Question

For the reaction, $N_{2}(g)+3 H_{2}(g) \rightleftharpoons 2 N H_{3}(g)$ at $400\, K,\, K_{p}=41 .$ Find the value for the following reaction, $\frac{1}{2} N _{2}(g)+\frac{3}{2} H _{2}(g) \rightleftharpoons NH _{3}(g)$

• A. 6.4
• B. 0.02
• C. 50
• D. 4.6

Answer 1

Answer: (A)

6.4

Answer 2

Given, $N_{2}(g)+3 H_{2}(g) \rightleftharpoons 2 N H_{3}(g) ;K_{P}=41\,$
$K_{p}=\frac{p_{N H_{3}}^{2}}{p_{N_{2}} p_{H_{2}}^{3}}=41$ ... (i)
For reaction,
$\frac{1}{2} N_{2}(g)+\frac{3}{2} H_{2}(g) \rightleftharpoons N H_{3}(g),$
$K_{p}'=\frac{p_{N H_{3}}}{p_{N_{2}}^{1 / 2} p_{H_{2}}^{3 / 2}}$ ... (ii)
On squaring both sides, we get
$\left(K_{p}'\right)^{2}=\frac{p_{N H_{3}}^{2}}{p_{N_{2}} p_{H_{2}}^{3}}$ ...(iii)
On dividing E (i) by E (iii),
we get $K_{p}'=\sqrt{K_{p}}=\sqrt{41}=6.4$