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$^{-1}$]}
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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!
Step 1: Understanding the Question:
This chemical kinetics problem asks us to find the temperature ($T_2$) at which a reaction will have a half-life of $152\text{ s}$, given its rate constant ($k_1$) at $T_1 = 600\text{ K}$ and its activation energy ($E_a$). Step 2: Key Formula or Approach:
1. Identify the reaction order: The unit of the rate constant $k_1$ is $\text{s}^{-1}$, which indicates a first-order reaction.
2. Calculate the required rate constant ($k_2$) at the new temperature ($T_2$) using the first-order half-life relationship:
\[ k_2 = \frac{\ln 2}{t_{1/2}} \]
3. Relate rate constants at different temperatures using the integrated 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 3: Detailed Explanation:
Let's carry out the calculations step-by-step:
• 1. Calculate the rate constant $k_2$ at temperature $T_2$:
- Given $t_{1/2} = 152\text{ s}$ at $T_2$.
\[ k_2 = \frac{\ln 2}{152\text{ s}} \approx \frac{0.69315}{152} \approx 4.56 \times 10^{-3}\text{ s}^{-1} \]
• 2. Calculate the ratio of the rate constants:
- Given $k_1 = 5.0 \times 10^{-5}\text{ s}^{-1}$ at $T_1 = 600\text{ K}$.
\[ \frac{k_2}{k_1} = \frac{4.56 \times 10^{-3}\text{ s}^{-1}}{5.0 \times 10^{-5}\text{ s}^{-1}} = 91.2 \]
- Taking the natural logarithm:
\[ \ln(91.2) \approx 4.513 \]
• 3. Calculate the $E_a / R$ ratio:
- Given $E_a = 191.47\text{ kJ/mol} = 191470\text{ J/mol}$.
- Given $R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}$.
\[ \frac{E_a}{R} = \frac{191470}{8.314} \approx 23030\text{ K} \]
• 4. Solve for $T_2$ using Arrhenius equation:
- Substitute the computed values into the Arrhenius expression:
\[ 4.513 = 23030 \left( \frac{1}{600} - \frac{1}{T_2} \right) \]
- Divide both sides by 23030:
\[ \frac{4.513}{23030} = \frac{1}{600} - \frac{1}{T_2} \]
\[ 0.00019596 \approx 0.00166667 - \frac{1}{T_2} \]
- Rearrange to solve for $1/T_2$:
\[ \frac{1}{T_2} = 0.00166667 - 0.00019596 \approx 0.00147071 \]
- Calculate the final temperature $T_2$:
\[ T_2 = \frac{1}{0.00147071} \approx 679.94\text{ K} \approx 680\text{ K} \]
Step 4: Final Answer:
The temperature at which the half-life becomes 152 s is approximately $680\text{ K}$, which corresponds to option (A).
