Rate of two reaction whose rate constants are`k_(1)& k_(2)` are equal at 300 k such that: So caculate `ln(A_(2))/(A_(1))=?" "Ea_(2)-Ea_(1)=2RT.`

Options

A. a. ln4
B. b. 2
C. c. log 2
D. d. 2-ln2

Explanation

To solve the problem, we will use the Arrhenius equation and the information provided in the question. Let's break down the steps:

### Step 1: Write the Arrhenius equation for both reactions
The Arrhenius equation states:
\[ k = A e^{-\frac{E_a}{RT}} \]

For the two reactions, we can write:
1. For reaction 1:
\[ k_1 = A_1 e^{-\frac{E_{a1}}{RT}} \]
2. For reaction 2:
\[ k_2 = A_2 e^{-\frac{E_{a2}}{RT}} \]

### Step 2: Set the rate constants equal
Given that the rates of the two reactions are equal at 300 K, we have:
\[ k_1 = k_2 \]

### Step 3: Take the natural logarithm of both equations
Taking the natural logarithm of both sides of the equations for \( k_1 \) and \( k_2 \):
1. For reaction 1:
\[ \ln k_1 = \ln A_1 - \frac{E_{a1}}{RT} \]
2. For reaction 2:
\[ \ln k_2 = \ln A_2 - \frac{E_{a2}}{RT} \]

### Step 4: Equate the two logarithmic expressions
Since \( k_1 = k_2 \), we can set the two logarithmic expressions equal to each other:
\[ \ln A_1 - \frac{E_{a1}}{RT} = \ln A_2 - \frac{E_{a2}}{RT} \]

### Step 5: Rearrange the equation
Rearranging the equation gives:
\[ \ln A_1 - \ln A_2 = \frac{E_{a2}}{RT} - \frac{E_{a1}}{RT} \]

### Step 6: Factor out the common terms
This can be simplified to:
\[ \ln \frac{A_1}{A_2} = \frac{E_{a2} - E_{a1}}{RT} \]

### Step 7: Use the given information
We are given that:
\[ E_{a2} - E_{a1} = 2RT \]
Substituting this into the equation gives:
\[ \ln \frac{A_1}{A_2} = \frac{2RT}{RT} \]

### Step 8: Simplify the equation
This simplifies to:
\[ \ln \frac{A_1}{A_2} = 2 \]

### Step 9: Rearranging for \( \ln \frac{A_2}{A_1} \)
Taking the reciprocal of the fraction gives:
\[ \ln \frac{A_2}{A_1} = -2 \]

### Final Answer
Thus, the final answer is:
\[ \ln \frac{A_2}{A_1} = -2 \]

---