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

N2(g)+3H2(g)leftharpoons 2NH3(g) From law of mass action, K=([N H3]2/[N2][H2]3) ldots (i) Similarly, for reaction (1/2)N2+(3/2)H2 leftharpoons NH3 K' =([N H3]/[N2]1 / 2[H2]3 / 2) ...(ii) So, from Eqs.(i) and (ii) K'=√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

$K = K^{'}$

$K^{'} = K$

$K = K^{'}$

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

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

N2​(g)+3H2​(g)⇌2NH3​(g) From law of mass action, K=[N2​][H2​]3[NH3​]2​… (i) Similarly, for reaction 21​N2​+23​H2​⇌NH3​ K′=[N2​]1/2[H2​]3/2[NH3​]​...(ii) So, from Eqs.(i) and (ii) K′=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

$K = K^{'}$

$K^{'} = K$

$K = K^{'}$

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