Question

A certain reaction has an activation energy of 51.02 kJ/mol. At what Kelvin temperature will the reaction proceed 4.50 times faster than it did at 365 K

Answer 1

Answer:

$T_2=400.73K$

Explanation:

From the question we are told that:

Activation energy $E_a= 51.02 kJ/mol.=>51.02*10^3J/mol$

Reaction Ratio $\triangle K=4.50$

Initial Temperature $T_1=365K$

Generally the equation for Final Temperature $T_2$ is mathematically given by

 $log \triangle K=\frac{E_a}{2.303R}*(\frac{T_2-T_1}{T_1T_2})$

Where

 $R=Gas constant$

 $R =8.3143$

Therefore

 $log 4.50=\frac{51.2*10^3}{2.303*8.31432}*(\frac{T_2-365}{365*T_2})$

 $log 4.50=7.328*\frac{T_2-365}{T_2}$

 $0.0892=\frac{T_2-365}{T_2}$

 $0.0892T_2=T_2-365$

 $365=T_2-0.0892T_2$

 $365=0.91T_2$

 $T_2=\frac{365}{0.91}$

 $T_2=400.73K$