1. Tardigrade
  2. Question
  3. Chemistry
  4. The value of Kp/KC for the following reactions at 300 K are, respectively :(At 300 K, RT = 24.62 dm3 atm mol-1) N2(g) + O2 (g) rightharpoons 2NO(g) N2O4(g) rightharpoons 2NO2(g) N2 (g) + 3H2 (g) rightharpoons 2NH3 (g)

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Q. The value of $K_p/K_C$ for the following reactions at $300\,K$ are, respectively :(At $300\,K, RT = 24.62$ $dm^3$ atm $mol^{-1}$)
$N_2(g) + O_2 (g) \leftrightharpoons 2NO(g)$
$N_2O_4(g) \leftrightharpoons 2NO_2(g)$
$N_2 (g) + 3H_2 (g) \leftrightharpoons 2NH_3 (g) $

JEE MainJEE Main 2019Equilibrium

Solution:

We know that, the relationship between $K_{p}$ and $K_{C}$ of a chemical equilibrium state (reaction) is
$K_{p}=K_{C}(R T)^{\Delta n_{g}}$
$\Rightarrow \frac{K_{p}}{K_{C}}=(R T)^{\Delta n_{g}}$
where, $\Delta n_{g}=\Sigma n_{\text {Products }}-\Sigma n_{\text {Reactants }}$
(i) $N _{2}(g)+ O _{2}(g) \rightleftharpoons 2 NO (g)$
$\Rightarrow(R T)^{2-(1+1)}=(R T)^{0}=1$
(ii) $N _{2} O _{4}(g) \rightleftharpoons 2 NO _{2}(g)$
$\Rightarrow(R T)^{2-1}=R T=24.62\, dm ^{3} \,atm \,mol ^{-1}$
(iii) $N _{2}(g)+3 H _{2}(g) \rightleftharpoons 2 NH _{3}(g)$
$\Rightarrow(R T)^{2-(3+1)}=(R T)^{-2}$
$=\frac{1}{\left(24.62\, dm ^{3} \,atm \,mol ^{-1}\right)^{2}}$
$=1.649 \times 10^{-3} \,dm ^{-6} \,atm ^{-2}\, mol ^{2}$