Question

For first order reaction, rate constant at $27^{\circ} C$ and $t^{\circ} C$ are $1.5 \times 10^3$ and $4.5 \times 10^3$ respectively. If activation energy of the reaction is $60 \text{ kJmol}^{-1}$, then ($R = 8.3 \text{ Jmol}^{-1}K^{-1}$) $\ln 3 = 1.1$
Find temperature 't' (in $^{\circ} C$).

Answer 1

This problem uses the logarithmic form of the Arrhenius equation to calculate the change in temperature required to triple the rate of a first-order reaction.
1. Ratio of rate constants:
$\frac{k_2}{k_1} = \frac{4.5 \times 10^3}{1.5 \times 10^3} = 3$

2. Logarithmic calculation:
$\ln 3 = 1.1$ (as provided in the problem).

3. Substituting into the Arrhenius formula:
$\ln 3 = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)$
$1.1 = \frac{60 \times 1000}{8.3} \left(\frac{1}{300} - \frac{1}{T_2}\right)$

4. Finding $T_2$:
$\frac{1.1 \times 8.3}{60000} = \frac{1}{300} - \frac{1}{T_2}$
$0.00015216 = 0.003333 - \frac{1}{T_2}$
$\frac{1}{T_2} = 0.003333 - 0.000152 = 0.003181$
$T_2 \approx 314.36 \text{ K}$

5. Final Answer:
$t = 314.36 - 273 = 41.36^{\circ} C$
The answer is approximately 41.