For first order reaction rate constant at 27°C and t°C is 1.5 × 10³ and 4.5 × 10³ respectively. If the activation energy of reaction is 60 kJ, then find temperature t. (R = 8.3 Jmol^–1 K^–1)

The correct answer is 41 (which corresponds to 41°C).

To find the temperature $t$, we use the Arrhenius equation in its logarithmic form, which relates the rate constants of a reaction at two different temperatures to its activation energy:

$\log \left(\frac{k_2}{k_1}\right)=\frac{E_a}{2.303R}(\frac{1}{T_1}-\frac{1}{T_2})$

Let us identify the given values from the problem statement:
- Initial rate constant, $k_1=1.5\times {10}^3{\text{s}}^{-1}$
- Initial temperature, $T_1=27°\text{C}=27+273=300\text{K}$
- Rate constant at temperature $t$, $k_2=4.5\times {10}^3{\text{s}}^{-1}$
- Activation energy, $E_a=60\text{kJ/mol}=60,000\text{J/mol}$
- Gas constant, $R=8.3{\text{J mol}}^{-1}{\text{K}}^{-1}$
- Final temperature, $T_2=(t+273)\text{K}$

First, calculate the ratio of the rate constants:
$\frac{k_2}{k_1}=\frac{4.5\times {10}^3}{1.5\times {10}^3}=3$

Taking the logarithm (base 10) of this ratio:
$\log \left(3\right)\approx 0.4771$

Substitute these values into the Arrhenius equation:

$0.4771=\frac{60,000}{2.303\times 8.3}(\frac{1}{300}-\frac{1}{T_2})$

Simplify the term before the parenthesis:
$2.303\times 8.3\approx 19.115$
$\frac{60,000}{19.115}\approx 3138.9$

Now, solve for the temperature terms:

$\frac{1}{300}-\frac{1}{T_2}=\frac{0.4771}{3138.9}\approx 1.52\times {10}^{-4}{\text{K}}^{-1}$

Convert $\frac{1}{300}$ to a decimal representation:
$\frac{1}{300}\approx 3.333\times {10}^{-3}{\text{K}}^{-1}=33.33\times {10}^{-4}{\text{K}}^{-1}$

Now, compute $\frac{1}{T_2}$:

$\frac{1}{T_2}=33.33\times {10}^{-4}-1.52\times {10}^{-4}=31.81\times {10}^{-4}{\text{K}}^{-1}$

Taking the reciprocal to find $T_2$:

$T_2=\frac{1}{3.181\times {10}^{-3}}\approx 314.36\text{K}$

Finally, convert the temperature back to Celsius:
$t=T_2-273=314.36-273\approx 41.36°\text{C}$

Rounding to the nearest whole integer, we get:
$t\approx 41°\text{C}$