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

The Arrhenius Equation describes the relationship between the rate constant of a chemical reaction and its temperature. For two different temperatures, the integrated form is used.
1. Identify the given parameters:
Initial temperature, $T_1 = 27 + 273 = 300 \text{ K}$
Initial rate constant, $k_1 = 1.5 \times 10^3$
Final rate constant, $k_2 = 4.5 \times 10^3$
Activation Energy, $E_a = 60 \text{ kJ/mol} = 60 \times 10^3 \text{ J/mol}$
Universal gas constant, $R = 8.3 \text{ J/mol}\cdot K$

2. Apply the Arrhenius Equation:
$\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left[\frac{T_2 - T_1}{T_1 T_2}\right]$

3. Substitute the values:
$\ln\left(\frac{4.5 \times 10^3}{1.5 \times 10^3}\right) = \frac{60 \times 10^3}{8.3} \left[\frac{T_2 - 300}{300 \cdot T_2}\right]$
$\ln 3 = \frac{60000}{8.3} \left[\frac{T_2 - 300}{300 \cdot T_2}\right]$

4. Solve for $T_2$:
Given $\ln 3 = 1.1$
$1.1 = \frac{200}{8.3} \left[\frac{T_2 - 300}{T_2}\right]$
$1.1 \times 8.3 = 200 \left(1 - \frac{300}{T_2}\right)$
$9.13 = 200 - \frac{60000}{T_2}$
$\frac{60000}{T_2} = 200 - 9.13 = 190.87$
$T_2 = \frac{60000}{190.87} \approx 314.35 \text{ K}$

5. Convert to Celsius:
$t = 314.35 - 273 = 41.35^{\circ} C$
Rounding to the nearest integer as per standard practice, we get 41.