Medium
The following concentrations were obtained for the formation of $NH_{3}$ from $N_{2}$ and $H_{2}$ at equilibrium at $500\, K$: $[N_{2}] = 1.5 \times 10^{-2}\, M$,$[H_{2}] = 3.0 \times 10^{-2}\, M$ and $[NH_{3}] = 1.2 \times 10^{-2}\, M$. Calculate the equilibrium constant.
(N/A) The balanced chemical equation for the formation of ammonia is: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
The expression for the equilibrium constant $K_{c}$ is given by: $K_{c} = \frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^{3}}$
Substituting the given values: $K_{c} = \frac{(1.2 \times 10^{-2})^{2}}{(1.5 \times 10^{-2}) \times (3.0 \times 10^{-2})^{3}}$
$K_{c} = \frac{1.44 \times 10^{-4}}{(1.5 \times 10^{-2}) \times (27 \times 10^{-6})}$
$K_{c} = \frac{1.44 \times 10^{-4}}{40.5 \times 10^{-8}}$
$K_{c} = \frac{1.44}{40.5} \times 10^{4} \approx 0.03555 \times 10^{4} = 3.555 \times 10^{2}$
The expression for the equilibrium constant $K_{c}$ is given by: $K_{c} = \frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^{3}}$
Substituting the given values: $K_{c} = \frac{(1.2 \times 10^{-2})^{2}}{(1.5 \times 10^{-2}) \times (3.0 \times 10^{-2})^{3}}$
$K_{c} = \frac{1.44 \times 10^{-4}}{(1.5 \times 10^{-2}) \times (27 \times 10^{-6})}$
$K_{c} = \frac{1.44 \times 10^{-4}}{40.5 \times 10^{-8}}$
$K_{c} = \frac{1.44}{40.5} \times 10^{4} \approx 0.03555 \times 10^{4} = 3.555 \times 10^{2}$
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