Chemical and Ionic Equilibrium 1 Question 20
21. For the reversible reaction,
$$ \mathrm{N}{2}(g)+3 \mathrm{H}{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$
at $500^{\circ} \mathrm{C}$, the value of $K_{p}$ is $1.44 \times 10^{-5}$ when partial pressure is measured in atmosphere. The corresponding value of $K_{c}$ with concentration in $\mathrm{mol} / \mathrm{L}$ is
$(2000,5,1 M)$
(a) $\frac{1.44 \times 10^{-5}}{(0.082 \times 500)^{-2}}$
(b) $\frac{1.44 \times 10^{-5}}{(8.314 \times 773)^{-2}}$
(c) $\frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{2}}$
(d) $\frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}}$
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Solution:
- $\mathrm{N}{2}(g)+3 \mathrm{H}{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$
$\Delta n=-2$
$$ \begin{aligned} K_{p} & =K_{c}(R T)^{\Delta n} \ K_{c} & =\frac{K_{p}}{(R T)^{\Delta n}}=\frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}} \end{aligned} $$
