One mole of A(g) is heated to 200°C in a 1 litre closed flask, till the following equilibrium is reached. A(g)⇔B(g)°. The rate of forward reaction at equilibrium is 0.02 mole.lit-1 ·min-1. What is the rate (in mole.lit · -1min-1) of the backward reaction at equilibrium ?

⇒kc=kfkb⇒2=0.02kb⇒kb=0.022⇒kb=0.01 mol/lit/minHence the correct option is (B) 0.01

One mole of A(g) is heated to 200°C in a 1 litre closed flask, till the following equilibrium is reached. A(g)⇔B(g)°. The rate of forward reaction at equilibrium is 0.02 mole.lit-1 ·min-1. What is the rate (in mole.lit · -1min-1) of the backward reaction at equilibrium ?

⇒kc=kfkb⇒2=0.02kb⇒kb=0.022⇒kb=0.01 mol/lit/minHence the correct option is (B) 0.01

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One mole of $A_{(g)}$ is heated to $200 ° C$ in a 1 litre closed flask, till the following equilibrium is reached. $A_{(g)} \Leftrightarrow B_{(g) °}$ . The rate of forward reaction at equilibrium is $0.02$ $m o l e . l i t^{- 1}$ $\cdot \min^{- 1}$ . What is the rate (in mole.lit ${}^{- 1}\cdot \min^{- 1}$ ) of the backward reaction at equilibrium ?

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0.02

0.01

0.04

answer is B.

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Detailed Solution

$\Rightarrow k_{c} = \frac{k_{f}}{k_{b}}$

$\Rightarrow 2 = \frac{0.02}{k_{b}} \Rightarrow k_{b} = \frac{0.02}{2}$

$\Rightarrow k_{b} = 0.01 mol / lit / \min$

Hence the correct option is (B) $0.01$

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One mole of A(g) is heated to 200°C in a 1 litre closed flask, till the following equilibrium is reached. A(g)⇔B(g)°. The rate of forward reaction at equilibrium is 0.02 mole.lit-1 ·min-1. What is the rate (in mole.lit · -1min-1) of the backward reaction at equilibrium ?

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