The plot of log $$k_f$$ versus $$1/T$$ for a reversible reaction A(g) $$\rightleftharpoons$$ 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 6, 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]

Since,
$$\log k = \log A - \frac{E_a}{2.303 R} \cdot \frac{1}{T}$$

For the forward reaction:

Given:

• Pre-exponential factor for the forward reaction, $$A_f = 10^{15}\text{ s}^{-1} \implies \log A_f = 15$$.
• From the given graph, at $$\frac{1}{T} = 0.002\text{ K}^{-1}$$ (which corresponds to $$T = 500\text{ K}$$), the value of $$\log k_f = 9$$.

$$9 = 15 - \frac{E_{a,f}}{2.303 R} \times 0.002$$
$$\frac{E_{a,f}}{2.303 R} \times 0.002 = 15 - 9 = 6$$
$$\frac{E_{a,f}}{2.303 R} = \frac{6}{0.002} = 3000\text{ K}$$

The relationship between the equilibrium constant ($K$) and the forward and backward rate constants ($k_f$ and $$k_b$$) is:

$$K = \frac{k_f}{k_b} \implies \log K = \log k_f - \log k_b$$

At $$T = 500\text{ K}$$:

• $$\log k_f = 9$$ (from the graph)
• $$\log K = 6$$ (given)

Substituting these values:

$$6 = 9 - \log k_b$$
$$\log k_b = 3 \quad (\text{at } 500\text{ K})$$

For the backward reaction:

Given:

• Pre-exponential factor for the backward reaction, $$A_b = 10^{11}\text{ s}^{-1} \implies \log A_b = 11$$.
• At $$T = 500\text{ K}$$, $$\frac{1}{T} = 0.002\text{ K}^{-1}$$ and $$\log k_b = 3$$.

Substituting these values:

$$3 = 11 - \frac{E_{a,b}}{2.303 R} \times 0.002$$
$$\frac{E_{a,b}}{2.303 R} \times 0.002 = 11 - 3 = 8$$
$$\frac{E_{a,b}}{2.303 R} = \frac{8}{0.002} = 4000\text{ K}$$

Now, Calculate $$|\log k_b|$$ at $$250\text{ K}$$

At $$T = 250\text{ K}$$, the value of $$\frac{1}{T}$$ is:

$$\frac{1}{T} = \frac{1}{250} = 0.004\text{ K}^{-1}$$

Using the Arrhenius equation for the backward reaction at $$250\text{ K}$$:

$$\log k_b = 11 - 4000 \times 0.004$$
$$\log k_b = 11 - 16 = -5$$

Taking the absolute value:

$$|\log k_b| = |-5| = 5$$