In the decomposition of `H_2O_2` at 300 K, the energy of activation was found to be `18 kcal//"mol"` white it decreases to `6 kcal//"mol"` when the decomposition was carried out in the presence of a catalyst at 300 K. The number of times the catalysed reaction faster than uncatalysed one is (Give you answer by multiplying `10^7`)

Explanation

To solve the problem, we will use the Arrhenius equation, which relates the rate constant (k) of a reaction to its activation energy (E_a) and temperature (T).

### Step-by-Step Solution:

1. **Identify the Activation Energies**:
- For the uncatalyzed reaction, the activation energy \( E_u = 18 \, \text{kcal/mol} = 18000 \, \text{cal/mol} \).
- For the catalyzed reaction, the activation energy \( E_c = 6 \, \text{kcal/mol} = 6000 \, \text{cal/mol} \).

2. **Write the Arrhenius Equation**:
The Arrhenius equation is given by:
\[ k = A \cdot e^{-\frac{E_a}{RT}} \]
where:
- \( k \) = rate constant
- \( A \) = pre-exponential factor (constant)
- \( E_a \) = activation energy
- \( R \) = gas constant (2 cal/(mol·K))
- \( T \) = temperature in Kelvin (300 K)

3. **Calculate the Rate Constants**:
- For the uncatalyzed reaction:
\[ k_u = A \cdot e^{-\frac{E_u}{RT}} = A \cdot e^{-\frac{18000}{2 \cdot 300}} \]
- For the catalyzed reaction:
\[ k_c = A \cdot e^{-\frac{E_c}{RT}} = A \cdot e^{-\frac{6000}{2 \cdot 300}} \]

4. **Take the Ratio of Rate Constants**:
The ratio of the rate constants for the catalyzed and uncatalyzed reactions is:
\[ \frac{k_c}{k_u} = \frac{A \cdot e^{-\frac{6000}{600}}}{A \cdot e^{-\frac{18000}{600}}} = e^{-\frac{6000}{600} + \frac{18000}{600}} = e^{\frac{12000}{600}} = e^{20} \]

5. **Calculate \( e^{20} \)**:
Using a calculator or approximation:
\[ e^{20} \approx 4.85 \times 10^8 \]

6. **Multiply by \( 10^7 \)**:
The problem asks for the number of times the catalyzed reaction is faster than the uncatalyzed reaction, multiplied by \( 10^7 \):
\[ 4.85 \times 10^8 \times 10^7 = 4.85 \times 10^{15} \]

### Final Answer:
The number of times the catalyzed reaction is faster than the uncatalyzed reaction, after multiplying by \( 10^7 \), is:
\[ \boxed{4.85 \times 10^{15}} \]