Explanation: Hints: সাম্যাবস্থ?? N\(_2\) ও H\(_2\) এর মধ্যে এবং \(K_C\) দেওয়া আছে। Solve: \(N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)\) \[ K_C = \frac{[NH_3]^2}{[N_2][H_2]^3} = 3 \times 10^{-3} \implies [NH_3]^2 = 3 \times 10^{-3} \cdot [3 \cdot (0.1)^3] \implies [NH_3] = 3 \times 10^{-3} \, \text{molL}^{-1} \] Ans. (E) ব্যাখ্যা: \(K_P\) এবং \(K_C\) এর তুলনা: \[ \begin{array}{|c|c|c|} \hline \Delta n & \text{টেকনিক} & \text{উদাহরণ} \\ \hline \Delta n = 0 & K_P = K_C & H_2 + I_2 \rightleftharpoons 2HI \\ \Delta n = +\text{ve} & K_P > K_C & 2SO_3 \rightleftharpoons 2SO_2 + O_2 \\ \Delta n = -\text{ve} & K_P < K_C & N_2 + 3H_2 \rightleftharpoons 2NH_3 \\ \hline \end{array} \] মনে রাখার জন্য: \[ \text{সমীকরণ } PCl_5 \rightleftharpoons PCl_3 \implies K_P = \frac{\alpha^2}{1 - \alpha^2} \cdot P \] \[ N_2O_4 \rightleftharpoons 2NO_2 \implies K_P = \frac{4\alpha^2}{1 - \alpha^2} \cdot P \] \[ 2HI \rightleftharpoons H_2 + I_2 \implies K_P = \frac{\alpha^2}{4(1 - \alpha)^2} \cdot P \]