The rate constant for the decompoistion of a certain reaction is described by the equation:
`log k(s^(-1)) = 14 - (1.25 xx 10^(4) K)/(T)`
What is the effect on the rate of reaction at `127^(@)C`, if in the presence of catalyst, energy of activation is lowered by `10 kJ mol^(-1)`?

Options

A. `5` times
B. `10` times
C. `7` times
D. `20` times

Explanation

To solve the problem step by step, we will use the Arrhenius equation and the information provided about the change in activation energy due to the presence of a catalyst.

### Step 1: Understand the Arrhenius Equation
The Arrhenius equation is given by:
\[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \( k \) is the rate constant,
- \( A \) is the pre-exponential factor,
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant (8.314 J/mol·K),
- \( T \) is the temperature in Kelvin.

### Step 2: Convert Temperature to Kelvin
The temperature given is \( 127^\circ C \). We need to convert this to Kelvin:
\[ T(K) = 127 + 273.15 = 400.15 \, K \approx 400 \, K \]

### Step 3: Define Activation Energies
Let:
- \( E_a \) = activation energy without catalyst,
- \( E_{ac} \) = activation energy with catalyst.

According to the problem, the catalyst lowers the activation energy by \( 10 \, kJ/mol \):
\[ E_{ac} = E_a - 10 \, kJ/mol \]
Converting \( 10 \, kJ/mol \) to J/mol gives:
\[ E_{ac} = E_a - 10,000 \, J/mol \]

### Step 4: Write the Rate Constants
The rate constants without and with the catalyst can be expressed as:
\[ k_1 = A e^{-\frac{E_a}{RT}} \]
\[ k_2 = A e^{-\frac{E_{ac}}{RT}} = A e^{-\frac{E_a - 10,000}{RT}} \]

### Step 5: Find the Ratio of Rate Constants
To find the ratio of the rate constants \( \frac{k_2}{k_1} \):
\[ \frac{k_2}{k_1} = \frac{A e^{-\frac{E_{ac}}{RT}}}{A e^{-\frac{E_a}{RT}}} = e^{-\frac{E_{ac}}{RT} + \frac{E_a}{RT}} = e^{\frac{10,000}{RT}} \]

### Step 6: Substitute Values
Substituting \( R = 8.314 \, J/mol·K \) and \( T = 400 \, K \):
\[ \frac{10,000}{RT} = \frac{10,000}{8.314 \times 400} \approx \frac{10,000}{3325.6} \approx 3.006 \]
Thus,
\[ \frac{k_2}{k_1} = e^{3.006} \approx 20.2 \]

### Step 7: Conclusion
The presence of the catalyst increases the rate of reaction by a factor of approximately 20. Therefore, the new rate with the catalyst will be about 20 times the original rate.

### Final Answer
The rate of reaction in the presence of the catalyst is approximately 20 times that without the catalyst.

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