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

Question

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 (A) 70.8 kJ (B) 7.08 kJ (C) 35.8 kJ (D) 14.85 kJ

Tardigrade Admin · Accepted Answer

log 2.5=( E a /2.303 × R ) × (10/300 × 310) 0.3979=( E a /2.303 × 8.314) × (10/300 × 310) E a =70.77 × 103 J =70.8 kJ

Q. When the temperature of a reaction increases from $2 7^{\circ} C$ to $3 7^{\circ} C$ , the rate increases by $2.5$ times, the activation energy in the temperature range is

70.8 kJ

7.08 kJ

35.8 kJ

14.85 kJ

$lo g 2.5 = \frac{E _{a}}{2.303 \times R} \times \frac{10}{300 \times 310}$
$0.3979 = \frac{E _{a}}{2.303 \times 8.314} \times \frac{10}{300 \times 310}$
$E_{a} = 70.77 \times 1 0^{3} J = 70.8 k J$

Q. 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