Milk turns sour at $40 ° C$ three times faster than at $0 ° C$ . Hence, $E_{a}$ in calories of process of turning of milk sour is:

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$\frac{2.303 \times 2 \times 313 \times 273}{40} \log (\frac{1}{3})$

$\frac{2.303 \times 2 \times 313 \times 273}{40} \log 3$

none of the above

$\frac{2.303 \times 2 \times 40}{273 \times 313} \log 3$

answer is A.

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Detailed Solution

Milk becomes sour three times as quickly at $40 ° C$ as at $0 ° C$ .

Then $\frac{rate at 40^{\circ} C}{rate at 0^{\circ} C} = \frac{k_{40^{\circ} C}}{k_{0^{\circ} C}}$

$E_{a} = \frac{2.303 R T_{1} T_{2}}{(T_{2} - T_{1})} \log (\frac{k_{40^{\circ} C}}{k_{0^{\circ} C}})$

$E_{a} = \frac{2.303 \times 2 \times 313 \times 273 \log (3)}{40}$

Hence, option A is correct.

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