The equilibrium constant for the reaction H2O(g)+CO(g)leftharpoons H2(g)+CO2(g) 64. The rate constant for the forward reaction is 160, the rate constant for the backward reaction is:

Question

The equilibrium constant for the reaction H2O(g)+CO(g)leftharpoons H2(g)+CO2(g) 64. The rate constant for the forward reaction is 160, the rate constant for the backward reaction is: (A) 0.4 (B) 2.5 (C) 6.2 (D)  10.24× 103

Tardigrade Admin · Accepted Answer

H2O+CO xrightarrowH2+CO2 Kc=64,kf=160,kb=? We know that, Equilibrium constant Kc text= ( textRate textconstant textof textforward textreaction/ textRate textconstant textof textbackward textreaction) textKc=(kf/kb) So, 64=(160/kb) kb=(160/64)=2.5

Q. The equilibrium constant for the reaction $H_{2} O (g) + CO (g) ⇌ H_{2} (g) + C O_{2} (g)$ 64. The rate constant for the forward reaction is 160, the rate constant for the backward reaction is:

0.4

2.5

6.2

$10.24 \times 10^{3}$

$H_{2} O + CO H_{2} + C O_{2}$ $K_{c} = 64, k_{f} = 160, k_{b} = ?$ We know that, Equilibrium constant $K_{c}$ $= \frac{Rate constant of forward reaction}{Rate constant of backward reaction}$ $K_{c} = \frac{k _{f}}{k _{b}}$ So, $64 = \frac{160}{k _{b}}$ $k_{b} = \frac{160}{64} = 2.5$

Q. The equilibrium constant for the reaction H2​O(g)+CO(g)⇌H2​(g)+CO2​(g) 64. The rate constant for the forward reaction is 160, the rate constant for the backward reaction is:

0.4

2.5

6.2

10.24×103

H2​O+CO​H2​+CO2​ Kc​=64,kf​=160,kb​=? We know that, Equilibrium constant Kc​ =RateconstantofbackwardreactionRateconstantofforwardreaction​ Kc​=kb​kf​​ So, 64=kb​160​ kb​=64160​=2.5

Q. The equilibrium constant for the reaction $H_{2} O (g) + CO (g) ⇌ H_{2} (g) + C O_{2} (g)$ 64. The rate constant for the forward reaction is 160, the rate constant for the backward reaction is:

$10.24 \times 10^{3}$

$H_{2} O + CO H_{2} + C O_{2}$ $K_{c} = 64, k_{f} = 160, k_{b} = ?$ We know that, Equilibrium constant $K_{c}$ $= \frac{Rate constant of forward reaction}{Rate constant of backward reaction}$ $K_{c} = \frac{k _{f}}{k _{b}}$ So, $64 = \frac{160}{k _{b}}$ $k_{b} = \frac{160}{64} = 2.5$