NEET 2024 — Chemistry solution

NEET2024Chemistry

The rate of a reaction quadruples when temperature changes from 27 $^{\circ}$ C to 57 $^{\circ}$ C. Calculate the energy of activation. Given $R = 8.314$ J K $^{- 1}$ mol $^{- 1}$ , $lo g 4 = 0.6021$ .

A
3.80 kJ/mol
B
3804 kJ/mol
C
38.04 kJ/mol
D
380.4 kJ/mol

Solution

$lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{T _{2} - T _{1}}{T _{1} T _{2}})$ .

$0.6021 = \frac{E _{a}}{2.303 \times 8.314} \cdot \frac{30}{300 \times 330}$ .

Solving: $E_{a} \approx 38040$ J/mol $\approx 38.04$ kJ/mol.

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