15: Equilibria

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15.1 Chemical Equilibrium

Textbook: Section 15.1

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Exercise \(\PageIndex{1a}\)

Which of the following statements is TRUE?

 

a)      Equilibrium is a static state in which both the forward and reverse reactions no longer occur.

b)      Equilibrium is a state in which the net reactant and product concentrations remain constant with time.         

c)      Equilibrium is a state in which the rates of the forward and reverse reactions are equal.

d)     (b) and (c) are true.

e)      (a), (b) and (c) are true.

Answer

d)     (b) and (c) are true.

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Exercise \(\PageIndex{1}\)

What is the relationship between the rate constants of the forward and reverse reactions and the equilibrium constant for a reaction?

 

a.  K=k_fk_r                    b.  K = k_f/k_r                c.  K = k_f  +  k_r                      d. K = k_f - k_r     
e.  there is no relationship

Answer

  b.  K = k_f/k_r  

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15.2 Equilibrium Constant and Reaction Quotient

Textbook: Section 15.2

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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

Answer

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).

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Exercise \(\PageIndex{2.b}\)

Use the following reaction to answer

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

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

Answer

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

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

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Exercise \(\PageIndex{2.c}\)

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

Answer

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.

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Exercise \(\PageIndex{2.d}\)

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

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

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Exercise \(\PageIndex{2.e}\)

Use the following reaction to answer

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

What is K_c for the system in question 15.2d?

Answer

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

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

\[K_{c}=3.68 \nonumber \]

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 K_c for the first reaction above?

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

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

Answer a

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

Answer b

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

Answer c

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

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

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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 K_eq=50 at 440°C.

Answer

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

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

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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. \(K=\frac{k_{f}}{k_{r}}\)

Answer

e. \(K=\frac{k_{f}}{k_{r}}\)

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Exercise \(\PageIndex{2i}\)

What is the Kc for the following gas phase reaction?

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

Answer

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

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Exercise \(\PageIndex{2j}\)

The value of K_c for the reaction below is 2.0*10^-10 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

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

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

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Exercise \(\PageIndex{2k}\)

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

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

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Exercise \(\PageIndex{2l}\)

What is the equilibrium constant expression for the following reaction?

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

Answer

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

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Exercise \(\PageIndex{2m}\)

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 K_c for this reaction?

Answer

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

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Exercise \(\PageIndex{2n}\)

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

Answer

c. lower the energy of the transition state

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Exercise \(\PageIndex{2o}\)

Nitrogen dioxide (NO₂) is involved in the formation of smog and acid rain. A reaction that is important in the formation of NO₂ is:

O₃ (g) + NO (g) <==> O₂ (g) + NO₂ (g)

K_c = 6.0 x 10³⁴

If the air over a section of Little Rock contained 1.0 x 10^-6 M O₃, 1.0 x 10^-5 M NO, 2.5 x 10^-4 M NO₂ and 8.2 x 10^-3 M O₂, what can we conclude?

 

                              a)            There will be a tendency to form more NO and O₃.

                              b)            There will be a tendency to form more NO₂ and O₂.

                              c)            There will be a tendency to form more NO₂ and O₃.

                              d)            There will be a tendency to form more NO and O₂.

                              e)            There will be no tendency for change because the reaction is at equilibrium.

Answer

There will be a tendency to form more NO₂ and O₂.

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Exercise \(\PageIndex{2p}\)

The relationship between forward and reverse rate constants with the equilibrium constant is,

a.  k_fk_r                 b.  k_f  - k_r                            c.  k_f  + k_r                           d.  k_f/k_r               e.  k_r/k_f

Answer

d.  k_f/k_r

Exercise \(\PageIndex{2q}\)

A reaction that is product loaded and product favored has:

 

a) Q>K  and K>>1                            b)  Q<K add K>>1

c) Q>K and K<<1                               d)  Q<K and  K<<1

e. Q=K and K=1

Answer

a) Q>K  and K>>1                           

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Exercise \(\PageIndex{2r}\)

What is the equilibrium constant for the reaction,

2 H₂ (g) + S₂ (g) <==> 2 H₂S (g)

if a mixture of  the gaseous H₂, S₂ and H₂S is at equilibrium at 700^oC has the following concentrations:  0.208 M H₂ (g), 1.12 x 10^-6 M S₂ (g) and 0.725 M H₂S (g)?

Answer

\[
\text{Reaction: } \quad 2\,\mathrm{H_2(g)} + \mathrm{S_2(g)} \;\rightleftharpoons\; 2\,\mathrm{H_2S(g)}
\]

\[
K_c \;=\; \frac{[\mathrm{H_2S}]^{2}}{[\mathrm{H_2}]^{2} \,[\mathrm{S_2}]}
\]

\[
K_c \;=\; \frac{(0.725)^{2}}{(0.208)^{2}\,(1.12 \times 10^{-6})}
\]

\[
K_c \;=\; \frac{0.525625}{0.043264 \times 1.12 \times 10^{-6}}
\]

\[
K_c \;=\; \frac{0.525625}{4.8446 \times 10^{-8}}
\]

\[
K_c \;\approx\; 1.08 \times 10^{7}
\]
 

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Exercise \(\PageIndex{2p}\)

Identify the correct equilibrium constant expression, K, for:

                       

C₆H₁₂O₆(s)  +  6O₂(g)  <==>   6CO₂(g)  +  6H₂O(l)

 

Answer

\[
\text{Reaction: } \quad \mathrm{C_6H_{12}O_6 (s)} + 6\,\mathrm{O_2 (g)} 
\;\rightleftharpoons\; 6\,\mathrm{CO_2 (g)} + 6\,\mathrm{H_2O (l)}
\]

\[
K \;=\; \frac{[\mathrm{CO_2}]^{6}}{[\mathrm{O_2}]^{6}}
\]

\[
\text{Note: Pure solids (}\mathrm{C_6H_{12}O_6}\text{) and pure liquids (}\mathrm{H_2O}\text{) do not appear in the expression.}
\]
 

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Exercise \(\PageIndex{2q}\)

The equilibrium for the following reaction is K

 

                                    SO₂ (g)  +  1/2O₂ (g)  çè SO₃ (g) 

What is K for:

SO₃ (g)  çè  SO₂ (g)  +  1/2O₂ (g)

a. K²                b. –K               c. 1/K              d. 1/K²             e. 1/2K

Answer

c. 1/K

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Exercise \(\PageIndex{5.2p}\)

Assume all species are in the gas phase, calculate K for.

CO₂ + H₂ <==> CO + H₂O

Given:

CO₂ <==> CO + 1/2O₂, K_c1 = 9.1 x 10^-12
H₂O <==> H₂ + 1/2O₂, K_c2 = 7.1 x 10^-12

Answer

Given reactions

\[
\text{(1)} \quad CO_2 \rightleftharpoons CO + \tfrac12 O_2
\]

\[
K_{c1} = 9.1 \times 10^{-12}
\]

\[
\text{(2)} \quad H_2O \rightleftharpoons H_2 + \tfrac12 O_2
\]

\[
K_{c2} = 7.1 \times 10^{-12}
\]

Target reaction

\[
CO_2 + H_2 \rightleftharpoons CO + H_2O
\]

---

Step 1: Reverse reaction (2) so that \(H_2\) appears as a reactant

\[
H_2 + \tfrac12 O_2 \rightleftharpoons H_2O
\]

When a reaction is reversed, the equilibrium constant is inverted.

\[
K = \frac{1}{K_{c2}}
\]

---

Step 2: Add the reactions

\[
CO_2 \rightleftharpoons CO + \tfrac12 O_2
\]

\[
H_2 + \tfrac12 O_2 \rightleftharpoons H_2O
\]

Adding and canceling \( \tfrac12 O_2 \) gives

\[
CO_2 + H_2 \rightleftharpoons CO + H_2O
\]

---

Step 3: Combine equilibrium constants

When reactions are added, the equilibrium constants multiply.

\[
K = K_{c1}\left(\frac{1}{K_{c2}}\right)
\]

\[
K = \frac{K_{c1}}{K_{c2}}
\]

---

Step 4: Substitute values

\[
K =
\frac{9.1\times10^{-12}}{7.1\times10^{-12}}
\]

\[
K = 1.28
\]

---

\[
\boxed{K \approx 1.3}
\]

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Exercise \(\PageIndex{5.2q}\)

At 200^oC, nitrogen oxide reacts with oxygen to form nitrogen dioxide as follows:

2NO + O₂ <==> 2NO₂, K_c = 3 x 10⁶

If a mixture of these three gases contains 0.10 M NO, 0.10 M NO₂, and 0.01 M O₂, then we can accurately predict that the reaction:

1. 1. is at equilibrium.
   2. is not at equilibrium and must proceed from left to right to reach equilibrium.
   3. is not at equilibrium and must proceed from right to left to reach equilibrium.
   4. is not at equilibrium but insufficient information is given to predict which direction the reaction must go to reach equilibrium.
   5. the rate from right to left is greater than the rate from left to right.

Answer

is not at equilibrium and must proceed from left to right to reach equilibrium.

 

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Exercise \(\PageIndex{5.2r}\)

 Identify the correct equilibrium constant expression, K, for:

                       

C₆H₁₂O₆(s)  +  6O₂(g)  <==>   6CO₂(g)  +  6H₂O(l)

Answer

Reaction

\[
C_6H_{12}O_6(s) + 6O_2(g) \rightleftharpoons 6CO_2(g) + 6H_2O(l)
\]

---

Step 1: Write the general equilibrium expression

\[
K = \frac{[CO_2]^6[H_2O]^6}{[C_6H_{12}O_6][O_2]^6}
\]

---

Step 2: Remove pure solids and liquids

The concentrations of pure solids and liquids are constants and are not included in the equilibrium expression.

\[
C_6H_{12}O_6(s) \quad \text{and} \quad H_2O(l)
\]

are omitted.

---

Step 3: Final equilibrium constant expression

\[
K = \frac{[CO_2]^6}{[O_2]^6}
\]

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Relating K_p and K_C

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

Answer

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

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Exercise \(\PageIndex{2.1b}\)

\(3A(g)+2B(s)\rightarrow 6C(g)\)

What is the Δn for the equation above?

Answer

 

\[\Delta n=6-3=3\nonumber \]

 

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Exercise \(\PageIndex{2.1c}\)

\(3A(g)+2B(s)\rightarrow 6C(g)\)

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

Answer

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

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

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

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Exercise \(\PageIndex{2.1d}\)

\(3A(g)+2B(s)\rightarrow 6C(g)\)

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

Answer

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

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

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

\[K_{c}=1.5\nonumber \]

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Exercise \(\PageIndex{2.1e}\)

\(3A(g)+2B(s)\rightarrow 6C(g)\)

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

Answer

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

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

\[K_{c}=\frac{3.1*10^{4}}{\left (0.08206*303\right )^{\left (6-3\right )}} \nonumber \]

\[K_{c}=2.01\nonumber \]

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Exercise \(\PageIndex{2.1f}\)

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

\[3A(g)+4B(g)\rightarrow 2C(g)\nonumber \]

Answer

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

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

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

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

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Exercise \(\PageIndex{2.1g}\)

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

\[A(g)+3B(g)\rightarrow 2C(g)\nonumber \]

Answer

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

\[\frac{K_p}{K_c} = (RT)^{\Delta n} \nonumber\]

\[(\frac{K_p}{K_c})^{\frac{1}{\Delta n}} = (RT)^{\frac{\cancel{ \Delta n}}{\cancel{\Delta n}}} \nonumber\]

\[T = \frac{1}{R}(\frac{K_p}{K_c})^{\frac{1}{\Delta n}} \nonumber\]

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

\[T=299K\nonumber \]

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Exercise \(\PageIndex{2.1h}\)

For the following reaction at 25^oC, K_c = 3.0x10⁵.  What is K_p?

 

            2H₂S(g)  + 3O₂ (g)    <= =>   2H₂O(g)   +  2SO₂ (g)

Answer

\[
\text{Reaction: } \quad 2\,\mathrm{H_2S(g)} + 3\,\mathrm{O_2(g)} 
\;\rightleftharpoons\; 2\,\mathrm{H_2O(g)} + 2\,\mathrm{SO_2(g)}
\]

\[
K_p = K_c \,(RT)^{\Delta n}
\]

\[
\Delta n \;=\; (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})
\]

\[
\Delta n \;=\; (2 + 2) - (2 + 3) = 4 - 5 = -1
\]

\[
K_p = K_c \,(RT)^{-1} = \frac{K_c}{RT}
\]

\[
K_p = \frac{3.0 \times 10^{5}}{(0.08206)\,(298)}
\]

\[
K_p = \frac{3.0 \times 10^{5}}{24.45}
\]

\[
K_p \;\approx\; 1.2 \times 10^{4}
\]


 

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Exercise \(\PageIndex{2.1i}\)

 For the reaction system N₂O₄(g)  <===>  2NO₂(g) at 90^oC,             Kc=0.27.  Determine Kp for this system.  R=0.08206L-atm/(mol-K)

Answer

Reaction

\[
N_2O_4(g) \rightleftharpoons 2NO_2(g)
\]

Given

\[
K_c = 0.27
\]

\[
T = 90^\circ C = 363\ K
\]

\[
R = 0.08206\ \text{L·atm mol}^{-1}\text{K}^{-1}
\]

---

Step 1: Use the relationship between \(K_p\) and \(K_c\)

\[
K_p = K_c(RT)^{\Delta n}
\]

where

\[
\Delta n = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})
\]

---

Step 2: Determine \(\Delta n\)

\[
\Delta n = 2 - 1 = 1
\]

---

Step 3: Substitute values

\[
K_p = 0.27(0.08206 \times 363)^1
\]

\[
K_p \approx 8.0
\]

---

\[
\boxed{K_p \approx 8.0}
\]

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15.3 Determining an Equilibrium Constant

Textbook: Section 15.3

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Exercise \(\PageIndex{3.a}\)

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

Given K =50, 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

Answer

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

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

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

\[4x^2-5x+1.5625=30x+50x^2 \nonumber \]

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

\[x=0.042M \nonumber \]

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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 if K = 50?

[H₂]= 0.60M, [HBr]=1.25M and [Br₂]=0

Answer

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

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

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

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

\[x=0.042M \nonumber \]

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

 

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Exercise \(\PageIndex{3.c}\)

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

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

Answer

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

\[0.8mol/2.0L=0.40M\nonumber \]

\[0.40M=2x \nonumber \]

\[x=0.2M \nonumber \]

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

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Exercise \(\PageIndex{3.d}\)

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

At a certain temperature, K_eq=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?

Answer

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

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

\[ K=\left ( \frac{1.5+2x}{1.5-x} \right )^2  \nonumber\]

\[\left ( \frac{1.5+2x}{1.5-x} \right )=\sqrt{K} \nonumber \]

\[1.5+2x= \sqrt K\left ( 1.5-x \right ) = 1.5\sqrt K -x\sqrt K \nonumber \]
\[2x + x\sqrt K  = 1.5\sqrt K -1.5 \nonumber\]
\[x\left ( 2+ \sqrt K \right )=1.5(\sqrt K-1) \nonumber \]

\[x=\frac{1.5\sqrt K -1}{2+ \sqrt K} = \frac{1.5(\sqrt{630}-1)}{2+\sqrt{630}}=1.33 \nonumber \]

\[x=1.33 M \nonumber \]

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

 

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Exercise \(\PageIndex{3.e}\)

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

Given K = 630, what is [HBr] after equilibrium is reached if 4.50mol of each of the three species were placed into a 3.00L flask?

Answer

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

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

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Exercise \(\PageIndex{3.f}\)

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

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

Answer

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

\[x=0.05-0.0155=0.0345M \nonumber \]

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

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Exercise \(\PageIndex{3g}\)

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

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 K_c for this reaction at this temperature?

Answer

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

\[1.08=1.26-x \nonumber\]

\[x=1.26-1.08=0.18 \nonumber\]

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

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Exercise \(\PageIndex{3h}\)

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?

Answer

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

\[0.64-2x=0.46 \nonumber\]

\[-2x=-0.18 \nonumber\]

\[x=0.09 \nonumber\]

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

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

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Exercise \(\PageIndex{5.3i}\)

I₂ (0.50 mol) and Br₂ (0.50mol) were placed into a 1.00 L flask and allowed to come to equilibrium.  At equilibrium the flask contains 0.84 mole of IBr.  The value of K_c for the reaction below is _____.

 

                        I₂(g) +  Br₂(g)  <==> 2IBr(g)

Answer

\[
\textbf{Reaction:}\quad \mathrm{I_2(g)} + \mathrm{Br_2(g)} \;\rightleftharpoons\; 2\,\mathrm{IBr(g)}
\]

\[
\text{Given: } n_0(\mathrm{I_2})=0.50\text{ mol},\; n_0(\mathrm{Br_2})=0.50\text{ mol},\; V=1.00\text{ L},\;
n_\text{eq}(\mathrm{IBr})=0.84\text{ mol}.
\]
\[
\text{Since }V=1.00\text{ L, moles }=\text{ molarities.}
\]

\[
\begin{array}{c|ccc}
 & \mathrm{I_2} & \mathrm{Br_2} & \mathrm{IBr} \\
\hline
\text{Initial (M)} & 0.50 & 0.50 & 0 \\
\text{Change (M)} & -x & -x & +2x \\
\text{Equilibrium (M)} & 0.50 - x & 0.50 - x & 2x
\end{array}
\]

\[
2x = [\mathrm{IBr}]_\text{eq} = 0.84 \;\;\Longrightarrow\;\; x = 0.42
\]

\[
[\mathrm{I_2}]_\text{eq} = 0.50 - 0.42 = 0.08,\qquad
[\mathrm{Br_2}]_\text{eq} = 0.50 - 0.42 = 0.08,\qquad
[\mathrm{IBr}]_\text{eq} = 0.84
\]

\[
K_c \;=\; \frac{[\mathrm{IBr}]^2}{[\mathrm{I_2}]\,[\mathrm{Br_2}]}
\;=\; \frac{(0.84)^2}{(0.08)(0.08)}
\;=\; \frac{0.7056}{0.0064}
\;=\; 1.1025 \times 10^{2}
\]

\[
\boxed{K_c \;\approx\; 1.10 \times 10^{2}}
\]
 

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Exercise \(\PageIndex{5.3j}\)

Calculate K for the reaction if 12 moles of N₂(g) and 20.0 moles of O₂(g) entered an empty 1.0 L cylinder and 14 moles of NO(g) were found at equilibrium. 

N₂  +  O₂  < === >  2NO

Answer

Reaction

\[
N_2(g) + O_2(g) \rightleftharpoons 2NO(g)
\]

Given (in a \(1.0\ \text{L}\) cylinder)

\[
n_{N_2,0}=12\ \text{mol}\qquad n_{O_2,0}=20.0\ \text{mol}\qquad n_{NO,0}=0
\]

\[
n_{NO,\text{eq}}=14\ \text{mol}
\]

Because the volume is \(1.0\ \text{L}\), moles \(=\) molarity numerically.

---

Step 1: Set up an ICE table

Let \(x\) be the amount (in M) of \(N_2\) and \(O_2\) consumed.

\[
\begin{array}{c|ccc}
 & N_2 & O_2 & NO \\
\hline
Initial & 12 & 20.0 & 0 \\
Change & -x & -x & +2x \\
Equilibrium & 12-x & 20.0-x & 2x
\end{array}
\]

Given \(2x = 14\), so

\[
x = 7
\]

Thus,

\[
[N_2]_{\text{eq}} = 12-7 = 5
\]

\[
[O_2]_{\text{eq}} = 20.0-7 = 13.0
\]

\[
[NO]_{\text{eq}} = 14
\]

---

Step 2: Write the equilibrium constant expression

\[
K_c = \frac{[NO]^2}{[N_2][O_2]}
\]

---

Step 3: Substitute equilibrium concentrations

\[
K_c = \frac{(14)^2}{(5)(13.0)}
\]

\[
K_c = \frac{196}{65} = 3.02
\]

---

\[
\boxed{K_c \approx 3.0}
\]

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15.4: Using Equilibrium Constants

Textbook: Section 15.4

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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 K_eq=50

Answer

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

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

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

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

\[x=0.042M \nonumber \]

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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 K_eq=50

Answer

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

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

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

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

\[x=0.042M \nonumber \]

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

 

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Exercise \(\PageIndex{4c}\)

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

At a certain temperature, K_eq=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?

Answer

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

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

\[x=1.33 M \nonumber \]

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

 

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Exercise \(\PageIndex{4d}\)

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

At a certain temperature, K_eq=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? 

Answer

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

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

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Exercise \(\PageIndex{4e}\)

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

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 K_p of 0.0121?

Answer

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.

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Exercise \(\PageIndex{4f}\)

The value of K_c 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?

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

Answer

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

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

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

\[0.070x-0.00245=0.001225 \nonumber\]

\[0.070x=0.003675 \nonumber\]

\[x=0.0525 \nonumber\]

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

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Exercise \(\PageIndex{4g}\)

The value of K_c for the following  condensation reaction is 4.69. What is the equilibrium concentration (M) of C₂H₅OH given initial concentrations of [C₄H₇OOH]_initial = 0.590M, [C2₅OH]_initial = 0.450M?

\(C_{4}H_{7}OOH(aq) +  C_{2}H_{5}OH(aq) \rightleftharpoons C_{4}H_{7}OOC_2H_5(aq) + H_2O(l)\)

Answer

R	C4H7OOH	C2H5OH	⇌	C4H7OOC2H5	H2O(l)
I	0.590	0.450		0	ignore (constant)
C	-x	-x		+x	"
E	0.590-x	0.450-x		X	"

 

\[K_{c}=\frac{[C_4H_7OOC_2H_5]}{C_4H_7OOH][C_2H_5OH]}=4.69 \nonumber \]

\[4.69= \frac{x}{(0.590-x)(0.450-x)}=\frac{x}{x^2-1.04x+0.2655} \nonumber \]

\[ 4.69(x^2-1.04x+0.2655)=x \nonumber \]

\[ 4.69x^2 -5.88x +1.25 = 0 \nonumber \]

using quadratic formula

x=0.98 and 0.27

0.27 makes sense as 0.98 would have a negative concentration for the two reactants at equilibrium (you would consume more than you have)

 

Complete the Square

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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?

Answer

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

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

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

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

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

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

\[1.25 + 2x = 7.07(0.60-x )\nonumber \]

\[2x + 7.07x = 7.07(0.60)-1.25 \nonumber\]

\[x(9.07)=4.24-1.25 \nonumber\]

\[x=0.33\,M\nonumber \]

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

(section 15.4.3.1.2)

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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

Answer

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

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

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

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

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

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

\[x=0.33\,M\nonumber \]

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

(section 15.4.3.1.2)

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Exercise \(\PageIndex{4.1c}\)

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

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

Answer

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

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

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

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

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

\[x=0.03 M\nonumber \]

\[0.05-0.03=0.02 M\nonumber \]

(section 15.4.3.1.2)

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Exercise \(\PageIndex{4.1d}\)

K_c for the reaction below at 25^o C is 4.8x10^-6.  Calculate the equilibrium concentration (mol/L) of Cl₂(g) if the initial concentration of ICl(g) is 1.33 mol/L.  There is no I₂ or Cl₂ initially present.

                                    2ICl(g)  <==>  I₂(g) +  Cl₂(g)

Answer

\[
\textbf{Reaction:}\quad 2\,\mathrm{ICl(g)} \;\rightleftharpoons\; \mathrm{I_2(g)} + \mathrm{Cl_2(g)}
\qquad (K_c = 4.8\times 10^{-6} \text{ at } 25^\circ\mathrm{C})
\]

\[
\text{Initial: } [\mathrm{ICl}]_0 = 1.33\ \mathrm{M},\quad [\mathrm{I_2}]_0 = 0,\quad [\mathrm{Cl_2}]_0 = 0
\]

\[
\begin{array}{c|ccc}
 & \mathrm{ICl} & \mathrm{I_2} & \mathrm{Cl_2} \\
\hline
\text{Initial (M)}      & 1.33 & 0 & 0 \\
\text{Change (M)}       & -2x  & +x & +x \\
\hline
\text{Equilibrium (M)}  & 1.33-2x & x & x
\end{array}
\]

\[
K_c \;=\; \frac{[\mathrm{I_2}]\, [\mathrm{Cl_2}]}{[\mathrm{ICl}]^{2}}
\;=\; \frac{x\cdot x}{(1.33 - 2x)^2}
\;=\; 4.8\times 10^{-6}.
\]

\[
K_c \;=\; \left(\frac{x^2}{(1.33-2x)^2}\right)
\;=\; \left(\frac{x}{1.33-2x}\right)^2
\]

\[
\frac{x}{1.33-2x} \;=\;  \sqrt{K_c}
\]

\[x \; = \; 1.33\sqrt{K_c}-2x\sqrt{K_c}\]

\[x +2x\sqrt{K_c} = 1.33\sqrt{K_c}\] 

\[x(1+2\sqrt{K_c}) = 1.33\sqrt{K_c}\]

Note, there are positive and negative solutions to the square root.

\[
\boxed{\,x \;=\; \frac{1.33(\pm\sqrt{K_c})}{\,1 \pm 2\sqrt{K_c}\,}\,}
\]

--- Now substitute numbers ---
\[
K_c = 4.8\times 10^{-6} \;\Rightarrow\; \sqrt{K_c} = \sqrt{4.8\times 10^{-6}} \approx 2.19089\times 10^{-3}
\]

\[
\text{(+) branch:}\quad
x_+ \;=\; \frac{1.33(2.19089\times 10^{-3})}{\,1 + 2(2.19089\times 10^{-3})\,}
\;=\; \frac{2.9139\times 10^{-3}}{1.0043818}
\;\approx\; 2.90\times 10^{-3}\ \mathrm{M}
\]

\[
\text{(--) branch:}\quad
x_- \;=\; \frac{1.33(-2.19089\times 10^{-3})}{\,1 - 2(2.19089\times 10^{-3})\,}
\;=\; \frac{-2.9139\times 10^{-3}}{0.9956182}
\;\approx\; -2.93\times 10^{-3}\ \mathrm{M}
\]

\[
\text{Only } x_+ > 0 \text{ is physically meaningful (concentrations cannot be negative).}
\qquad
\boxed{[\mathrm{Cl_2}]_\text{eq} = x \approx 2.90\times 10^{-3}\ \mathrm{M}}
\]
 

 

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Exercise \(\PageIndex{4.1e}\)

If Kc for the following reaction is 49, calculate [Br2] and [IBr] after reaction is complete if initially 4 moles of IBr(g) was placed in a 2.0 L container.

I2(g) + Br2(g) < === > 2IBr(g)

Answer

Reaction

\[
I_2(g) + Br_2(g) \rightleftharpoons 2IBr(g)
\]

Given

\[
K_c = 49
\]

Initially 4 moles of \(IBr\) are placed in a \(2.0\ \text{L}\) container.

\[
[IBr]_0 = \frac{4}{2.0} = 2.0\ \text{M}
\]

---

Step 1: Write the equilibrium expression

\[
K_c = \frac{[IBr]^2}{[I_2][Br_2]}
\]

---

Step 2: ICE setup (starting with only product, so the reaction shifts left)

\[
\begin{array}{c|ccc}
 & I_2 & Br_2 & IBr \\
\hline
Initial & 0 & 0 & 2.0 \\
Change & +x & +x & -2x \\
Equilibrium & x & x & 2.0 - 2x
\end{array}
\]

---

Step 3: Substitute into \(K_c\)

\[
49 = \frac{(2.0-2x)^2}{x^2}
\]

Rewrite as a square

\[
49 = \left(\frac{2.0-2x}{x}\right)^2
\]

Take the square root of both sides

\[
7 = \frac{2.0-2x}{x}
\]

(We take the positive root because concentrations must remain positive.)

---

Step 4: Solve for \(x\)

\[
7x = 2.0 - 2x
\]

\[
9x = 2.0
\]

\[
x = \frac{2.0}{9} = 0.222
\]

---

Step 5: Determine equilibrium concentrations

\[
[Br_2] = x = 0.222\ \text{M}
\]

\[
[IBr] = 2.0 - 2x = 2.0 - 0.444 = 1.56\ \text{M}
\]

---

\[
\boxed{[Br_2] = 0.222\ \text{M}}
\]

\[
\boxed{[IBr] = 1.56\ \text{M}}
\]

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Complete the Cube

 

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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?

Answer

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

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

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

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

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

\[2x=0.14\nonumber \]

(section 15.4.3.1.3)

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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?

Answer

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

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

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

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

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

\[0.20-2x=0.06M\nonumber \]

(section 15.4.3.1.3)

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Exercise \(\PageIndex{4.2c}\)

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

\[
Cl_2(g) + 2HBr(g) \rightleftharpoons Br_2(g) + 2HCl(g)
\]

Answer

Reaction

\[
Cl_2(g) + 2HBr(g) \rightleftharpoons Br_2(g) + 2HCl(g)
\]

Given

\[
[Cl_2]_0 = 0.10\ \text{M}
\qquad
[HBr]_0 = 0.20\ \text{M}
\qquad
[Br_2]_0 = [HCl]_0 = 0
\]

\[
K_c = 15
\]

---

Step 1: ICE table

Let \(x\) be the amount of \(Cl_2\) and \(Br_2\) that reacts/forms.

\[
\begin{array}{c|cccc}
 & Cl_2 & HBr & Br_2 & HCl \\
\hline
Initial & 0.10 & 0.20 & 0 & 0 \\
Change & -x & -2x & +x & +2x \\
Equilibrium & 0.10-x & 0.20-2x & x & 2x
\end{array}
\]

---

Step 2: Write \(K_c\)

\[
K_c=\frac{[Br_2][HCl]^2}{[Cl_2][HBr]^2}
\]

Substitute equilibrium concentrations

\[
15=\frac{x(2x)^2}{(0.10-x)(0.20-2x)^2}
\]

\[
15=\frac{4x^3}{(0.10-x)(0.20-2x)^2}
\]

---

Step 3: "Complete the cube" (cube-root technique)

Factor \(2\) from the \(HBr\) term in the denominator:

\[
0.20-2x = 2(0.10-x)
\]

Square it:

\[
(0.20-2x)^2 = [2(0.10-x)]^2 = 4(0.10-x)^2
\]

Substitute this into the denominator:

\[
15=\frac{4x^3}{(0.10-x)\,[4(0.10-x)^2]}
\]

The 4's cancel:

\[
15=\frac{x^3}{(0.10-x)^3}
\]

Rewrite as a perfect cube ratio:

\[
15=\left(\frac{x}{0.10-x}\right)^3
\]

Take the cube root of both sides:

\[
\sqrt[3]{15}=\frac{x}{0.10-x}
\]

---

Step 4: Solve for \(x\)

\[
x = \sqrt[3]{15}(0.10-x)
\]

\[
x = 0.10\sqrt[3]{15} - x\sqrt[3]{15}
\]

\[
x\left(1+\sqrt[3]{15}\right)=0.10\sqrt[3]{15}
\]

\[
x=\frac{0.10\sqrt[3]{15}}{1+\sqrt[3]{15}}
\]

Numerically,

\[
\sqrt[3]{15}\approx 2.47
\]

\[
x\approx \frac{0.10(2.47)}{1+2.47}=0.0712
\]

---

Step 5: Find \([HCl]_{\text{eq}}\)

\[
[HCl]_{\text{eq}} = 2x = 2(0.0712)=0.142\ \text{M}
\]

---

\[
\boxed{[HCl]_{\text{eq}} \approx 0.142\ \text{M}}
\]

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15.5 Equilibrium Constants and Coupled Reactions

Textbook: Section 15.5

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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

Answer

b. 1/K²

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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

Answer

d. 1/K³

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Exercise \(\PageIndex{5c}\)

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

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

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

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Exercise \(\PageIndex{5d}\)

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:

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

Answer

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.

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

 

 

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

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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?

Answer

Reactants, because of \(\Delta H<0\)

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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?

Answer

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

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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?

Answer

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

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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?

Answer

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

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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?

Answer

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

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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?

Answer

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

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Exercise \(\PageIndex{6g}\)

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. \(4Fe(s)+3O_{2}(g)\rightleftharpoons 2Fe_{2}O_{3}(s)\)

Answer

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

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Exercise \(\PageIndex{6h}\)

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

Answer

e. increasing the system temperature

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Exercise \(\PageIndex{6i}\)

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

Answer

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

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Exercise \(\PageIndex{6j}\)

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

Answer

d. at low temperature and low pressure

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Exercise \(\PageIndex{6k}\)

 If the reaction, 3H₂ + N₂ <=> 2NH₃, is at equilibrium, DECREASING the volume of the reaction vessel by a factor of two, at constant temperature, will cause:

a)      The amount of ammonia to increase.

b)      The amount of ammonia to decrease.

c)      The concentrations of all species to decrease.

d)      The amounts of hydrogen and nitrogen to increase.

e)      No change in the amounts of any reactants or products.

Answer

a)      The amount of ammonia to increase.

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Exercise \(\PageIndex{1}\)

Consider the reaction, which will increase the rate of Carbon dioxide production?

            C₆H₁₂O₆(s)  +  6O₂(g)  < === >  6CO₂(g)  +  6H₂O(l)   DH = -2816 kJ/mol

a. Increase pressure                 b. Decrease pressure               c. Increase temperature
d. decrease temperature       e. none of the above

Answer

d. decrease temperature , it moves in direction of exothermicity to warm the system in response to a decrease in T

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15.7: General Questions

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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? K_c = 0.07

Answer

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

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

(sec. 15.2.3)

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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?

Answer

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

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

\[x=0.026 M \nonumber \]

\[0.035-0.026=0.009 M \nonumber\]

(section 15.4.3.1.1)

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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?

 

Answer

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)

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Exercise \(\PageIndex{7d}\)

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

Calculate the value of K_c 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.

Answer

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

\[x=0.030-0.018=0.012 \nonumber\]

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

\[K_{c}=0.068 \nonumber\]

(section 15.3)

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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? K_p =0.64.

Answer

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

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

\[x=0.2 atm  \nonumber\]

\[1.2-0.2=1.0atm \nonumber\]

 

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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)

Answer

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

(section 15.2.4)

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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

Answer

c. products predominate

(section 15.2.3)

 

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Exercise \(\PageIndex{7h}\)

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

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

Answer

\[T = 25+273.15 = 298.15\nonumber \]

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

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

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

(section 15.2.7)

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Exercise \(\PageIndex{7i}\)

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

\[T = 25+273.15 = 298.15\nonumber \]

\[\Delta n = 3-1 = 2 \nonumber \]

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

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

(section 15.2.7)

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Exercise \(\PageIndex{7j}\)

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

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

Answer

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

(section 15.2.3)

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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

Answer

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

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

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

\[\left [ B \right ]=\left ( \frac{k\left [ A \right ]^{2}}{\left [ C \right ]} \right )^{1/2}=\left ( \frac{0.20*\left [ 0.35 \right ]^{2}}{\left [ 0.15 \right ]} \right )^{1/2}=0.40 M\nonumber \]

(section 15.2.3)

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

Answer

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

(section 15.2)

Textbook: 15: Equilibria

Robert E. Belford (University of Arkansas Little Rock; Department of Chemistry). The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, [e-mail]. You should contact him if you have any concerns. This material has both original contributions, and content built upon prior contributions of the LibreTexts Community and other resources, including but not limited to:

• Emily Choate
• Liliane Poirot