1. Tardigrade
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  3. Chemistry
  4. When the temperature of a reaction increases from 27° C to 37° C, the rate increases by 2.5 times, the activation energy in the temperature range is

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Q. When the temperature of a reaction increases from $27^{\circ} C$ to $37^{\circ} C$, the rate increases by $2.5$ times, the activation energy in the temperature range is

Chemical Kinetics

Solution:

$\log 2.5=\frac{ E _{ a }}{2.303 \times R } \times \frac{10}{300 \times 310}$
$0.3979=\frac{ E _{ a }}{2.303 \times 8.314} \times \frac{10}{300 \times 310}$
$E _{ a }=70.77 \times 10^{3} J =70.8\, kJ$