The half time of a first order reaction is `6.93 xx 10^(-3) min` at `27^(@)C`. At this temeprature, `10^(-8)%` of the reactant molecules are able to cross-over the energy barrier. The pre-exponential factor `A` in the Arrhenius equation is equal to

Options

A. `10^(4) min^(-1)`
B. `10^(8) min^(-1)`
C. `10^(10) min^(-1)`
D. `10^(12) min^(-1)`

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant equations.

### Step 1: Understand the given data
We are given:
- Half-life (t₁/₂) of a first-order reaction = \(6.93 \times 10^{-3}\) min
- Percentage of reactant molecules crossing the energy barrier = \(10^{-8}\%\)

### Step 2: Convert the percentage to a fraction
The percentage of molecules that can cross the energy barrier can be expressed as a fraction:
\[ X = \frac{10^{-8}}{100} = 10^{-10} \]

### Step 3: Use the half-life formula for first-order reactions
For first-order reactions, the half-life is related to the rate constant (k) by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
We can rearrange this to find k:
\[ k = \frac{0.693}{t_{1/2}} \]

### Step 4: Calculate the rate constant (k)
Substituting the value of \(t_{1/2}\):
\[ k = \frac{0.693}{6.93 \times 10^{-3}} \approx 100 \text{ min}^{-1} \]

### Step 5: Use the Arrhenius equation
The Arrhenius equation is given by:
\[ k = A e^{-\frac{E_a}{RT}} \]
Where:
- \(A\) is the pre-exponential factor,
- \(E_a\) is the activation energy,
- \(R\) is the universal gas constant (approximately \(8.314 \, \text{J/mol·K}\)),
- \(T\) is the temperature in Kelvin.

### Step 6: Relate k to A and X
From the information given, we know:
\[ X = e^{-\frac{E_a}{RT}} = 10^{-10} \]
Taking the natural logarithm of both sides:
\[ \ln(X) = -\frac{E_a}{RT} \]
Thus, we can express \(E_a\) as:
\[ E_a = -RT \ln(X) \]

### Step 7: Calculate E_a
Convert the temperature from Celsius to Kelvin:
\[ T = 27 + 273.15 = 300.15 \, K \]
Now substituting the values:
\[ E_a = - (8.314 \, \text{J/mol·K})(300.15 \, K) \ln(10^{-10}) \]
Calculating \(\ln(10^{-10})\):
\[ \ln(10^{-10}) = -10 \ln(10) \approx -10 \times 2.303 = -23.03 \]
Now substituting this back into the equation for \(E_a\):
\[ E_a \approx - (8.314)(300.15)(-23.03) \approx 5740.3 \, J/mol \]

### Step 8: Substitute E_a back into the Arrhenius equation to find A
Now we can substitute \(E_a\) back into the Arrhenius equation to find \(A\):
\[ k = A e^{-\frac{E_a}{RT}} \Rightarrow A = \frac{k}{e^{-\frac{E_a}{RT}}} \]
We already know \(k = 100 \, \text{min}^{-1}\) and \(X = 10^{-10}\):
\[ A = \frac{100}{10^{-10}} = 100 \times 10^{10} = 10^{12} \, \text{min}^{-1} \]

### Final Answer
The pre-exponential factor \(A\) is:
\[ A = 10^{12} \, \text{min}^{-1} \]

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