In the decomposition of `H_(2)O_(2)` at a given temperature T, the energy of activation decreases from `E_(1)` to `E_(2)` by the use of a catalyst. How many times does the rate of the catalysed reaction increase?

Options

A. `(E_(1)-E_(2))`
B. `E_(2)/E_(1)`
C. `(E_(1)-E_(2))/R`
D. `e^((E_(1)-E_(2))//RT)`

Answer

To solve the problem of how many times the rate of the catalyzed reaction increases when the activation energy decreases from \(E_1\) to \(E_2\), we can use the Arrhenius equation, which relates the rate constant \(k\) of a reaction to the activation energy \(E_a\) and temperature \(T\):

\[ k = A e^{-\frac{E_a}{RT}} \]

Where:
- \(k\) is the rate constant,
- \(A\) is the pre-exponential factor (a constant),
- \(E_a\) is the activation energy,
- \(R\) is the universal gas constant,
- \(T\) is the temperature in Kelvin.

### Step 1: Write the rate constants for both reactions
For the uncatalyzed reaction (with activation energy \(E_1\)):
\[ k_1 = A e^{-\frac{E_1}{RT}} \]

For the catalyzed reaction (with activation energy \(E_2\)):
\[ k_2 = A e^{-\frac{E_2}{RT}} \]

### Step 2: Find the ratio of the rate constants
To determine how many times the rate of the reaction increases, we need to find the ratio of \(k_2\) to \(k_1\):
\[ \frac{k_2}{k_1} = \frac{A e^{-\frac{E_2}{RT}}}{A e^{-\frac{E_1}{RT}}} \]

### Step 3: Simplify the expression
Since \(A\) cancels out, we have:
\[ \frac{k_2}{k_1} = e^{-\frac{E_2}{RT}} \cdot e^{\frac{E_1}{RT}} = e^{-\frac{E_2 - E_1}{RT}} \]

### Step 4: Interpret the result
The increase in the rate of the reaction due to the catalyst can be expressed as:
\[ \frac{k_2}{k_1} = e^{-\frac{E_1 - E_2}{RT}} \]
This means that the rate of the catalyzed reaction \(k_2\) is \(e^{-\frac{E_1 - E_2}{RT}}\) times the rate of the uncatalyzed reaction \(k_1\).

### Final Answer
The rate of the catalyzed reaction increases by a factor of \(e^{-\frac{E_1 - E_2}{RT}}\).