Properties of the Equilibrium Constant

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AP Chemistry › Properties of the Equilibrium Constant

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

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

$2 C(g) \Leftrightarrow D (g) + E (g)$

CORRECT

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

Explanation

$2 C \Leftrightarrow A + B$

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

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

$NOCl(g) + \frac{1}{2} O_2 (g) \Leftrightarrow NO_2Cl$

$NO_2 (g) + \frac{1}{2} Cl_2 \Leftrightarrow NO_2Cl$

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

CORRECT

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

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

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

Explanation

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

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

$2 (NO_2 + 1/2 Cl_2 \Leftrightarrow NO_2Cl)$

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

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

$N_2 (g) + 3 H_2 (g) \Leftrightarrow 2 NH_3(g)$

$H_2 (g) + I_2 (g) \Leftrightarrow 2 HI (g)$

$3.2 x 10^{-3}$

CORRECT

Explanation

$2 NH_3 \Leftrightarrow N_2 + 3 H_2$

$3 ( H_2 + I_2 \Leftrightarrow 2 HI)$

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

$H_2(g) + I_2 (g) \Leftrightarrow 2 HI (g)$

CORRECT

Explanation

$2 HI \Leftrightarrow H_2 + I_2$

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

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

$2 C(g) \Leftrightarrow D (g) + E (g)$

25.3

15.8

50.5

CORRECT

73.2

0.143

Explanation

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

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

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

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