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Get Answers to all your Questions
Q.89) For the reaction $\mathrm{A}(\mathrm{g}) \rightleftharpoons 2 \mathrm{~B}(\mathrm{~g})$, the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500 , at 1000 K . [Given : $\mathrm{R}=0.0831{\mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1}\left(\mathrm{~K}^{-1} \text { ] }\right] ~}$ $\mathrm{K}_{\mathrm{p}}$ for the reaction at 1000 K is
A) 0.021
B) 83.1
C) $2.077 \times 10^5$
D) 0.033
Answers (1)
Q.89) We are given the reaction:
$$
\mathrm{A}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{~B}(\mathrm{~g})
$$
And told that:
- Rate constant of backward reaction is 2500 times the forward rate constant:
$$
k_{\text {backward }}=2500 \cdot k_{\text {forward }}
$$
- Temperature $T=1000 \mathrm{~K}$
- Gas constant $R=0.0831 \mathrm{~L}$ atm $\mathrm{mol}^{-1} \mathrm{~K}^{-1}$
We are asked to find $K_p$ for the reaction.
Step 1: Use relationship between rate constants and equilibrium constant
For a reversible reaction, the equilibrium constant in terms of rate constants is:
$$
K_c=\frac{k_{\text {forward }}}{k_{\text {backward }}}=\frac{1}{2500}
$$
Step 2: Convert $K_c$ to $K_p$
For gas-phase reactions, the relation is:
$$
K_p=K_c(R T)^{\Delta n}
$$
Here:
- $\Delta n=2-1=1$
- $R=0.0831 \mathrm{~L}$ atm ctibu $\mathrm{mol}^{-1} \mathrm{~K}^{-1}$
- $T=1000 \mathrm{~K}$
- $K_c=\frac{1}{2500}=0.0004$
So:
$$
K_p=0.0004 \times(0.0831 \times 1000)^1=0.0004 \times 83.1=0.03324
$$
Hence, the correct answer is option (4).
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