Kp and Kp' are the equilibrium constants of the two reactions, given below (1/2)N2 (g) + (1/2)H2 (g) leftharpoons NH3 (g) N2 (g) + 3H2 (g) leftharpoons 2NH3 (g) Therefore, Kp and Kp' are related by

Question

Kp and Kp' are the equilibrium constants of the two reactions, given below (1/2)N2 (g) + (1/2)H2 (g) leftharpoons NH3 (g) N2 (g) + 3H2 (g) leftharpoons 2NH3 (g) Therefore, Kp and Kp' are related by (A) (Kp')2 (B) Kp = √Kp' (C) Kp = 2Kp' (D) Kp = Kp'

Tardigrade Admin · Accepted Answer

When a balanced equation having equilibrium constant Kp is multiplied by a certain value n, the equilibrium constant for the new equation will be equal to (Kp)n. In the given equation, n = 2 .So, Kp'=(Kp)2 or Kp=√ Kp'

Q. $K_{p}$ and $K_{p}^{'}$ are the equilibrium constants of the two reactions, given below
$\frac{1}{2} N_{2} (g) + \frac{1}{2} H_{2} (g) ⇌ N H_{3} (g)$
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$
Therefore, $K_{p}$ and $K_{p}^{'}$ are related by

$(K_{p}^{'})^{2}$

$K_{p} = K_{p}^{'}$

$K_{p} = 2 K_{p}^{'}$

$K_{p} = K_{p}^{'}$

When a balanced equation having equilibrium constant $K_{p}$ is multiplied by a certain value n, the equilibrium constant for the new equation will be equal to $(K_{p})^{n}$ . In the given equation, n = 2 .So, $K_{p}^{'} = (K_{p})^{2}$ or $K_{p} = K_{p}^{'}$

Q. Kp​ and Kp′​ are the equilibrium constants of the two reactions, given below 21​N2​(g)+21​H2​(g)⇌NH3​(g) N2​(g)+3H2​(g)⇌2NH3​(g) Therefore, Kp​ and Kp′​ are related by

(Kp′​)2

Kp​=Kp′​​

Kp​=2Kp′​

Kp​=Kp′​

When a balanced equation having equilibrium constant Kp​ is multiplied by a certain value n, the equilibrium constant for the new equation will be equal to (Kp​)n. In the given equation, n = 2 .So, Kp′​=(Kp​)2 or Kp​=Kp′​​

Q. $K_{p}$ and $K_{p}^{'}$ are the equilibrium constants of the two reactions, given below
$\frac{1}{2} N_{2} (g) + \frac{1}{2} H_{2} (g) ⇌ N H_{3} (g)$
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$
Therefore, $K_{p}$ and $K_{p}^{'}$ are related by

$(K_{p}^{'})^{2}$

$K_{p} = K_{p}^{'}$

$K_{p} = 2 K_{p}^{'}$

$K_{p} = K_{p}^{'}$

When a balanced equation having equilibrium constant $K_{p}$ is multiplied by a certain value n, the equilibrium constant for the new equation will be equal to $(K_{p})^{n}$ . In the given equation, n = 2 .So, $K_{p}^{'} = (K_{p})^{2}$ or $K_{p} = K_{p}^{'}$