The rate constant for a first order reaction is 20 min⁻¹. The time required for the initial concentration of the reactant to reduce to its 1/32 level is _______ × 10⁻² min. (Nearest integer)
(Given : ln 10=2.303 and log 2=0.3010 )

Solution

Chemistry ma’am · 4 steps

One thirty-second is 1/2⁵, so the reactant has gone through exactly five half-lives.

1. Count half-lives. Each half-life halves the concentration. Since 1/32 = (1/2)⁵, the time needed is 5 t_(1/2).

   t = 2.303/k log[A]₀/[A] = 2.303/20 log 32 = (2.303 × 5 × 0.3010)/20

2. Put in numbers. log 32 = 5 × 0.3010 = 1.505. Then t = (2.303 × 1.505)/20 = 3.466/20 = 0.1733 min.
3. Check. With half-lives instead: t_(1/2) = 0.693/20 = 0.03465 min, and 5 × 0.03465 = 0.1733 min. Both routes agree.
4. So. t = 0.1733 min = 17.33 × 10⁻² min, which rounds to 17.

The common slip: 8. Using log 32 without the factor 2.303 gives 1.505/20 = 0.075 min, which is 8. In t = 1/k ln[A]₀/[A] the logarithm is natural, so it needs the 2.303.

The idea to keep. For a first order reaction, whenever the fraction left is 1/2ⁿ, the time is just n half-lives. The starting concentration does not matter.

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