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

`T_2=400.73K`

Step-by-step 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=(E_a)/(2.303R)*((T_2-T_1)/(T_1T_2))`

Where

`R=Gas constant`

`R =8.3143`

Therefore

`log 4.50=(51.2*10^3)/(2.303*8.31432)*((T_2-365)/(365*T_2))`

`log 4.50=7.328*(T_2-365)/(T_2)`

`0.0892=(T_2-365)/(T_2)`

`0.0892T_2=T_2-365`

`365=T_2-0.0892T_2`

`365=0.91T_2`

`T_2=(365)/(0.91)`

`T_2=400.73K`