JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Chemical Kinetics

An organic compound undergoes first order decomposition. If the time taken for the 60% decomposition is 540 s, then the time required for

90%

decomposition will be is ____ s. (Nearest integer).

Given :

ln 10 = 2.3; lo g 2 = 0.3

Answer: 1350

Solution

For a first-order change the time to reach a given extent depends only on the fraction left, through

k t = ln (a_{0} / a)

. When

60%

has decomposed, two-fifths remains, so

k (540) = ln \frac{1}{0.4} = ln 2.5.

When

90%

has decomposed, one-tenth remains, so

k t = ln \frac{1}{0.1} = ln 10.

Dividing the second relation by the first,

t = 540 \times \frac{ln 10}{ln 2.5} .

With

ln 2.5 = ln 10 - ln 4 = 2.3 - 2 (2.3 \times 0.3) = 0.92

, the time for

90%

decomposition is

t = 540 \times \frac{2.3}{0.92} = 1350 s .

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Integer answer