• # question_answer For the reaction ${{N}_{2}}(g)+3{{H}_{2}}(g)\rightleftharpoons 2N{{H}_{3}}(g),$at $400\,K,\,{{K}_{p}}=41$. Find the value of ${{K}_{p}}$ for each of the following reactions at the same temperature: (i) $2N{{H}_{3}}(g)\,\rightleftharpoons {{N}_{2}}(g)+3{{H}_{2}}(g)$ (ii) $\frac{1}{2}{{N}_{2}}(g)+\frac{3}{2}{{H}_{2}}(g)\rightleftharpoons N{{H}_{3}}(g)$ (iii)$2{{N}_{2}}(g)+6{{H}_{2}}(g)\rightleftharpoons 4N{{H}_{3}}(g)$

(i) It is the reverse of the given reaction. Hence, ${{K}_{p}}=\frac{1}{41}$ (ii) It is obtained by dividing the given equation by 2. Hence, ${{K}_{p}}=\sqrt{41}$ (iii) It is obtained by multiplying the given equation by 2. Hence, ${{K}_{p}}={{(41)}^{2}}$.