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Standard XII

Chemistry

Derivation of Kp and Kc

For the react...

Question

For the reaction, $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g), K_{p} = 0.157 a t m$ at $300 K$ . Calculate the value of $K_{c}$ for the same reaction at the same temperature.

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Solution

The given reaction is :-
$N_{2} O_{4} (g) ⇌ 2 {N O}_{2} (g); K_{P} = 0.157 a t m$
$T = 300 K$
We have, $K_{P} = K_{C} {(R T)}^{△ n}; △ n = 2 - 1 = 1$
$\Rightarrow K_{P} = K_{C} (R T) \Rightarrow K_{C} = \frac{K_{P}}{R T} = \frac{0.157}{0.0821 \times 300}$
$\Rightarrow K_{C} = 0.00637 M$

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For the reaction, N2O4(g)⇌2NO2(g),Kp=0.157 atm at 300 K. Calculate the value of Kc for the same reaction at the same temperature.

The given reaction is :-N2O4(g)⇌2NO2(g);KP=0.157atmT=300KWe have, KP=KC(RT)△n;△n=2−1=1⇒KP=KC(RT)⇒KC=KPRT=0.1570.0821×300⇒KC=0.00637M

For the reaction, $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g), K_{p} = 0.157 a t m$ at $300 K$ . Calculate the value of $K_{c}$ for the same reaction at the same temperature.

The given reaction is :-
$N_{2} O_{4} (g) ⇌ 2 {N O}_{2} (g); K_{P} = 0.157 a t m$
$T = 300 K$
We have, $K_{P} = K_{C} {(R T)}^{△ n}; △ n = 2 - 1 = 1$
$\Rightarrow K_{P} = K_{C} (R T) \Rightarrow K_{C} = \frac{K_{P}}{R T} = \frac{0.157}{0.0821 \times 300}$
$\Rightarrow K_{C} = 0.00637 M$