Chemical Equilibrium Question 4
Question 4 - 29 January - Shift 2
At $298 K$
$N_2(g)+3 H_2(g) \leftrightharpoons 2 NH_3(g), K_1=4 \times 10^{5}$
$N_2(g)+O_2(g) \leftrightharpoons 2 NO(g), K_2=1.6 \times 10^{12}$
$H_2(g)+\frac{1}{2} O_2(g) \leftrightharpoons H_2 O(g), K_3=1.0 \times 10^{-13}$
Based on above equilibria, the equilibrium constant of the reaction,
$2 NH_3(g)+\frac{5}{2} O_2(g) \leftrightharpoons 2 NO(g)+3 H_2 O(g)$
is $\quad \times 10^{-33} \quad$ (Nearest integer)
Show Answer
Answer: (4)
Solution:
Formula: Equilibrium constant
$ \begin{aligned} & N_2(g)+3 H_2(g) \leftrightharpoons 2 NH_3(g), K_1=4 \times 10^{5} \quad……(i) \\ & N_2(g)+O_2(g) \leftrightharpoons 2 NO(g), K_2=1.6 \times 10^{12}\quad……(ii) \\ & H_2(g)+\frac{1}{2} O_2(g) \leftrightharpoons H_2 O(g), K_3=1.0 \times 10^{-13}\quad……(iii) \\ & \text{ (ii) }+3 \times(\text{ iii) }- \text{ (i) } \\ & 2 NH_3(g)+\frac{5}{2} O_2(g) \leftrightharpoons 2 NO(g)+3 H_2 O(g) \\ & k _{eq}=\frac{k_2 \times k_3^{3}}{k_1}=\frac{1.6 \times 10^{12} \times(10^{-13})^{3}}{4 \times 10^{5}} \\ & =\frac{1.6}{4} \times 10^{-32}=4 \times 10^{-33} \end{aligned} $
