Question

If for a first-order reaction, the value of \( A = 4 \times 10^{13} \, \text{s}^{-1} \) and \( E_a = 98.6 \, \text{kJ mol}^{-1} \), then at what temperature will its half-life be 10 minutes?

• A. 330 K
• B. 300 K
• C. 330.95 K
• D. 311.15 K

Hint:

For first-order reactions, half-life is independent of concentration. The Arrhenius equation relates rate constant to temperature.

Answer 1

The Correct Option is D

To determine the temperature at which a first-order reaction's half-life is 10 minutes, the Arrhenius equation and the first-order half-life formula are employed.

The half-life (\( t_{1/2} \)) of a first-order reaction is defined as:

\[ t_{1/2} = \frac{\ln(2)}{k} \]

Here, \( k \) represents the rate constant. Given a half-life of \( t_{1/2} = 10 \) minutes, this duration is converted to seconds:

\( t_{1/2} = 10 \times 60 = 600 \) s

Substituting this value yields:

\[ 600 = \frac{\ln(2)}{k} \]

Solving for the rate constant \( k \):

\[ k = \frac{\ln(2)}{600} \]

The Arrhenius equation is stated as:

\[ k = A e^{-E_a/(RT)} \]

The parameters are as follows:

• \( A = 4 \times 10^{13} \) s\(^{-1}\)
• \( E_a = 98.6 \) kJ/mol (equivalent to \( 98600 \) J/mol)
• \( R = 8.314 \) J/(mol K) (the universal gas constant)

The calculated \( k \) is substituted into the Arrhenius equation:

\[ \frac{\ln(2)}{600} = 4 \times 10^{13} \cdot e^{-98600/(8.314 \cdot T)} \]

Applying the natural logarithm to both sides results in:

\[ \ln\left(\frac{\ln(2)}{600}\right) = \ln(4 \times 10^{13}) - \frac{98600}{8.314 \cdot T} \]

The temperature \( T \) is then solved for through these steps:

1. Compute \( \ln\left(\frac{\ln(2)}{600}\right) \).
2. Compute \( \ln(4 \times 10^{13}) \).
3. Rearrange the equation to isolate \( T \).

The numerical computations are as follows:

Step	Calculation
1	\(\ln\left(\frac{\ln(2)}{600}\right) \approx -6.3927\)
2	\(\ln(4 \times 10^{13}) \approx 30.5904\)

The equation now stands as:

\[ -6.3927 = 30.5904 - \frac{98600}{8.314 \cdot T} \]

Rearranging to solve for the term with \( T \):

\[ \frac{98600}{8.314 \cdot T} = 30.5904 + 6.3927 \]

\[ \frac{98600}{8.314 \cdot T} = 36.9831 \]

Solving for \( T \):

\[ T = \frac{98600}{8.314 \cdot 36.9831} \]

The final calculation yields \( T \approx 311.15 \) K.

Consequently, the temperature at which the reaction's half-life is 10 minutes is determined to be 311.15 K.