Chemical and Ionic Equilibrium - Result Question 77

####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:

  1. $N _2(g)+3 H _2(g) \rightleftharpoons 2 NH _3(g)$

$\Delta n=-2$

$$ \begin{aligned} K _p & =K _c(R T)^{\Delta n} \ K _c & =\frac{K _p}{(R T)^{\Delta n}}=\frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}} \end{aligned} $$



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