Question

A given sample of milk turns sour at room temperature $({{27}^{\circ }}C)$ in 5 hours. In a refrigerator at $-{{3}^{\circ }}C$, it can be stored 10 times longer. The energy of activation for the souring of milk is:
(A)- $2.303\times 10R\,kJ.\,mo{{l}^{-1}}$
(B)- $2.303\times 5R\,kJ.\,mo{{l}^{-1}}$
(C)- $2.303\times 3R\,kJ.\,mo{{l}^{-1}}$
(D)- $2.303\times 2.7R\,kJ.\,mo{{l}^{-1}}$

Answer 1

In the souring of milk, the rate of reaction is related to the activation energy from the Arrhenius equation. At different temperatures and rate constants, the activation energy is calculated using the Arrhenius theory.

Complete step by step solution:
In the given that the sample of milk kept at room temperature, ${{T}_{1}}={{27}^{\circ }}C=27+273=300K$ does not get sour for time, ${{t}_{1}}=5\,hours$. Then, its rate constant, ${{k}_{1}}$ from the First order rate law, will be:
${{k}_{1}}=\dfrac{2.303}{{{t}_{1}}}log\left( \dfrac{a}{a-x} \right)=\dfrac{2.303}{5}log\left( \dfrac{a}{a-x} \right)$ ------- (a)
This milk when kept in the refrigerator at temperature, ${{T}_{2}}=-{{3}^{\circ }}C=-3+273=270K$, it can be stored for 10 times longer, that is, ${{t}_{2}}=10\,{{t}_{1}}\,=50\,hours$, without getting sour than at the room temperature. Then, its rate constant, ${{k}_{2}}$ will be:
${{k}_{2}}=\dfrac{2.303}{{{t}_{2}}}log\left( \dfrac{a}{a-x} \right)=\dfrac{2.303}{50}log\left( \dfrac{a}{a-x} \right)$ ------- (b)
Now, in order to calculate the energy of activation for the souring of milk, we will use the Arrhenius equation that relates the activation energy, at two different temperatures and rate constants of the reaction. It is given as follows:
$\log \dfrac{{{k}_{2}}}{{{k}_{1}}}=\dfrac{{{E}_{a}}}{2.303R}\left[ \dfrac{{{T}_{2}}-{{T}_{1}}}{{{T}_{2}}{{T}_{1}}} \right]$ -------- (c)
So, from equation (a) and (b), we will have,
$\dfrac{{{k}_{2}}}{{{k}_{1}}}=\dfrac{\dfrac{2.303}{50}log\left( \dfrac{a}{a-x} \right)}{\dfrac{2.303}{5}log\left( \dfrac{a}{a-x} \right)}=\dfrac{5}{50}={{10}^{-1}}$ --------- (d)
Substituting this value of equation (d) in equation (c), we will get,
$\log {{10}^{-1}}=\dfrac{{{E}_{a}}}{2.303R}\left[ \dfrac{{{T}_{2}}-{{T}_{1}}}{{{T}_{2}}{{T}_{1}}} \right]$
$\log {{10}^{-1}}=\dfrac{{{E}_{a}}}{2.303\times 8.314}\left[ \dfrac{270-{{300}_{1}}}{270\times 300} \right]$
$\log {{10}^{-1}}=\dfrac{{{E}_{a}}}{2.303\,R}\left[ \dfrac{-1}{2700} \right]$
${{E}_{a}}=2.303\,R\times 2700\,J\,mo{{l}^{-1}}$ (since $\log {{10}^{-1}}=-\log 10=(-1)$ )
or, ${{E}_{a}}=2.303\,R\times 2.7kJ\,mo{{l}^{-1}}$

Therefore, the energy of activation for souring of milk is option (D)- $2.303\times 2.7R\,kJ.\,mo{{l}^{-1}}$.

Note: We see that with the decrease in temperature, the rate of reaction slows down, that is, the souring of milk is prolonged by 10 times than when kept in the refrigerator. Thus, causing an increase in the activation energy.
So, with the help of the Arrhenius equation we understand the rate of reaction and also the effect of temperature.