Question

the rate of a reaction at 293 kelvin is 1.7 ×10^5 s^-1.when the temperature is increased by 20 Kelvin ,the rate constant is increased to 2.57×10^6s-1. calculate Ea and A of the reaction.

Answer 1

Answer:

The value of activation energy is 103.556 kJ.

A = $4.89\times 10^{23} s^{-1}$

Explanation:

Rate of the reaction at temperature =$k_1=1.7 \times 10^5 s^{-1}$

$T_1= 293K$

Rate of the reaction at temperature =$k_2=2.57 \times 10^6 s^{-1}$

$T_1= 293K+20K=313 K$

$\log\frac{k_2}{k_1}=\frac{E_a}{2.303\times r}(\frac{T_2-T_1}{T_2T_1})$

$E_a=103,556.6 J=103.556 kJ$

$k=Ae^{\frac{-E_a}{RT}}$

$\log k_1=\log A-\frac{E_a}{2.303RT_1}$

$A = 4.89\times 10^{23} s^{-1]$