In the start of summer, a given sample of milk turns sour at room temperature `(27^(@)C)` in `48` hours. In a refrigerator at `2^(@)C`, milk can be stored there times longer before it sours.
The time taken by the milk to sour at `37^(@)C`
Given: Activation energy:: 30146 J/mol
(a) `35.2 hr`
(b)`32.5 hr`
(c) `35.3 hr`
(d) `32.3 hr`

Options

A. `35.2 hr`
B. `32.5 hr`
C. `35.3 hr`
D. `32.3 hr`

Answer

B

Explanation

Now `T_(1) = 300 K, T_(2) = 37^(@)C = 310 K`
`E_(a) = 30146 J mol^(-1)`
`log.(k_(310))/(k_(300)) = (30146 J mol^(-1))/(2.303 xx 8.314) xx (310 - 300)/(300 xx 310) = 0.1693`
`:. (k_(310))/(k_(300)) = "antilog"(0.1693) = 1.477`
Higher the rate constant, faster is the reaction, i.e., lesser is the time taken, hence
`(t_(310))/(t_(300)) = (k_(300))/(k_(310)) = (1)/(1.477)`
`:. t_(310) = t_(300) xx(1)/(1.477) = (48)/(1.477) = 32.5 hr`