Chemical Equilibrium Question 2
Question 2 - 2024 (29 Jan Shift 2)
The following concentrations were observed at $500 K$ for the formation of $NH _3$ from $N _2$ and $H _2$. At equilibrium : $\left[N _2\right]=2 \times 10^{-2} M,\left[H _2\right]=3 \times 10^{-2} M$ and $\left[NH _3\right]=1.5 \times 10^{-2} M$. Equilibrium constant for the reaction is __________
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Answer (417)
Solution
$K _C=\frac{\left[NH _3\right]^{2}}{\left[N _2\right]\left[H _2\right]^{3}}$
$K _C=\frac{\left(1.5 \times 10^{-2}\right)^{2}}{\left(2 \times 10^{-2}\right) \times\left(3 \times 10^{-2}\right)^{3}}$
$K _C=417$
