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