Question

Consider the following reaction at 600K: A reaction mixture at equilibrium at 600K contains 0.170 bar H_2, 0.120 bar I_2 and 1.26 bar HI. Calculate the equilibrium constant (K_p) for this reaction A second reaction mixture, also at 600 K contains 0.135 bar of H_2, 0.135 bar I_2 and 1.10 bar of HI. Is the mixture at equilibrium? If the mixture in part b is not at equilibrium, what will be the partial pressures of H_2, I_2 and HI when the reaction mixture reaches equilibrium? What will be the total pressure in the flask at equilibrium?

Answer 1

i) the equilibrium constant is given by

The equilibrium constant is given by K_p = [P_HI]^2/[P_H2] [P_I2] K_p = [1.26]^2/[0.17] [0.12] = 77.82 Therefore k_p = 77.82 �for the second reaction mixture Reaction Quotient Q = [P_HI]^2/[P_H2] [P_I2] Q = [1.1]^2/[0.135] [0.135] = 66.39 ince Q < K the reaction is not equilibrium.

$K_{p}=\frac{\left [1.26 \right ]^{2}}{\left [0.17\right ]\left [0.12\right ]}= 77.82$

$\therefore K_{p}= 77.82$

ii) for the second reaction mixture

$Reaction\: Quotient \: Q=\frac{\left [P_{ HI} \right ]^{2}}{\left [P_{ H_{2} }\right ]\left [P_{ I_{2}} \right ]}$

$Q=\frac{\left [1.1 \right ]^{2}}{\left [0.135\right ]\left [0.135 \right ]}=66.39$

Since Q<K the reaction is not equilibrium.