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

Calculation of \( \log k_b \) at 500 K 

Given: The equation for the equilibrium constant \( K \) is related to the forward and backward rate constants \( k_f \) and \( k_b \) as:

\(\log K = \log \left( \frac{k_f}{k_b} \right)\)

Thus, we can express \( K \) as:

\(K = \frac{k_f}{k_b}\)

Step 1: Solving for \( \log k_b \)

We are given that \( 6 = \log k_f - \log k_b \), and also that \( \log k_f = 9 \) at 500 K. Therefore, we can substitute these values into the equation:

\(6 = 9 - \log k_b\)

Solving for \( \log k_b \):

\(\log k_b = 3 \quad \text{at} \, 500K\)

Step 2: Arrhenius Equation

The Arrhenius equation gives the rate constant \( k_b \) as:

\(k_b = A_b e^{\frac{-E_{ab}}{RT}}\)

Taking the natural logarithm of both sides:

 

\(\ln k_b = \ln A_b - \frac{E_{ab}}{RT}\)

Multiplying both sides by 2.303 to convert to base 10 logarithms:

 

\(2.303 \log k_b = 2.303 \log A_b - \frac{E_{ab}}{500R}\)

Step 3: Solving for \( E_a \)

Now, we use the equation for the logarithmic difference of the rate constants at two temperatures \( T_1 = 500 K \) and \( T_2 = 250 K \):

\(\frac{E_a}{500R} = 2.303 \left( \log 10^{11} - 3 \right)\)

Simplifying this expression:

 

\(E_a = 2.303 \times R \times 500\)

Step 4: Finding \( \log K_{250K} \)

Next, we calculate the logarithmic ratio of the rate constants at 250 K and 500 K:

\(\ln \left( \frac{k_{250K}}{k_{500K}} \right) = \frac{-2.303 \times R \times 500}{R} \left( \frac{1}{500} - \frac{1}{250} \right)\)

This simplifies to:

 

\(\log K_{250K} - 3 = -8\)

Step 5: Final Calculation

Solving for \( \log K_{250K} \), we get:

 

\(\log K_{250K} = -5\)

Taking the absolute value, we find:

 

\(| \log K_{250K} | = 5\)

Conclusion

The 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:
\[\boxed{5}\]