The following concentrations were observed at 500 K for the formation of from and . At equilibrium: Equilibrium constant for the reaction is

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Solution

Q.
The following concentrations were observed at 500 K for the formation of ${NH}_{3}$ from $N_{2}$ and $H_{2}$ . At equilibrium; $[N_{2}] = 2 \times 10^{- 2} M, [H_{2}] = 3 \times 10^{- 2} M and [{NH}_{3}] = 1.5 \times 10^{- 2} M .$ Equilibrium constant for the reaction is _____________ . [2024]

Ans.
(417)

$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)$

$K_{c} = \frac{{[{NH}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}} = \frac{{(1.5 \times 10^{- 2})}^{2}}{[2 \times 10^{- 2}] {[3 \times 10^{- 2}]}^{3}} = 417$

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