Question

A sample of milk kept at 25 °C is found to sour 40 times as rapidly as when it is kept at 4 °C. Estimate the activation energy for the souring process.

Answer 1

Answer:

120.575 kJ is the activation energy for the souring process.

Explanation:

The formula for an activation energy is given as:

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant at $25^oC$ = $40k$

$K_2$ = rate constant at $4^oC$ = $k$

$Ea$ = activation energy for the reaction = ?

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $25^oC=273+25=298 K$

$T_2$ = final temperature = $4^oC=273+4=277 K$

Now put all the given values in this formula, we get:l

$\log (\frac{k}{40k})=\frac{Ea}{2.303\times 8.314 J/mol K}[\frac{1}{298K}-\frac{1}{277 K}]$

$E_a=120,575.61J=120.575 kJ$

120.575 kJ is the activation energy for the souring process.