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Q.

The plot of $\log k_{f}$ versus $\frac{1}{T}$ for a reversible reaction A (g) $⇌$ P (g) is shown.

Pre-exponential factors for the forward and backward reactions are $10^{15} s^{- 1}$ and $10^{11} s^{- 1}$ , respectively. If the value of log K for the reaction at 500 K is 5, the value of $| \log k_{b} |$ at 250 K is
[K = equilibrium constant of the reaction
$k_{f}$ = rate constant of forward reaction
$k_{b}$ = rate constant of backward reaction]

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Detailed Solution

For reaction $A (g) ⇌ P (g)$
$\log k_{f} = \frac{- E_{f}}{2.303 R T} + \log A_{f}$ [Arrheniusequation for forward reaction]
From plot when, $\frac{1}{T} = 0.002, \log k_{f} = 9 \Rightarrow 9 = \frac{- E_{f}}{2.303 R} (0.002) + \log (A_{f})$
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The plot of log kf versus 1T for a reversible reaction A (g) ⇌P (g) is shown.Pre-exponential factors for the forward and backward reactions are 1015s−1 and 1011s−1, respectively. If the value of log K for the reaction at 500 K is 5, the value of |logkb| at 250 K is[K = equilibrium constant of the reactionkf = rate constant of forward reactionkb = rate constant of backward reaction]