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AP Chemistry : Properties of the Equilibrium Constant

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Example Questions

Example Question #136 : Reactions And Equilibrium

Calculate the equilibrium constant for $2 C(g) \Leftrightarrow A (g) + B (g)$ $\displaystyle 2 C(g) \Leftrightarrow A (g) + B (g)$ given the following information:

$A (g) + B (g) \Leftrightarrow 2 C (g)$ $\displaystyle A (g) + B (g) \Leftrightarrow 2 C (g)$ $K_1 = 4.39 x 10^{-3}$ $\displaystyle K_1 = 4.39 x 10^{-3}$

$2 C(g) \Leftrightarrow D (g) + E (g)$ $\displaystyle 2 C(g) \Leftrightarrow D (g) + E (g)$ $\displaystyle K_2 = 1.15 x 10^4$

Possible Answers:

$3.14 \, x \, 10^{-3}$ $\displaystyle 3.14 \, x \, 10^{-3}$

$1.14\, x\, 10^2$ $\displaystyle 1.14\, x\, 10^2$

11.4 $\displaystyle 11.4$

22.8 $\displaystyle 22.8$

$2.28 \, x \, 10^2$ $\displaystyle 2.28 \, x \, 10^2$

Correct answer:

$2.28 \, x \, 10^2$ $\displaystyle 2.28 \, x \, 10^2$

Explanation:

$2 C \Leftrightarrow A + B$ $\displaystyle 2 C \Leftrightarrow A + B$ $\displaystyle K = 1/K_1= 228$

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Example Question #1 : Properties Of The Equilibrium Constant

Calculate the equilibrium constant for $A (g) + B (g) \Leftrightarrow D (g) + E (g)$ $\displaystyle A (g) + B (g) \Leftrightarrow D (g) + E (g)$ given the following information:

$A (g) + B (g) \Leftrightarrow 2 C (g)$ $\displaystyle A (g) + B (g) \Leftrightarrow 2 C (g)$ $K_1 = 4.39 \, x \, 10^{-3}$ $\displaystyle K_1 = 4.39 \, x \, 10^{-3}$

$2 C(g) \Leftrightarrow D (g) + E (g)$ $\displaystyle 2 C(g) \Leftrightarrow D (g) + E (g)$ $K_2 = 1.15 \, x \, 10^4$ $\displaystyle K_2 = 1.15 \, x \, 10^4$

Possible Answers:

15.8

0.143

50.5

73.2

25.3

Correct answer:

50.5

Explanation:

$A + B \Leftrightarrow 2 C$ $\displaystyle A + B \Leftrightarrow 2 C$ $K_1 = 4.39 x 10^{-3}$ $\displaystyle K_1 = 4.39 x 10^{-3}$

$2C \Leftrightarrow D + E$ $\displaystyle 2C \Leftrightarrow D + E$ $K_2 = 1.15 x 10^{-4}$ $\displaystyle K_2 = 1.15 x 10^{-4}$

$K = K_1 K_2 = (4.39 x 10^{-3}) (1.15 x 10^{-4}) = 50.5$ $\displaystyle K = K_1 K_2 = (4.39 x 10^{-3}) (1.15 x 10^{-4}) = 50.5$

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Example Question #1 : Properties Of The Equilibrium Constant

Calculate the equilibrium constant for $N_2 (g) + O_2 (g) + Cl_2 (g) \Leftrightarrow 2 \: NOCl$ $\displaystyle N_2 (g) + O_2 (g) + Cl_2 (g) \Leftrightarrow 2 \: NOCl$given the following information:

$\frac{1}{2}N_2 + O_2(g) \Leftrightarrow NO_2 (g)$ $\displaystyle \frac{1}{2}N_2 + O_2(g) \Leftrightarrow NO_2 (g)$ $K_1 = 1.0 \, x \, 10^{-9}$ $\displaystyle K_1 = 1.0 \, x \, 10^{-9}$

$NOCl(g) + \frac{1}{2} O_2 (g) \Leftrightarrow NO_2Cl$ $\displaystyle NOCl(g) + \frac{1}{2} O_2 (g) \Leftrightarrow NO_2Cl$ $K_2 = 1.1 \, x\, 10^2$ $\displaystyle K_2 = 1.1 \, x\, 10^2$

$NO_2 (g) + \frac{1}{2} Cl_2 \Leftrightarrow NO_2Cl$ $\displaystyle NO_2 (g) + \frac{1}{2} Cl_2 \Leftrightarrow NO_2Cl$ $\displaystyle K_3 = 0.3$

Possible Answers:

$7\, x\, 10^{-24}$ $\displaystyle 7\, x\, 10^{-24}$

$5 \, x \, 10^{-10}$ $\displaystyle 5 \, x \, 10^{-10}$

$7\, x\, 10^{-8}$ $\displaystyle 7\, x\, 10^{-8}$

$3\, x\, 10^{-3}$ $\displaystyle 3\, x\, 10^{-3}$

$\displaystyle 11$

Correct answer:

$7\, x\, 10^{-24}$ $\displaystyle 7\, x\, 10^{-24}$

Explanation:

$2 (1/2 N_2 + O_2 \Leftrightarrow 2 NO_2 )$ $\displaystyle 2 (1/2 N_2 + O_2 \Leftrightarrow 2 NO_2 )$ $K = K_1^2 = 1 x 10^{-18}$ $\displaystyle K = K_1^2 = 1 x 10^{-18}$

$2 (NO_2Cl \Leftrightarrow NOCl + 1/2 O_2)$ $\displaystyle 2 (NO_2Cl \Leftrightarrow NOCl + 1/2 O_2)$ $K = (1/K_2)^2 = 8.3 x 10^{-5}$ $\displaystyle K = (1/K_2)^2 = 8.3 x 10^{-5}$

$2 (NO_2 + 1/2 Cl_2 \Leftrightarrow NO_2Cl)$ $\displaystyle 2 (NO_2 + 1/2 Cl_2 \Leftrightarrow NO_2Cl)$ $\displaystyle K = K_3^2 = 0.09$

$K = (1 x 10^{-18}) (8.3 x 10^{-5}) (0.09) = 7 x 10^{-24}$ $\displaystyle K = (1 x 10^{-18}) (8.3 x 10^{-5}) (0.09) = 7 x 10^{-24}$

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Example Question #141 : Reactions And Equilibrium

Calculate the equilibrium constant for $2 NH_3 (g) + 3 I_2 (g) \Leftrightarrow N_2 (g) + 6 HI (g)$ $\displaystyle 2 NH_3 (g) + 3 I_2 (g) \Leftrightarrow N_2 (g) + 6 HI (g)$ given the following information:

$N_2 (g) + 3 H_2 (g) \Leftrightarrow 2 NH_3(g)$ $\displaystyle N_2 (g) + 3 H_2 (g) \Leftrightarrow 2 NH_3(g)$ $\displaystyle K_1 = 0.50$

$H_2 (g) + I_2 (g) \Leftrightarrow 2 HI (g)$ $\displaystyle H_2 (g) + I_2 (g) \Leftrightarrow 2 HI (g)$ $\displaystyle K_2 = 50$

Possible Answers:

$3.2 x 10^{-3}$ $\displaystyle 3.2 x 10^{-3}$

$\displaystyle 2.8 x 10^6$

$\displaystyle 5.6 x 10^4$

$\displaystyle 1.5 x 10^3$

$\displaystyle 2.5 x 10^5$

Correct answer:

$\displaystyle 2.5 x 10^5$

Explanation:

$2 NH_3 \Leftrightarrow N_2 + 3 H_2$ $\displaystyle 2 NH_3 \Leftrightarrow N_2 + 3 H_2$ $\displaystyle K = 1/K_1 = 2$

$3 ( H_2 + I_2 \Leftrightarrow 2 HI)$ $\displaystyle 3 ( H_2 + I_2 \Leftrightarrow 2 HI)$ $\displaystyle K = K_2^3 = 1.25 x 10^5$

$\displaystyle K = (2) (1.25 x 10^5) = 2.5 x 10^5$

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Example Question #3 : Properties Of The Equilibrium Constant

Calculate the equilibrium constant for $2 HI (g) \Leftrightarrow H_2 (g) + I_2 (g)$ $\displaystyle 2 HI (g) \Leftrightarrow H_2 (g) + I_2 (g)$ given the following information:

$H_2(g) + I_2 (g) \Leftrightarrow 2 HI (g)$ $\displaystyle H_2(g) + I_2 (g) \Leftrightarrow 2 HI (g)$ $\displaystyle K_1 = 50$

Possible Answers:

0.05 $\displaystyle 0.05$

0.02 $\displaystyle 0.02$

0.01 $\displaystyle 0.01$

1.21 $\displaystyle 1.21$

0.25 $\displaystyle 0.25$

Correct answer:

0.02 $\displaystyle 0.02$

Explanation:

$2 HI \Leftrightarrow H_2 + I_2$ $\displaystyle 2 HI \Leftrightarrow H_2 + I_2$ $\displaystyle K = 1/K_1 = 0.02$

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AP Chemistry : Properties of the Equilibrium Constant

Study concepts, example questions & explanations for AP Chemistry

All AP Chemistry Resources

Example Questions

Example Question #136 : Reactions And Equilibrium

Example Question #1 : Properties Of The Equilibrium Constant

Example Question #1 : Properties Of The Equilibrium Constant

Example Question #141 : Reactions And Equilibrium

Example Question #3 : Properties Of The Equilibrium Constant

All AP Chemistry Resources

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