The rate constant for the forward reaction is at . If moles of A and 100 moles of B are present in a 10 litre vessel at equilibrium, then the rate constant for the backward reaction at this temperature is

Question

The rate constant for the forward reaction $A (g) \to 2 B (g)$ is $1.5 \times 10^{- 3} s^{- 1}$ at $100 K$ . If $10^{- 5}$ moles of A and 100 moles of B are present in a 10 litre vessel at equilibrium, then the rate constant for the backward reaction at this temperature is

Smart Achievers · Accepted Answer

(4)

$A ⇌ 2 B K_{c} = \frac{{[B]}^{2}}{[A]}$

$\frac{K_{f}}{K_{b}} = \frac{{(\frac{100}{10})}^{2}}{(\frac{10^{- 5}}{10})} \Rightarrow \frac{1.5 \times 10^{- 3}}{K_{b}} = \frac{100}{(10^{- 6})} \Rightarrow K_{b} = 1.5 \times 10^{- 11}$

Solution

Q.
The rate constant for the forward reaction $A (g) \to 2 B (g)$ is $1.5 \times 10^{- 3} s^{- 1}$ at $100 K$ . If $10^{- 5}$ moles of A and 100 moles of B are present in a 10 litre vessel at equilibrium, then the rate constant for the backward reaction at this temperature is

1 $1.50 \times 10^{4} L m o l^{- 1} s^{- 1}$

2 $1.5 \times 10^{11} L m o l^{- 1} s^{- 1}$

3 $1.5 \times 10^{10} L m o l^{- 1} s^{- 1}$

4 $1.5 \times 10^{- 11} L m o l^{- 1} s^{- 1}$

Ans.
(4)

$A ⇌ 2 B K_{c} = \frac{{[B]}^{2}}{[A]}$

$\frac{K_{f}}{K_{b}} = \frac{{(\frac{100}{10})}^{2}}{(\frac{10^{- 5}}{10})} \Rightarrow \frac{1.5 \times 10^{- 3}}{K_{b}} = \frac{100}{(10^{- 6})} \Rightarrow K_{b} = 1.5 \times 10^{- 11}$

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Solution

Q. The rate constant for the forward reaction A(g)→2B(g) is 1.5×10-3 s-1 at 100 K. If 10-5 moles of A and 100 moles of B are present in a 10 litre vessel at equilibrium, then the rate constant for the backward reaction at this temperature is

1 1.50×104 L mol-1 s-1

2 1.5×1011 L mol-1 s-1

3 1.5×1010 L mol-1 s-1

4 1.5×10-11 L mol-1 s-1

Ans. (4) A⇌2B Kc=[B]2[A] KfKb=(10010)2(10-510)⇒1.5×10-3Kb=100(10-6)⇒Kb=1.5×10-11