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, we need to determine how much a catalyst increases the rate of the reaction based on the given activation energies. Here’s a step-by-step solution:

### Step 1: Understand the Given Data
- The enthalpy change (ΔH) for the reaction \( Ah + Arr \rightarrow B + C \) is \(-10 \text{ kcal}\).
- The activation energy of the backward reaction (from products to reactants) is \( E_{AB} = 15 \text{ kcal/mol} \).
- The activation energy of the forward reaction in the presence of a catalyst is \( E_{AF} = 3 \text{ kcal/mol} \).

### Step 2: Calculate the Activation Energy of the Forward Reaction Without Catalyst
Using the relationship between the activation energies and the enthalpy change:
\[ E_{AF} + \Delta H = E_{AB} \]
Substituting the known values:
\[ E_{AF} + (-10 \text{ kcal}) = 15 \text{ kcal} \]
Rearranging gives:
\[ E_{AF} = 15 \text{ kcal} + 10 \text{ kcal} = 25 \text{ kcal/mol} \]

### Step 3: Use the Arrhenius Equation to Relate the Rate Constants
The Arrhenius equation is given by:
\[ k = A e^{-\frac{E_a}{RT}} \]
Where:
- \( k \) = rate constant
- \( A \) = pre-exponential factor
- \( E_a \) = activation energy
- \( R \) = universal gas constant (approximately \( 1.987 \text{ cal/(mol K)} \))
- \( T \) = temperature in Kelvin

### Step 4: Calculate the Rate Constants
1. **Without Catalyst**:
\[ k = A e^{-\frac{E_{AF}}{RT}} = A e^{-\frac{25 \text{ kcal/mol}}{(1.987 \text{ cal/(mol K)})(300 \text{ K})}} \]
Converting kcal to cal:
\[ 25 \text{ kcal/mol} = 25000 \text{ cal/mol} \]
Thus,
\[ k = A e^{-\frac{25000}{596.1}} \approx A e^{-41.9} \]

2. **With Catalyst**:
\[ k' = A e^{-\frac{E_{AF}}{RT}} = A e^{-\frac{3 \text{ kcal/mol}}{(1.987 \text{ cal/(mol K)})(300 \text{ K})}} \]
Converting kcal to cal:
\[ 3 \text{ kcal/mol} = 3000 \text{ cal/mol} \]
Thus,
\[ k' = A e^{-\frac{3000}{596.1}} \approx A e^{-5.03} \]

### Step 5: Calculate the Ratio of Rate Constants
Now, we can find the ratio of the rate constants:
\[ \frac{k'}{k} = \frac{A e^{-5.03}}{A e^{-41.9}} = e^{-5.03 + 41.9} = e^{36.87} \]

### Step 6: Calculate the Increase in Rate
The increase in the rate of reaction due to the catalyst is given by:
\[ \text{Increase in rate} = e^{36.87} \]

### Step 7: Approximate the Value
Using a calculator or approximation:
\[ e^{36.87} \approx 3.2 \times 10^{16} \]

### Final Answer
Thus, the catalyst will increase the rate of reaction at 300 K by approximately \( 3.2 \times 10^{16} \) times.

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