The activation energy for a reaction is 9 kcal/mol. The increase in late (XI 'slant when its temperature is increased from 298K to 308 K is `(R = 2 cal K^(-1)mot^(-1), 10^(0.213) = 1.63)`

Options

A. 0.75
B. 0.25
C. 1
D. 0.63

Answer

To solve the problem of how the rate constant changes when the temperature is increased from 298 K to 308 K, we will use the Arrhenius equation. Here’s a step-by-step breakdown of the solution:

### Step 1: Write the Arrhenius Equation
The Arrhenius equation in logarithmic form is given by:
\[ \log k = \log A - \frac{E_A}{2.303RT} \]
where:
- \( k \) is the rate constant,
- \( A \) is the pre-exponential factor,
- \( E_A \) is the activation energy,
- \( R \) is the gas constant,
- \( T \) is the temperature in Kelvin.

### Step 2: Set Up the Equation for Two Temperatures
We want to compare the rate constants at two different temperatures, \( T_1 = 298 \, K \) and \( T_2 = 308 \, K \). Therefore, we can write:
\[ \log k_2 - \log k_1 = -\frac{E_A}{2.303R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]
This can be rearranged to:
\[ \log \frac{k_2}{k_1} = -\frac{E_A}{2.303R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]

### Step 3: Convert Activation Energy to Appropriate Units
The activation energy \( E_A \) is given as 9 kcal/mol. We need to convert this to calories:
\[ E_A = 9 \, \text{kcal/mol} \times 1000 \, \text{cal/kcal} = 9000 \, \text{cal/mol} \]

### Step 4: Substitute Known Values
Now we substitute the values into the equation. The gas constant \( R \) is given as \( 2 \, \text{cal K}^{-1} \text{mol}^{-1} \):
\[ \log \frac{k_2}{k_1} = -\frac{9000}{2.303 \times 2} \left( \frac{1}{308} - \frac{1}{298} \right) \]

### Step 5: Calculate the Temperature Difference
Calculate \( \frac{1}{T_2} - \frac{1}{T_1} \):
\[ \frac{1}{308} - \frac{1}{298} = \frac{298 - 308}{308 \times 298} = \frac{-10}{91784} \approx -0.000109 \]

### Step 6: Calculate the Logarithm
Now substitute this value back into the equation:
\[ \log \frac{k_2}{k_1} = -\frac{9000}{2.303 \times 2} \times (-0.000109) \]
Calculating the right-hand side:
\[ \log \frac{k_2}{k_1} \approx \frac{9000 \times 0.000109}{4.606} \approx 0.2129 \]

### Step 7: Find the Ratio of Rate Constants
Now we can find \( \frac{k_2}{k_1} \):
\[ \frac{k_2}{k_1} = 10^{0.2129} \approx 1.63 \]

### Step 8: Calculate the Increase in Rate Constant
The increase in rate constant can be calculated as:
\[ \text{Increase} = \frac{k_2 - k_1}{k_1} = \frac{1.63 k_1 - k_1}{k_1} = 1.63 - 1 = 0.63 \]

### Step 9: Convert to Percentage
To express this increase as a percentage:
\[ \text{Percentage Increase} = 0.63 \times 100 = 63\% \]

### Final Answer
The increase in the rate constant when the temperature is increased from 298 K to 308 K is approximately 63%.

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