Chemical and Ionic Equilibrium - Result Question 21
####21. For the reversible reaction,
$$ N _2(g)+3 H _2(g) \rightleftharpoons 2 NH _3(g) $$
at $500^{\circ} C$, 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 $mol / L$ 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:
- $N _2(g)+3 H _2(g) \rightleftharpoons 2 NH _3(g)$
$\Delta n=-2$
$$ \begin{aligned} K _p & =K _c(R T)^{\Delta n} \ K _c & =\frac{K _p}{(R T)^{\Delta n}}=\frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}} \end{aligned} $$
