Question

Half life of a first order reaction is $900\text{ minute}$ at $400\text{ K}$, find its half life at $300\text{ K}$ ?
$\left( \frac{\text{E}_\text{a}}{2.303\text{R}} = 1.3056 \times 10^3 \right)$

• A. $5512.5\text{ minute}$
• B. $11025.0\text{ minute}$
• C. $8314.3\text{ minute}$
• D. $2303.1\text{ minute}$

Hint:

Temperature down $\rightarrow$ Rate down $\rightarrow$ Half-life up.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For a first-order reaction, the rate constant $k$ is inversely proportional to the half-life $t_{1/2}$. We use the Arrhenius equation to relate rate constants at different temperatures.
Step 2: Key Formula or Approach:
$\log\left(\frac{t_{1/2(1)}}{t_{1/2(2)}}\right) = \frac{\text{E}_\text{a}}{2.303\text{R}} \left[\frac{\text{T}_2 - \text{T}_1}{\text{T}_1\text{T}_2}\right]$
Note: $\frac{k_2}{k_1} = \frac{t_{1/2(1)}}{t_{1/2(2)}}$ where index 2 is for higher temp.
Step 3: Detailed Explanation:
Given:
$\text{T}_2 = 400\text{ K}$, $t_{1/2(2)} = 900\text{ min}$
$\text{T}_1 = 300\text{ K}$, $t_{1/2(1)} = ?$
$\frac{\text{E}_\text{a}}{2.303\text{R}} = 1.3056 \times 10^3$
\[ \log\left(\frac{t_{1/2(1)}}{900}\right) = 1305.6 \times \left[\frac{400 - 300}{400 \times 300}\right] \]
\[ \log\left(\frac{t_{1/2(1)}}{900}\right) = 1305.6 \times \frac{100}{120000} \]
\[ \log\left(\frac{t_{1/2(1)}}{900}\right) = \frac{1305.6}{1200} = 1.088 \]
\[ \frac{t_{1/2(1)}}{900} = \text{antilog}(1.088) \approx 12.25 \]
\[ t_{1/2(1)} = 900 \times 12.25 = 11025\text{ minutes} \]
Step 4: Final Answer:
The half life at $300\text{ K}$ is $11025.0\text{ minute}$.