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

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).

 

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

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

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

 

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.

 

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

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

 

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

 

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

 

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. \(K=\frac{k_{f}}{k_{r}}\)

 

Exercise \(\PageIndex{2i}\)

What is the Kc for the following gas phase reaction?

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

Answer

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

 

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

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

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

 

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

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

 

Exercise \(\PageIndex{2l}\)

What is the equilibrium constant expression for the following reaction?

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

Answer

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

 

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

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

 

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

15.7

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

 

Exercise \(\PageIndex{2.1b}\)

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

What is the Δn for the equation above?

Answer

 

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

 

 

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

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

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

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

 

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

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

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

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

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

 

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

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

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

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

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

 

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

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

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

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

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

 

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

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

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

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

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

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

\[T=299K\nonumber \]

 

15.3 Determining an Equilibrium Constant

Textbook: Section 15.3

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

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

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

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

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

\[x=0.042M \nonumber \]

 

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

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

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

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

\[x=0.042M \nonumber \]

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

 

 

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

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

\[0.40M=2x \nonumber \]

\[x=0.2M \nonumber \]

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

 

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

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

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

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

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

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

\[x=1.33 M \nonumber \]

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

 

 

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

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

 

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

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

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

 

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

\[1.08=1.26-x \nonumber\]

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

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

 

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

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

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

\[x=0.09 \nonumber\]

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

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

 

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 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

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

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

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

\[x=0.042M \nonumber \]

 

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

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

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

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

\[x=0.042M \nonumber \]

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

 

 

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

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

\[x=1.33 M \nonumber \]

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

 

 

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

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

 

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.

 

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

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

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

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

\[0.070x=0.003675 \nonumber\]

\[x=0.0525 \nonumber\]

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

 

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

 

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)

 

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)

 

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

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

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)

 

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?

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)

 

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)

 

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

Answer

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

Answer

d. 1/K³

 

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

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

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

 

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

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 \(\Delta H<0\)

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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? 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)

 

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)

 

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)

 

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)

 

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

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

\[x=0.2 atm  \nonumber\]

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

 

 

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)

 

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)

 

 

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)

 

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)

 

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)

 

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)

 

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