Given : `AhArrB+C, DeltaH=-"10 kcals"`, the energy of activation of backward reaction is `"15 kcal mol"^(-1)`. If the energy of activation of forward reaction in the presence of a catalyst is `"3 kcal mol"^(-1)` the catalyst will increase the rate of reaction at 300 K by the number of times equal to

Options

A. `e^(3.33)`
B. `e^(4.21)`
C. `e^(-2.7)`
D. `e^(2.303)`

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Understand the Given Information
We are given:
- Reaction: A → B + C
- ΔH = -10 kcal (indicating an exothermic reaction)
- Activation energy of the backward reaction (EAB) = 15 kcal/mol
- Activation energy of the forward reaction in the presence of a catalyst (EAF') = 3 kcal/mol

### Step 2: Calculate the Activation Energy of the Forward Reaction (EAF)
Using the relationship between the activation energies and the enthalpy change (ΔH):
\[ EAF - EAB = \Delta H \]

Substituting the known values:
\[ EAF - 15 \text{ kcal/mol} = -10 \text{ kcal/mol} \]

Now, solve for EAF:
\[ EAF = -10 + 15 \]
\[ EAF = 5 \text{ kcal/mol} \]

### Step 3: Use the Arrhenius Equation to Find the Increase in Rate
The formula that relates the rate constants with and without the catalyst is:
\[ \frac{K_C}{K} = e^{\frac{E_A - E_{A'}}{RT}} \]

Where:
- \( K_C \) = rate constant with catalyst
- \( K \) = rate constant without catalyst
- \( E_A \) = activation energy without catalyst (EAF = 5 kcal/mol)
- \( E_{A'} \) = activation energy with catalyst (EAF' = 3 kcal/mol)
- \( R \) = gas constant = 1.987 cal/(mol·K) = 0.001987 kcal/(mol·K)
- \( T \) = temperature = 300 K

### Step 4: Substitute Values into the Equation
Now we substitute the values into the equation:
\[ \frac{K_C}{K} = e^{\frac{5 \text{ kcal/mol} - 3 \text{ kcal/mol}}{(0.001987 \text{ kcal/(mol·K)})(300 \text{ K})}} \]

Calculating the exponent:
\[ \frac{K_C}{K} = e^{\frac{2 \text{ kcal/mol}}{0.5961 \text{ kcal/mol}}} \]
\[ \frac{K_C}{K} = e^{3.354} \]

### Step 5: Calculate the Final Value
Now we can calculate \( e^{3.354} \):
Using a calculator or mathematical software:
\[ e^{3.354} \approx 28.7 \]

### Conclusion
The catalyst will increase the rate of reaction at 300 K by approximately **28.7 times**.

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