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 ________ s. (Nearest integer).
Given: ln 10=2.3 ; log 2=0.3

Solution

Chemistry ma’am · 4 steps

For first order, divide the two time equations: k cancels and only a ratio of logs is left.

1. First-order time. For a first-order reaction, t = 2.303/k log[A]₀/[A]. Write it for 60% and for 90% decomposition and divide one by the other, so k cancels.

   t₉₀/t₆₀ = (log 100/10)/(log 100/40) = (log 10)/(log 2.5)

2. Find the logs. After 60% decomposition, 40% is left, so the ratio is 100/40 = 2.5. Then log 2.5 = log 10 - log 4 = 1 - 2(0.3) = 0.4. After 90%, 10% is left, and log 10 = 1.
3. Put in numbers. t₉₀/540 = 1/0.4 = 2.5, so t₉₀ = 540 × 2.5 = 1350 s.
4. So. The time for 90% decomposition is 1350 s.

The common slip: 810. Treating time as proportional to the percentage decomposed gives 540 × 90/60 = 810. First-order decay is logarithmic, not linear.

The idea to keep. In first-order problems the log uses the amount left, not the amount used up. Comparing two times? Take their ratio, and k disappears.

Answer from the NTA final answer key. Solution by PrepWiser.