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: Concept For first order reactions, $k = \frac{0.693}{t_{1/2}}$. The Arrhenius equation relates $k$ and $T$: $\log(\frac{k_2}{k_1}) = \log(\frac{t_{1/2(1)}}{t_{1/2(2)}}) = \frac{E_a}{2.303R} (\frac{T_2 - T_1}{T_1 T_2})$.

Step 2: Meaning Let $T_1 = 400\text{ K}$ and $T_2 = 300\text{ K}$. $t_{1/2(1)} = 900$.

Step 3: Analysis $\log(\frac{900}{t_{1/2(2)}}) = 1.3056 \times 10^3 \times (\frac{300 - 400}{400 \times 300}) = 1305.6 \times (\frac{-100}{120000})$.
$\log(\frac{900}{t_{1/2(2)}}) \approx -1.088$.
$\frac{900}{t_{1/2(2)}} = 10^{-1.088} \approx 0.0816$.

Step 4: Conclusion $t_{1/2(2)} = \frac{900}{0.0816} \approx 11025.0\text{ minutes}$. Final Answer: (B)