If the equilibrium constant for the given reaction is 0.25 NO leftharpoons (1/2)N2 +(1/2)O2, then the equilibrium constant for the reaction (1/2)N2 +(1/2)O2leftharpoons NO will be

Question

If the equilibrium constant for the given reaction is 0.25 NO leftharpoons (1/2)N2 +(1/2)O2, then the equilibrium constant for the reaction (1/2)N2 +(1/2)O2leftharpoons NO will be (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

NO leftharpoons (1/2)N2 +(1/2)O2 ; Kc=0.25 (1/2)N2 +(1/2)O2leftharpoons NO ; Kc = (1/0.25)=4

Q. If the equilibrium constant for the given reaction is $0.25$
$NO ⇌ \frac{1}{2} N_{2} + \frac{1}{2} O_{2}$ , then the equilibrium constant for the reaction $\frac{1}{2} N_{2} + \frac{1}{2} O_{2} ⇌ NO$ will be

$1$

$2$

$3$

$4$

$NO ⇌ \frac{1}{2} N_{2} + \frac{1}{2} O_{2}$ ; $K_{c} = 0.25$
$\frac{1}{2} N_{2} + \frac{1}{2} O_{2} ⇌ NO$ ;
$K_{c} = \frac{1}{0.25} = 4$

Q. If the equilibrium constant for the given reaction is 0.25 NO⇌21​N2​+21​O2​, then the equilibrium constant for the reaction 21​N2​+21​O2​⇌NO will be

1

2

3

4

NO⇌21​N2​+21​O2​ ; Kc​=0.25 21​N2​+21​O2​⇌NO ; Kc​=0.251​=4

Q. If the equilibrium constant for the given reaction is $0.25$
$NO ⇌ \frac{1}{2} N_{2} + \frac{1}{2} O_{2}$ , then the equilibrium constant for the reaction $\frac{1}{2} N_{2} + \frac{1}{2} O_{2} ⇌ NO$ will be

$1$

$2$

$3$

$4$

$NO ⇌ \frac{1}{2} N_{2} + \frac{1}{2} O_{2}$ ; $K_{c} = 0.25$
$\frac{1}{2} N_{2} + \frac{1}{2} O_{2} ⇌ NO$ ;
$K_{c} = \frac{1}{0.25} = 4$