Chemistry - Chemical Kinetics Question with Solution | TestHub

Using the Arrhenius equation:

$\log \left(\frac{k_2}{k_1}\right)=\frac{E_a}{2.303R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$

Let $T_2=400$ K and $T_1=300$ K. Substituting the given values:

$\log \left(\frac{k_{400}}{k_{300}}\right)=1.3056\times 10^3\times \left(\frac{400-300}{400\times 300}\right)$

$\log \left(\frac{k_{400}}{k_{300}}\right)=1.3056\times 10^3\times \frac{100}{120000}=\frac{1305.6}{1200}=1.088$

Taking antilog on both sides:

$\frac{k_{400}}{k_{300}}=\text{antilog}(1.088)\approx 12.25$

For a first order reaction, half-life is inversely proportional to the rate constant, $t_{1/2}=\frac{0.693}{k}$.

$\frac{(t_{1/2}{)}_{300}}{(t_{1/2}{)}_{400}}=\frac{k_{400}}{k_{300}}=12.25$

$(t_{1/2}{)}_{300}=12.25\times 900=11025.0$ minute

Answer: $11025.0$ minute