Solveeit Logo
Login

Question

Chemistry Question on Equilibrium

K1K_1 and K2K_2 are equilibrium constant for reactions (i) and (ii)
(i) N2(g)+O2(g)⇋2NO(g)N_2 ( g ) + O_2 ( g ) \leftrightharpoons 2 NO ( g)
(ii) NO(g)⇋12N2(g)+12O2(g)NO ( g ) \leftrightharpoons \frac{1}{2} N_2 ( g ) + \frac{ 1}{ 2} O_2 ( g ).Then,

A

K1=[1K2]2K_1 = \bigg [ \frac{1}{ K_2 } \bigg ] ^2

B

K1=K22 K_1 = K_2^2

C

K1=1K2K_1 = \frac{1}{ K_2 }

D

K1=(K2)0K_1 = ( K_2 )^0

Answer

K1=[1K2]2K_1 = \bigg [ \frac{1}{ K_2 } \bigg ] ^2

Explanation

Solution

The correct answer is A:K1=[1K2]2K_1 = \bigg [ \frac{1}{ K_2 } \bigg ] ^2
Consider reaction (i),
N2(g)+O2(g)⇋2NO(g)N_2 ( g ) + O_2 ( g ) \leftrightharpoons 2 NO ( g) ... ( i )
K1=[NO]2[N2][O2]K_1 = \frac{[ NO]^2 }{ [ N_2 ] [ O_2 ] }
Now, consider reaction (ii),
NO ( g) ⇋12N2(g)+12O2(g)\leftrightharpoons \frac{1}{2} N_2 ( g) + \frac{1}{2} O_2 ( g )
K2=[N2]1/2[O2]1/2[NO]K_2 = \frac{ [ N_2 ]^{1/2} [ O_2]^{1/2}}{ [ NO] }
1K2=1[N2]1/2[O2]1/2[NO]=[NO][N2]1/2[O2]1/2\frac{1}{ K_2 } = \frac{1}{ \frac{ [ N_2 ]^{1/2} [ O_2 ]^{1/2}}{ [ NO] } } = \frac{ [ NO ] }{ [ N_2]^{1/2} [ O_2 ]^{1/2} }
\bigg( \frac{ 1}{ K_2 } \bigg)^2 = \bigg \\{ \frac{ [ NO ] }{ [ N_2 ]^{1/2} [ O_2]^{1/2}} \bigg \\}^2 = \frac{ [ NO]^2 }{ [ N_2 ] [ O_2 ] } = K_1