The equilibrium constant for the reversible reaction, N2+3H2 leftharpoons 2NH3 is K and for the reaction (1/2)N2+(3/2)H2leftharpoons NH3 the equilibrium constant is K'.K and K' will be related as

Question

The equilibrium constant for the reversible reaction, N2+3H2 leftharpoons 2NH3 is K and for the reaction (1/2)N2+(3/2)H2leftharpoons NH3 the equilibrium constant is K'.K and K' will be related as (A) K=K' (B) K'=√K (C) K=√[⋅]K' (D) K× K'=1

Tardigrade Admin · Accepted Answer

K' =Kn ; Hence n=(1/2) ∴ K' =K1/2=√K

Q. The equilibrium constant for the reversible reaction, $N_{2} + 3 H_{2} ⇌ 2 N H_{3}$ is K and for the reaction $\frac{1}{2} N_{2} + \frac{3}{2} H_{2} ⇌ N H_{3}$ the equilibrium constant is $K^{'} . K$ and $K^{'}$ will be related as

$K = K^{'}$

$K^{'} = K$

$K = \cdot K^{'}$

$K \times K^{'} = 1$

$K^{'} = K^{n};$ Hence $n = \frac{1}{2}$
$∴ K^{'} = K^{1/2} = K$

Q. The equilibrium constant for the reversible reaction, N2​+3H2​⇌2NH3​ is K and for the reaction 21​N2​+23​H2​⇌NH3​ the equilibrium constant is K′.K and K′ will be related as

K=K′

K′=K​

K=⋅K′​

K×K′=1

K′=Kn; Hence n=21​ ∴K′=K1/2=K​