Question

The rate constant of a reaction at 600 K with an activation energy of 191.47 kJ mol$^{-1}$ is 5.0 $\times$ 10$^{-5}$ s$^{-1}$. What is the temperature at which the half-life of the reaction becomes 152 s? [Consider pre-exponential factor and activation energy to be independent of temperature. R = 8.314 J K$^{-1}$mol$^{-1}$]}

• A. 680 K
• B. 640 K
• C. 760 K
• D. 720 K

Hint:

Always ensure units are consistent before calculating:
Activation energy ($E_a$) is often given in $\text{kJ/mol}$ while the gas constant ($R$) is in $\text{J K}^{-1}\text{ mol}^{-1}$.
Always convert $E_a$ to Joules ($1\text{ kJ} = 1000\text{ J}$) before substituting into the equation!

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The unit of the rate constant ($s^{-1}$) indicates a first-order reaction.
We need to use the Arrhenius equation to find the temperature $T_{2}$.
Key Formula or Approach:
1. First-order half-life: \( k = \frac{\ln 2}{t_{1/2}} \).
2. Arrhenius equation: \( \ln\left(\frac{k_{2}}{k_{1}}\right) = \frac{E_{a}}{R} \left( \frac{1}{T_{1}} - \frac{1}{T_{2}} \right) \).

Step 2: Detailed Explanation:
1. Find $k_{2$ at temperature $T_{2}$:}
$t_{1/2} = 152 \text{ s}$.
$k_{2} = \frac{0.693}{152} \approx 4.56 \times 10^{-3} \text{ s}^{-1}$.
2. Ratio of rate constants:
$k_{1} = 5.0 \times 10^{-5} \text{ s}^{-1}$.
$\frac{k_{2}}{k_{1}} = \frac{4.56 \times 10^{-3}}{5.0 \times 10^{-5}} = 91.2$.
$\ln(91.2) \approx 4.513$.
3. Apply Arrhenius equation:
$E_{a} = 191470 \text{ J/mol}$; $R = 8.314$; $T_{1} = 600 \text{ K}$.
\[ 4.513 = \frac{191470}{8.314} \left( \frac{1}{600} - \frac{1}{T_{2}} \right) \]
\[ 4.513 = 23030 \left( 0.001666 - \frac{1}{T_{2}} \right) \]
\[ 0.000196 = 0.001666 - \frac{1}{T_{2}} \implies \frac{1}{T_{2}} \approx 0.00147 \]
\[ T_{2} \approx \frac{1}{0.00147} \approx 680 \text{ K} \]

Step 3: Final Answer:
The required temperature is approximately 680 K.
This matches option (A).