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 find the temperature at which the half-life of a first-order reaction is 10 minutes, we use the Arrhenius equation and the half-life formula for first-order reactions.

The half-life (\( t_{1/2} \)) for a first-order reaction is given by:

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

where \( k \) is the rate constant. For \( t_{1/2} = 10 \) minutes, we convert this to seconds:

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

Thus:

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

Solving for \( k \):

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

The Arrhenius equation is:

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

Where:

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

Substitute \( k \) into the Arrhenius equation:

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

Take the natural logarithm on both sides:

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

Solve for \( T \):

1. Calculate \( \ln\left(\frac{\ln(2)}{600}\right) \).
2. Calculate \( \ln(4 \times 10^{13}) \).
3. Rearrange to solve for \( T \).

Performing these calculations gives:

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

Solve:

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

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

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

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

Calculating yields \( T \approx 311.15 \) K.

Thus, the temperature at which the half-life of the reaction is 10 minutes is 311.15 K.

Answer 2

Step 1: Understanding the Arrhenius Equation
The Arrhenius equation is: \[ \log k = \log A - \frac{E_a}{2.303RT} \] Step 2: Finding the Rate Constant \( k \)
For a first-order reaction, the half-life is given by: \[ k = \frac{0.693}{t_{1/2}} \] Substituting \( t_{1/2} = 10 \) minutes \( = 600 \) s: \[ k = \frac{0.693}{600} = 1.1 \times 10^{-3} { s}^{-1} \] Step 3: Substituting into the Arrhenius Equation
\[ \log (1.1 \times 10^{-3}) = \log (4 \times 10^{13}) - \frac{98.6 \times 10^3}{2.303 \times 8.314 \times T} \] Step 4: Solving for \( T \)
\[ T = 311.15 { K} \] Thus, the correct answer is (D).