If the value of equilibrium constant Kc for the reaction, N2 + 3H2 leftharpoons 2NH3 is 7. The equilibrium constant for the reaction 2N2 + 6H2 leftharpoons 4NH3 will be

Question

If the value of equilibrium constant Kc for the reaction, N2 + 3H2 leftharpoons 2NH3 is 7. The equilibrium constant for the reaction 2N2 + 6H2 leftharpoons 4NH3 will be (A) 49 (B) 7 (C) 14 (D) 28

Tardigrade Admin · Accepted Answer

N 2+3 H 2 leftharpoons 2 NH 3, given that K c =7 2 N 2+6 H 2 leftharpoons 4 NH 3 If the first reaction is multiplied by 2 , the equilibrium constant becomes square of the first reaction K c . So, K c prime= K c 2=72=49

Q. If the value of equilibrium constant $K_{c}$ for the reaction, $N_{2} + 3 H_{2} ⇌ 2 N H_{3}$ is $7$ . The equilibrium constant for the reaction $2 N_{2} + 6 H_{2} ⇌ 4 N H_{3}$ will be

$49$

$7$

$14$

$28$

$N_{2} + 3 H_{2} ⇌ 2 N H_{3}$ , given that $K_{c} = 7$
$2 N_{2} + 6 H_{2} ⇌ 4 N H_{3}$
If the first reaction is multiplied by 2 , the equilibrium constant becomes square of the first reaction $K_{c}$ .
So, $K_{c}^{'} = K_{c}^{2} = 7^{2} = 49$

Q. If the value of equilibrium constant Kc​ for the reaction, N2​+3H2​⇌2NH3​ is 7. The equilibrium constant for the reaction 2N2​+6H2​⇌4NH3​ will be

49

7

14

28

N2​+3H2​⇌2NH3​, given that Kc​=7 2N2​+6H2​⇌4NH3​ If the first reaction is multiplied by 2 , the equilibrium constant becomes square of the first reaction Kc​. So, Kc′​=Kc2​=72=49

Q. If the value of equilibrium constant $K_{c}$ for the reaction, $N_{2} + 3 H_{2} ⇌ 2 N H_{3}$ is $7$ . The equilibrium constant for the reaction $2 N_{2} + 6 H_{2} ⇌ 4 N H_{3}$ will be

$49$

$7$

$14$

$28$

$N_{2} + 3 H_{2} ⇌ 2 N H_{3}$ , given that $K_{c} = 7$
$2 N_{2} + 6 H_{2} ⇌ 4 N H_{3}$
If the first reaction is multiplied by 2 , the equilibrium constant becomes square of the first reaction $K_{c}$ .
So, $K_{c}^{'} = K_{c}^{2} = 7^{2} = 49$