At `227^(@)C` , the presence of catalyst causes the activation energy of a reaction to decrease by `4.606 K cal. `The rate of the reaction will be increased by `: -`

Options

A. 2 times
B. 10 times
C. 100 times
D. 1000 times

Explanation

To solve the problem, we will use the Arrhenius equation and the relationship between activation energy and the rate constant of a reaction. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the Problem
The problem states that at 227°C, the presence of a catalyst decreases the activation energy (Ea) of a reaction by 4.606 kcal. We need to determine how much the rate of the reaction will increase due to this decrease in activation energy.

### Step 2: Convert Temperature to Kelvin
First, we need to convert the temperature from Celsius to Kelvin:
\[ T(K) = T(°C) + 273 = 227 + 273 = 500 \, K \]

### Step 3: Use the Arrhenius Equation
The Arrhenius equation relates the rate constant (k) to the activation energy (Ea):
\[ k = A e^{-\frac{E_a}{RT}} \]
Where:
- \( k \) = rate constant
- \( A \) = pre-exponential factor (constant)
- \( E_a \) = activation energy
- \( R \) = universal gas constant (approximately 1.987 cal/(mol·K))
- \( T \) = temperature in Kelvin

### Step 4: Set Up the Ratio of Rate Constants
Let \( E_{a1} \) be the original activation energy and \( E_{a2} \) be the new activation energy after the catalyst is added:
\[ E_{a2} = E_{a1} - 4.606 \, \text{kcal} \]
The ratio of the rate constants before and after the addition of the catalyst can be expressed as:
\[ \frac{k_2}{k_1} = \frac{e^{-\frac{E_{a2}}{RT}}}{e^{-\frac{E_{a1}}{RT}}} = e^{-\frac{E_{a2} - E_{a1}}{RT}} = e^{-\frac{-4.606 \times 1000}{(1.987)(500)}} \]

### Step 5: Calculate the Exponent
Now we need to calculate the exponent:
\[ \Delta E_a = -4.606 \, \text{kcal} = -4606 \, \text{cal} \]
Substituting the values:
\[ \frac{k_2}{k_1} = e^{\frac{4606}{(1.987)(500)}} \]
Calculating the denominator:
\[ 1.987 \times 500 = 993.5 \]
Now calculate the exponent:
\[ \frac{4606}{993.5} \approx 4.64 \]
Thus:
\[ \frac{k_2}{k_1} = e^{4.64} \]

### Step 6: Calculate \( e^{4.64} \)
Using a calculator:
\[ e^{4.64} \approx 104.1 \]
This means:
\[ \frac{k_2}{k_1} \approx 104.1 \]

### Step 7: Conclusion
The rate of the reaction will increase by approximately 104 times due to the presence of the catalyst.

### Final Answer
The rate of the reaction will be increased by approximately **100 times**.

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