A catalyst lowers the activation energy of a reaction form `20 kJ mol^(-1)` to 10 `kJ mol^(-1)`. The temperature at which the uncatalyzed reaction will have the same rate as that of the catalyzed at `27^(@)C` is

Options

A. `-123^(@)C`
B. `327^(@)C`
C. `32.7^(@)C`
D. `+23^(@)C`

Explanation

To solve the problem, we need to find the temperature at which the uncatalyzed reaction (with activation energy \(E_A = 20 \, \text{kJ mol}^{-1}\)) will have the same rate as the catalyzed reaction (with activation energy \(E_{A_C} = 10 \, \text{kJ mol}^{-1}\)) at a given temperature of \(T_{A_C} = 27^\circ C\).

### Step-by-Step Solution:

1. **Convert the given temperature from Celsius to Kelvin**:
The temperature at which the catalyzed reaction occurs is given as \(27^\circ C\). To convert this to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
Therefore,
\[ T_{A_C} = 27 + 273 = 300 \, K \]

2. **Set up the ratio of activation energies and temperatures**:
According to the Arrhenius equation, if the rates of two reactions are the same, the ratio of their activation energies is equal to the ratio of their temperatures:
\[ \frac{E_A}{E_{A_C}} = \frac{T_A}{T_{A_C}} \]
Here, \(E_A = 20 \, \text{kJ mol}^{-1}\) and \(E_{A_C} = 10 \, \text{kJ mol}^{-1}\).

3. **Substitute the known values into the equation**:
Substituting the values we have:
\[ \frac{20}{10} = \frac{T_A}{300} \]
Simplifying the left side gives:
\[ 2 = \frac{T_A}{300} \]

4. **Solve for \(T_A\)**:
To find \(T_A\), we multiply both sides by 300:
\[ T_A = 2 \times 300 = 600 \, K \]

5. **Convert the temperature back to Celsius**:
Finally, we convert \(T_A\) from Kelvin back to Celsius:
\[ T_A(°C) = T_A(K) - 273 = 600 - 273 = 327^\circ C \]

### Final Answer:
The temperature at which the uncatalyzed reaction will have the same rate as that of the catalyzed reaction at \(27^\circ C\) is \(327^\circ C\).

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