The half-life period for first order reaction having activation energy 39.3 k cal `mol^(-1)` at `300^@` C and frequency constant `1.11xx10^(11) s^-1` will be

Options

A. 1 h
B. 1.68 h
C. 1.28 h
D. 1.11 h

Answer

To solve the problem of finding the half-life period for a first-order reaction given the activation energy and frequency constant, we will follow these steps:

### Step 1: Convert Activation Energy to Appropriate Units
The activation energy (Ea) is given as 39.3 kcal/mol. We need to convert this to calories per mole since the gas constant (R) is typically expressed in calories.

\[ 1 \text{ kcal} = 1000 \text{ cal} \]
\[ Ea = 39.3 \text{ kcal/mol} \times 1000 \text{ cal/kcal} = 39300 \text{ cal/mol} \]

### Step 2: Use the Arrhenius Equation to Find the Rate Constant (k)
The Arrhenius equation is given by:

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

Where:
- \( A = 1.11 \times 10^{11} \text{ s}^{-1} \) (frequency factor)
- \( R = 1.987 \text{ cal/(mol K)} \) (gas constant)
- \( T = 300 \text{ °C} = 300 + 273 = 573 \text{ K} \)

Now substituting the values into the equation:

\[ k = 1.11 \times 10^{11} \text{ s}^{-1} \times e^{-\frac{39300 \text{ cal/mol}}{1.987 \text{ cal/(mol K)} \times 573 \text{ K}}} \]

Calculating the exponent:

\[ -\frac{39300}{1.987 \times 573} \approx -34.32 \]

Now calculating \( e^{-34.32} \):

\[ e^{-34.32} \approx 3.1 \times 10^{-16} \]

Thus,

\[ k \approx 1.11 \times 10^{11} \times 3.1 \times 10^{-16} \approx 3.44 \times 10^{-5} \text{ s}^{-1} \]

### Step 3: Calculate the Half-Life (t1/2)
For a first-order reaction, the half-life is given by the formula:

\[ t_{1/2} = \frac{0.693}{k} \]

Substituting the value of \( k \):

\[ t_{1/2} = \frac{0.693}{3.44 \times 10^{-5}} \approx 20100 \text{ seconds} \]

### Step 4: Convert Seconds to Hours
To convert seconds to hours:

\[ \text{Hours} = \frac{20100 \text{ seconds}}{3600 \text{ seconds/hour}} \approx 5.58 \text{ hours} \]

### Final Answer
The half-life period for the first-order reaction is approximately **20100 seconds** or **5.58 hours**.

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