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:
$$\boxed{5}$$