Question

the rate of a reaction increases by 4 times when the temperature of the reaction is raised from 340 Kelvin to 360 Kelvin calculate the energy of activation of the reaction given R = 8.314J/K/mol

Answer 1

log (k2/k1)=Ea/2.303R (T2-T1/T1.T2)
here k2/k1=4,R=8.314,T1=340,T2=360
putting these values
log4=Ea/2.303×8.314 (360-340/360×340)
2log2=Ea×20/2.303×8.314×340×360
2×0.3010=Ea×20/2343610.18
0.602=Ea/0.00000853
Ea=0.00000514

Answer 2

Answer: $67.5\times 10^{3}J$

Explanation:

$log\frac{k_2}{k_1}=\frac{E_a}{2.303\times R}\left(\frac{1}{T_1}-\frac{1}{T_2} \right )$

Where,  $k_2$ = rate constant at temperature $T_2$

$k_1$ = rate constant at temperature $T_1$

$E_a$= activation energy

R= gas constant= 8.314J/Kmol

$T_1$= 340K

$T_2$= 360K

if $k_1$=k, thus $k_2$=4k

$log\frac{4k}{k}=\frac{E_a}{2.303\times 8.314J/Kmol}\left(\frac{1}{340}-\frac{1}{360} \right )$

$E_a=67.5\times 10^{3}J$