Milk turns sour at 40 ∘ C three times as faster as 0 ∘ C. Hence, E a in the process of turning of milk sour is Options: A . &nbsp 2.303 × 2 × 313 × 273 40 l o g 3 B . &nbsp 2.303 × 2 × 313 × 273 40 l o g ( 1 / 3 ) c a l C . &nbsp 2.303 × 2 × 40 273 × 313 l o g 3 c a l D . &nbsp N o n e o f t h e s e

Answer: Option A : A K 40 ∘ C = A e − E a / R × 313 K 0 ∘ C = A e − E a / R × 273 3 = e − E a / R × ( 1 / 313 − 1 / 273 ) On taking logs on both sides, you get the required answer

Milk turns sour at 40∘C three times as faster as 0∘C. Hence, Ea

Discussion Forum : Chemical Kinetics

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

Options:

A .

\frac{2.303 \times 2 \times 313 \times 273}{40} l o g 3

B .

\frac{2.303 \times 2 \times 313 \times 273}{40} l o g (1 / 3) c a l

C .

\frac{2.303 \times 2 \times 40}{273 \times 313} l o g 3 c a l

D .

N o n e o f t h e s e

Answer: Option A
:
A

K_{40^{\circ} C} = A e^{- E a / R \times 313} K_{0^{\circ} C} = A e^{- E a / R \times 273} 3 = e^{- E a / R \times (1 / 313 - 1 / 273)}

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

Was this answer helpful ?

Next Question

Discussion Forum : Chemical Kinetics

Submit Your Solution hear: