right ]\left [ HCl \right ]^{2}}{\left [ HBr \right ]^{2}\left [ Cl \right ]}=\frac{\left [ x \right ]\left [ 2x \right ]^{2}}{\left [ 0.2-2x \right ]^{2}\left [ 0.1-x \right ]}=\frac{\left [ x \right ]\left(4\left [ x \right ]^{2}\right)}{\left(2\left [ 0.1-x \right ]^{2}\right)\left [ 0.1-x \right ]}\]

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

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

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

\[0.20-2x=0.06M\]

Exercise \(\PageIndex{ae}\)

For the reaction, What is [O_{2}] at equilibrium if 0.050mol of N_{2}O_{2}, O_{2}, and NO_{2} are mixed in a 1.0L container? k=25

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

**Answer**-
R N _{2}O_{2}(g) +O _{2}(g)⇌ 2NO _{2}(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 \[K=\frac{\left [ 0.05+2x \right ]^{2}}{\left [ 0.05-x \right ]\left [ 0.05-x \right ]}=25\]

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

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

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

\[x=0.03 M\]

\[0.05-0.03=0.02 M\]

### Relating K_{p} and K_{c}

Exercise \(\PageIndex{af}\)

What is the relationship between Kp and Kc?

- \(K_{p}=K_{c}\left ( RT \right )^{\Delta n}\)
- \(K_{c}=K_{p}\left ( RT \right )^{\Delta n}\)
- \(K_{p}=K_{c}\)
- None of the above

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

Exercise \(\PageIndex{ag}\)

Use the reaction from Figure 15.7 to answer:

What is the Δn for the equation above?

**Answer**-
\[3A_{2}(g)+2B_{3}(s)\rightarrow 6A_{3}B_{2}(g)\]

\[\Delta n=6-(3+2)\]

\[\Delta n=6-5=1\]

Exercise \(\PageIndex{ah}\)

Use the reaction from Figure 15.7 to answer:

Solve for K_{p}, given K_{c} = 2.3*20^{4} at 30°C for the equation above?

**Answer**-
\[K_{p}=K_{c}\left ( RT \right )^{\Delta n}\]

\[K_{p}=2.3*10^{4}\left (0.08206*303\right )^{\left(6-\left (3+2\right )\right)}\]

\[K_{p}=5.7*10^{5}\]

Exercise \(\PageIndex{ai}\)

Use the reaction from Figure 15.7 to answer:

Solve for K_{c}, given K_{p} = 2.3*10^{4} at 30°C for the equation above?

**Answer**-
\[K_{p}=K_{c}\left ( RT \right )^{\Delta n}\]

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

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

\[K_{c}=1.5\]

Exercise \(\PageIndex{aj}\)

Use the reaction from Figure 15.7 to answer:

Solve for K_{c}, given K_{p} = 3.1*10^{4} at 30°C for the equation above?

**Answer**-
\[K_{p}=K_{c}\left ( RT \right )^{\Delta n}\]

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

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

\[K_{c}=2.1\]

Exercise \(\PageIndex{ak}\)

Solve for Kc, given K_{p} = 3.2*10^{4} at 30°C for the following equation:

\[3A_{2}(g)+4B_{3}(g)\rightarrow 2A_{3}B_{2}(g)\]

**Answer**-
\[K_{p}=K_{c}\left ( RT \right )^{\Delta n}\]

\[3.2*10^{4}=K_{c}\left (0.08206*303\right )^{\left (2-4\right )} \]

\[K_{c}=\frac{3.2*10^{4}}{\left (0.08206*303\right )^{\left (2-4\right )}} \]

\[K_{c}=2.0*10^{7}\]

Exercise \(\PageIndex{al}\)

What is the temperature, given K_{c} = 3.5*10^{8}, and K_{p} = 5.8*10^{5} for the following equation?

\[A_{2}(g)+3B_{3}(g)\rightarrow 2AB_{3}(g)\]

**Answer**-
\[K_{p}=K_{c}\left ( RT \right )^{\Delta n}\]

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

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

\[T=299K|]

### General Questions

Exercise \(\PageIndex{am}\)

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

- \(K= k_{f} + k_{r}\)
- \(K= k_{f}k_{r}\)
- \(K=k_{f}-k_{r}\)
- \(K= \frac{1}{\left ( k_{f}k_{r} \right )}\)
- \(K=\frac{k_{f}}{k_{r}}\)

**Answer**-
e. \(K=\frac{k_{f}}{k_{r}}\)

Exercise \(\PageIndex{an}\)

A flask of an aqueous equilibrium mixture of \(CoCl_{4}^{2-}\), \(CoBr_{4}^{2-}\), \(Cl^{-}\), and \(Br^{-}\) is at 25°C. Which of the following actions will change the value of the equilibrium constant from that which currently describes the concentration relationships of the four species above?

- add more \(CoBr_{4}^{2-}\) to the solution
- add more \(CoCl_{4}^{2-}\) to the solution
- add more \(Br^{-}\) to the solution
- add more \(Cl^{-}\) to the solution
- put the flask into an 80°C water bath

**Answer**-
e. put the flask into an 80°C water bath

Exercise \(\PageIndex{ao}\)

What is the Kc for the following reaction?

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

**Answer**-
\[K_{c}=\frac{\left [ CH_{4} \right ]\left [ H_{2}O \right ]}{\left [ CO \right ]\left [ H_{2} \right ]^{3}}\]

Exercise \(\PageIndex{ap}\)

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)

**Answer**-
\[k_{eq1}=\frac{\left [ D \right ]^{2}\left [ E \right ]}{\left [ A \right ]^{3}\left [ B \right ]^{2}}\]

\[k_{eq2}=\frac{\left [ A \right ]^{3}\left [ B \right ]^{2}}{\left [ D \right ]^{2}\left [ E \right ]}\]

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

Exercise \(\PageIndex{aq}\)

The reaction

\(A+B \rightleftharpoons X+Y\)

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

- only products exist
- only reactants exist
- products predominate
- reactants predominate
- roughly equal molar amounts of products and reactants are present

**Answer**-
c. products predominate

Exercise \(\PageIndex{ar}\)

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

- 1/2K
- 1/K
^{2} - 2K
- K
^{2} - -K
^{2}

**Answer**-
b. 1/K

^{2}

Exercise \(\PageIndex{as}\)

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

- K
^{3} - 3K
- K/3
- 1/K
^{3} - -K
^{3}

**Answer**-
d. 1/K

^{3}

Exercise \(\PageIndex{at}\)

If the value of K_{c} for the following reaction is 0.25:

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

What is the value of K_{c} for the reaction below?

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

**Answer**-
\[K_{c1}=\frac{\left [ SO_{3} \right ]\left [ NO \right ]}{\left [ SO_{2} \right ]\left [ NO_{2} \right ]}=0.25\]

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

Exercise \(\PageIndex{au}\)

What is K_{p} for the following reaction at 25°C, K_{c} = 3.0*10^{5}?

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

**Answer**-
\[T = 25+273.15 = 298.15 \]

\[\Delta n = 4-5 = -1 \]

\[K_{p}=K_{c}\left ( RT \right )^{\Delta n} \]

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

Exercise \(\PageIndex{av}\)

The value of K_{c} 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 K_{c} for the reverse reaction at 100°C?

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

**Answer**-
\[K_{c1}=2.0*10^{-10} \]

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

Exercise \(\PageIndex{aw}\)

The value of K_{c} for the following reaction is 1.10 at 25.0°C. What is the value of K_{p} for this reaction?

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

**Answer**-
\[T = 25+273.15 = 298.15 \]

\[\Delta n = 7-5 = 2 \]

\[K_{p}=K_{c}\left ( RT \right )^{\Delta n} \]

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

Exercise \(\PageIndex{ax}\)

What is the value of K_{c} for a flask at equilibrium that contains 0.0114 M HCl, 0.0931 M Cl_{2}, and 0.0154 M H_{2} at a certain temperature?

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

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

Exercise \(\PageIndex{ay}\)

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 |

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

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

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

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

Exercise \(\PageIndex{az}\)

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

\(\frac{\left [IBr\right ]}{\left [I_{2}\right ]\left [Br_{2}\right ]}\)

- \(I_{2}(g)+Br_{2}(g)\rightleftharpoons IBr(g)\)
- \(I_{2}(g)+Br_{2}(g)\rightleftharpoons 2 IBr(g)\)
- \(2 IBr(g)\rightleftharpoons I_{2}(g)+Br_{2}(g)\)
- \(2 I_{2}(g)+2 Br_{2}(g)\rightleftharpoons IBr(g)\)
- \(IBr(g)\rightleftharpoons 2I_{2}(g)+2Br_{2}(g)\)

**Answer**-
b. \(I_{2}(g)+Br_{2}(g)\rightleftharpoons 2 IBr(g)\)

Exercise \(\PageIndex{ba}\)

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

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

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

\(HI(g)\rightleftharpoons \frac{1}{2}H_{2}(g)+\frac{1}{2}I_{2}(g)\)

**Answer**-
\[K_{c}^{1}=54.0 \]

\[K_{c}^{-1}=\frac{1}{\left (K_{c}^{1}\right )^{n}} \]

n is the factor of difference between the coefficients of the two reactions. So in this case it is 1/2.

\[K_{c}^{-1}=\frac{1}{54.0^{1/2}}=\frac{1}{7.35}=0.136\]

Exercise \(\PageIndex{bb}\)

What is the equilibrium constant expression for the following reaction?

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

**Answer**-
\[K_{c}=\frac{\left [SO_{2}\right ]^{3}}{\left [HCl\right ]^{6}}\]

Exercise \(\PageIndex{bc}\)

What is the equilibrium constant expression for the following reaction?

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

**Answer**-
\[K_{c}=\frac{\left [SO_{3}\right ]^{2}}{\left [SO_{2}\right ]^{3}}\]

Exercise \(\PageIndex{bd}\)

Consider the following chemical reaction:

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

At equilibrium, the concentrations of H_{2}, I_{2}, and HI were 0.15M, 0.033 M, and 0.55 M, respectively. What is the K_{c} for this reaction?

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

Exercise \(\PageIndex{be}\)

Use the reaction from Figure 15.8 to answer:

Initially, 1.26 mol of PCl_{5}(g) was placed in a 1.0 L flask. At equilibrium, 1.08 mol of PCl_{5}(g) was present. What is the value of K_{c} for this reaction at this temperature?

**Answer**-
R PCl _{5}⇌ PCl _{3}+Cl _{2}I 1.26 0 0 C -x +x +x E 1.08 x x \[1.08=1.26-x \]

\[x=1.26-1.08=0.18 \]

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

Exercise \(\PageIndex{bf}\)

Use the reaction from Figure 15.8 to answer:

What is the equilibrium partial pressure of PCl_{3}? If in a 3.00 L vessel that was charged with 0.123 atm of PCl_{5} has a K_{p} of 0.0121?

**Answer**-
R PCl _{5}⇌ PCl _{3}+Cl _{2}I 0.123 0 0 C -x +x +x E 0.123-x x x \[K_{p}=\frac{\left [PCl_{3}\right ]\left [Cl_{2}\right ]}{\left [PCl_{5}\right ]}=\frac{x^{2}}{0.123-x}=0.0121\]

\[\frac{x^{2}}{0.123-x}=0.0121 \]

\[x^{2}=\left (0.123-x\right )*0.0121 \]

\[x^{2}=0.001488-0.0121x \]

\[0=x^{2}+0.0121x-0.001488\]

Use the quadratic formual to solve for x

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

\[x=0.0330\]

The negtive square is rejected because pressure cannot be negative.

Exercise \(\PageIndex{bg}\)

The value of K_{c} for the following reaction is 0.070. What is the equilibrium concentration (M) of C_{4}H_{10} if the equilibrium concentrations of C_{2}H_{6} and C_{2}H_{4} are both 0.035M?

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

**Answer**-
R C _{4}H_{10}⇌ C _{2}H_{6}+C _{2}H_{4}I x 0 0 C -y +y +y E x-y y y \[\left [C_{2}H_{6}\right ]=\left [C_{2}H_{4}\right ]=0.035\,M=y \]

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

\[0.070x-0.00245=0.001225 \]

\[0.070x=0.003675 \]

\[x=0.0525 \]

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

Exercise \(\PageIndex{bh}\)

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 Br_{2}. At equilibrium, the flask contained 0.46 mol of NOBr. How many moles of NO and Br_{2} are in the flask at equilibrium?

**Answer**-
R 2NOBr ⇌ 2NO + Br _{2}I 0.64 mol 0 0 C -2x +2x +x E 0.64-2x 2x x \[0.64-2x=0.46 \]

\[-2x=-0.18 \]

\[x=0.09 ]

\[mol_{NO}=2x=2*0.09=0.18\,mol \]

\[mol_{Br_{2}}=x=0.09\,mol\]

Exercise \(\PageIndex{bi}\)

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

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

**Answer**-
c. \(2SO_{3}(g)\rightleftharpoons 2SO_{2}(g)+O_{2}(g)\)

Exercise \(\PageIndex{bj}\)

For the __endothermic__ reaction

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

only _____ would favor shifting the equilibrium position to form more CO_{2} gas.

__both__decreasing the system temperature__and__increasing the system pressure- decreasing the system temperature
- increasing
__both__the system temperature__and__the system pressure - increasing the system pressure
- increasing the system temperature

**Answer**-
e. increasing the system temperature

Exercise \(\PageIndex{bk}\)

Which of the following reactions would increase pressure at constant temperature __not__ change the concentration of reactants and products?

- \(2N_{2}(g)+O_{2}(g)\rightleftharpoons 2N_{2}O(g)\)
- \(N_{2}(g)+2O_{2}(g)\rightleftharpoons 2NO_{2}(g)\)
- \(N_{2}(g)+3H_{2}(g)\rightleftharpoons 2NH_{3}(g)\)
- (N_{2}O_{4}(g)\rightleftharpoons 2NO_{2}(g)\)
- \(N_{2}(g)+O_{2}(g)\rightleftharpoons 2NO(g)\)

**Answer**-
e. \(N_{2}(g)+O_{2}(g)\rightleftharpoons 2NO(g)\)

Exercise \(\PageIndex{bl}\)

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 _____.

- at high temperature and high pressure
- at high temperature and low pressure
- at low temperature and high pressure
- at low temperature and low pressure
- in the presence of solid carbon

**Answer**-
d. at low temperature and low pressure

Exercise \(\PageIndex{bm}\)

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

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

**Answer**-
c. lower the energy of the transition state