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 10¹⁵ s⁻¹ and 10¹¹ s⁻¹, respectively. If the value of log $K$ for the reaction at 500 K is 6, the value of | log $k_b$| at 250 K is:

The correct answer is 5.

Given the reversible reaction:

$\text{A(g)}\rightleftharpoons \text{P(g)}$

We are given the following information:

• Pre-exponential factor for the forward reaction, $A_f={10}^{15}{\text{s}}^{-1}$
• Pre-exponential factor for the backward reaction, $A_b={10}^{11}{\text{s}}^{-1}$
• At $T=500\text{K}$, $\text{log}K=6$

From the given image plot of $\text{log}{k}_f$ versus $\frac{1}{T}$, we observe the specific point where:

At $\frac{1}{T}=0.002{\text{K}}^{-1}$ (which corresponds to $T=\frac{1}{0.002}=500\text{K}$), the value of $\text{log}{k}_f=9$.

Step 1: Determine the activation energy of the forward reaction ($E_{a,f}$)

According to the Arrhenius equation for the forward rate constant:

$k_f=A_f·e^{-\frac{E_{a,f}}{RT}}$

Taking base-10 logarithm on both sides:

$\text{log}{k}_f=\text{log}{A}_f-\frac{E_{a,f}}{2.303RT}$

Substituting the values at $T=500\text{K}$ ($\frac{1}{T}=0.002{\text{K}}^{-1}$) where $\text{log}{k}_f=9$ and $A_f={10}^{15}$:

$9=\text{log}\left({10}^{15}\right)-\frac{E_{a,f}}{2.303R}$

$9=15-\frac{E_{a,f}}{2.303R}$

$\frac{E_{a,f}}{2.303R}$

$\frac{E_{a,f}}{2.303R}$

Step 2: Determine the rate constant of the backward reaction ($k_b$) at 500 K

The equilibrium constant $K$ for a reversible reaction is given by:

$K=\frac{k_f}{k_b}$

Taking log on both sides at $T=500\text{K}$:

$\text{log}K=\text{log}{k}_f-\text{log}{k}_b$

Given $\text{log}K=6$ and $\text{log}{k}_f=9$:

$6=9-\text{log}{k}_b$

$\text{log}{k}_b\left(\text{at 500 K}\right)=3$

Step 3: Determine the activation energy of the backward reaction ($E_{a,b}$)

Using the Arrhenius equation for backward reaction at $T=500\text{K}$:

$\text{log}{k}_b=\text{log}{A}_b-\frac{E_{a,b}}{2.303RT}$

Substitute $\text{log}{k}_b=3$, $A_b={10}^{11}$, and $\frac{1}{T}=0.002$:

$3=\text{log}\left({10}^{11}\right)-\frac{E_{a,b}}{2.303R}$

$3=11-\frac{E_{a,b}}{2.303R}$

Step 4: Calculate $\text{log}{k}_b$ at $T=250\text{K}$

At $T=250\text{K}$, we have $\frac{1}{T}=\frac{1}{250}=0.004{\text{K}}^{-1}$.

Substitute these values into the logarithmic form of the Arrhenius equation for the backward reaction:

$\text{log}{k}_b=\text{log}{A}_b-\frac{E_{a,b}}{2.303R}$

$\text{log}{k}_b=11-\left(4000\right)·\left(0.004\right)$

$\text{log}{k}_b=11-16=-5$

Taking the absolute value as requested:

$\left|\text{log}{k}_b\right|=|-5|=5$