K_1, K_2 and K_3 are the equilibrium constant | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For reaction $(I): N_2 + 2O_2 \rightleftharpoons 2NO_2$,the equilibrium constant is $K_1 = \frac{[NO_2]^2}{[N_2][O_2]^2} \dots (i)$For reaction $(

Vedclass · Accepted Answer

$K_1 = \frac{1}{K_2} = \frac{1}{(K_3)^2}$

Vedclass · Answer

(B) For reaction $(I): N_2 + 2O_2 \rightleftharpoons 2NO_2$,the equilibrium constant is $K_1 = \frac{[NO_2]^2}{[N_2][O_2]^2} \dots (i)$For reaction $(II): 2NO_2 \rightleftharpoons N_2 + 2O_2$,which is the reverse of $(I)$,the equilibrium constant is $K_2 = \frac{[N_2][O_2]^2}{[NO_2]^2} = \frac{1}{K_1} \dots (ii)$For reaction $(III): NO_2 \rightleftharpoons \frac{1}{2} N_2 + O_2$,the equilibrium constant is $K_3 = \frac{[N_2]^{1/2}[O_2]}{[NO_2]}$Squaring $K_3$,we get $(K_3)^2 = \frac{[N_2][O_2]^2}{[NO_2]^2} = K_2 = \frac{1}{K_1} \dots (iii)$Thus,$K_1 = \frac{1}{K_2} = \frac{1}{(K_3)^2}$.

Similar Questions

For the reaction $A + 2B \rightleftharpoons C$,the expression for the equilibrium constant is.........

The rate of forward reaction is two times that of reverse reaction at a given temperature and identical concentration. $K_{equilibrium}$ is

If the equilibrium constants for the given reactions are $K_1$ and $K_2$ respectively,find the relationship between $K_2$ and $K_1$.
$2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$
$SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$

For the formation of $NH_{3(g)}$ from its constituent elements,which of the relations between the reaction quotient $(Q)$ and equilibrium constant $(K_C)$ is correct for the backward reaction?

Vedclass Test Series

Exam Paper Generator

Online Exam Module