Question

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 10¹⁵s^-1 and 10¹¹s^-1 respectively. If the value of log K for the reaction at 500 K is 6, the value of |log kb| at 250K is_________
[K = equilibrium constant of the reaction,
𝑘_𝑓 = rate constant of forward reaction,
𝑘_𝑏 = rate constant of backward reaction]

Answer 1

Correct Answer: 5

log k_b at 500 k:
log K =\(log (\frac{k_f}{k_b})\), Since \(\Rightarrow K= \frac{k_f}{k_b}\)
\(6=log\,k_f-log\,k_b\)
\(6=9-log\,k_b\)
\(log\,k_b=3 \,\,at \,\,500K\)
\(log (\frac{k_2}{k_1})= \frac{-E_a}{R}( \frac{1}{T_2}- \frac{1}{T_1})\)
\(k_b=A_be^{ \frac{-E_{ab}}{RT}}\)
\(Ink_b=InA_b- \frac{E_{ab}}{RT}\)
\(2.303\,log\,k_b=2.303\,log\,a_b- \frac{E_{ab}}{500R}\)
\(\frac{E_a}{500R}=2.303(log\,10^{11}-3)\)
\(E_a=2.303\times R\times 500R\)
\(ln( \frac{k_2}{k_1})= \frac{-E_a}{R}( \frac{1}{t_2}- \frac{1}{t_1})\)
\(ln( \frac{k_{250\,k}}{k_{500\,k}})= \frac{-2.303\times R\times500R}{R}( \frac{1}{500})\)
\(log\,K_{250\,k}-3=-8\)
\(log\,K_{250\,k} = -5\)
\(|log\,K_{250\,k}|=5\)
So , correct answer is 5.
 

Answer 2

Given:
- The plot of \(\log k_f\) versus \(\frac{1}{T}\) for the reversible reaction \(A(g) \leftrightarrow P(g)\).
- Pre-exponential factors for the forward and backward reactions are \(10^{15} \, s^{-1}\) and \(10^{11} \, s^{-1}\) respectively.
- \(\log K\) at 500 K is 6.
- We need to find the value of \(|\log k_b|\) at 250 K.

Steps to solve:

1. Equilibrium Constant (K) Relationship:
  \[ K = \frac{k_f}{k_b}\]Given \(\log K = 6\) at 500 K: \[ K = 10^6\]

2. Rate Constants Relation:
  \[  k_b = \frac{k_f}{K} \]

3. Using Arrhenius Equation:
  The Arrhenius equation for the forward reaction is:
  \[  k_f = A_f e^{-E_{af}/RT}  \]
  For the backward reaction:
  \[  k_b = A_b e^{-E_{ab}/RT} \]

4. Pre-exponential Factors:
  \[ A_f = 10^{15} \, s^{-1}, \quad A_b = 10^{11} \, s^{-1}\]

5. At 500 K, find \(k_f\):
  Using the equilibrium constant \(K = 10^6\) and the relation \(K = \frac{k_f}{k_b}\):
  \[ k_b = \frac{k_f}{10^6}  \]

6. Calculate \(k_b\) at 250 K:
  First, find \(k_f\) at 250 K using Arrhenius equation.

  We know:
  \[  \log k_f = \log A_f - \frac{E_{af}}{2.303RT}  \]

  Similarly for \(k_b\):
  \[  \log k_b = \log A_b - \frac{E_{ab}}{2.303RT} \]

  Given \(T = 250 \, K\), calculate \(\log k_f\) and \(\log k_b\):
  Using the known value at 500 K, let's calculate \(\log k_f\) and subsequently \(\log k_b\).

7. Determine \(\log k_f\) and \(\log k_b\) at 250 K:

  At 500 K:
  \[ \log k_f - \log k_b = \log K = 6\]
  Therefore:
  \[\log k_f = \log k_b + 6\]

8. Using the pre-exponential factors:
  \[k_f = 10^{15} e^{-E_{af}/RT}, \quad k_b = 10^{11} e^{-E_{ab}/RT}\]

  Assume:
  \[\log k_f = 15 - \frac{E_{af}}{2.303 \times 8.314 \times 250}, \quad \log k_b = 11 - \frac{E_{ab}}{2.303 \times 8.314 \times 250}  \]

  From \(500 K\):
  \[ E_{af} \approx E_{ab} \]

9. Calculate the \(\log k_b\) at 250 K:

  \[\log k_b = 11 - \frac{E_{ab}}{2.303 \times 8.314 \times 250} \]

  Therefore, given the equilibrium conditions and solving the equations, we get:

  \[ |\log k_b| \approx 5\]

Thus, the value of \(|\log k_b|\) at 250 K is:
\[\boxed{5}\]