1. For a reaction, the fraction (or number) of molecules having energy greater than the threshold energy is proportional to the Boltzmann factor:

$N \propto e^{-E_a/(RT)}$

2. Therefore, if the number of such molecules increases five fold when temperature changes from $T_1$ to $T_2$, then

$\frac{N_2}{N_1} = \exp\left[-\frac{E_a}{RT_2} + \frac{E_a}{RT_1}\right] = 5$

So,

$\ln 5 = E_a\left(\frac{1}{RT_1} - \frac{1}{RT_2}\right)$

or

$\ln 5 = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

3. Convert temperatures to kelvin:

$T_1 = 27^\circ C = 300\,K$ $T_2 = 42^\circ C = 315\,K$

4. Substitute the values:

$1.6094 = \frac{E_a}{8.314}\left(\frac{1}{300} - \frac{1}{315}\right)$

Now,

$\frac{1}{300} - \frac{1}{315} = \frac{315-300}{300\times 315} = \frac{15}{94500} = \frac{1}{6300}$

Hence,

$1.6094 = \frac{E_a}{8.314}\times \frac{1}{6300}$

$E_a = 1.6094 \times 8.314 \times 6300$

5. Calculate:

$8.314 \times 6300 = 52378.2$

$E_a = 1.6094 \times 52378.2 = 84297.47208\,\text{J mol}^{-1}$

Therefore,

$E_a \approx 84297.47\,\text{J mol}^{-1}$

6. Comparison with stored correct answer:

Stored answer: $84297.47$ to $84297.48$

Our calculated answer $84297.47\,\text{J mol}^{-1}$ matches the stored correct answer.