Question

Milk turns sour at ${40}^{\circ }$C three times as faster as $0^{\circ }$C. Hence, $E_a$ in the process of turning of milk sour is

• A. $\frac{2.303\times 2\times 313\times 273}{40}log3$
• B. $\frac{2.303\times 2\times 313\times 273}{40}log\left(1/3\right)cal$
• C. $\frac{2.303\times 2\times 40}{273\times 313}log3cal$
• D. $Noneofthese$

Answer 1

The correct option is A $\frac{2.303\times 2\times 313\times 273}{40}log3$

$K_{{40}^{\circ }C}=A{e}^{-Ea/R\times 313}{K}_{0^{\circ }C}=A{e}^{-Ea/R\times 273}3=e^{-Ea/R\times (1/313-1/273)}$

On taking logs on both sides, you get the required answer