A catalyst lowers the enrgy of activation by `25%`, temperature at which rate of uncatalysed reaction will be equal to that of the catalyst one at `27^(@)C` is `:`

Options

A. `400^(@)C`
B. `127^(@)C`
C. `300^(@)C`
D. `227^(@)C`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between the rate constants and activation energies.
The rate of a reaction is directly proportional to its rate constant (k). According to the Arrhenius equation, the rate constant is given by:
\[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \( A \) = pre-exponential factor (constant for a given reaction)
- \( E_a \) = activation energy
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

### Step 2: Set up the equations for both the uncatalyzed and catalyzed reactions.
Let:
- \( k_1 \) = rate constant for the uncatalyzed reaction
- \( k_2 \) = rate constant for the catalyzed reaction

For the uncatalyzed reaction:
\[ k_1 = A e^{-\frac{E_{a1}}{RT_1}} \]

For the catalyzed reaction, where the activation energy is lowered by 25%:
\[ E_{a2} = E_{a1} - 0.25E_{a1} = 0.75E_{a1} \]
Thus,
\[ k_2 = A e^{-\frac{E_{a2}}{RT_2}} = A e^{-\frac{0.75E_{a1}}{RT_2}} \]

### Step 3: Set the rates equal to each other.
Since the rates are equal:
\[ k_1 = k_2 \]
This leads to:
\[ A e^{-\frac{E_{a1}}{RT_1}} = A e^{-\frac{0.75E_{a1}}{RT_2}} \]

### Step 4: Cancel out the common terms.
Canceling \( A \) from both sides gives:
\[ e^{-\frac{E_{a1}}{RT_1}} = e^{-\frac{0.75E_{a1}}{RT_2}} \]

### Step 5: Equate the exponents.
Taking the natural logarithm of both sides results in:
\[ -\frac{E_{a1}}{RT_1} = -\frac{0.75E_{a1}}{RT_2} \]
Cancelling the negative signs and \( E_{a1} \) (assuming \( E_{a1} \neq 0 \)):
\[ \frac{1}{T_1} = \frac{0.75}{T_2} \]

### Step 6: Solve for \( T_1 \).
Rearranging gives:
\[ T_1 = \frac{T_2}{0.75} \]

### Step 7: Substitute the known temperature.
Given \( T_2 = 27^\circ C = 300 \, K \):
\[ T_1 = \frac{300}{0.75} = 400 \, K \]

### Step 8: Convert \( T_1 \) back to Celsius.
To convert Kelvin to Celsius:
\[ T_1 = 400 \, K - 273 = 127^\circ C \]

### Final Answer:
The temperature at which the rate of the uncatalyzed reaction will be equal to that of the catalyzed one is:
\[ \boxed{127^\circ C} \]

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