For the given first order reaction
A → B
the half life of the reaction is 0.3010 min. The ratio of the initial concentration of reactant to the concentration of reactant at time 2.0 min will be equal to ___________. (Nearest integer)

Solution

Chemistry ma’am · 4 steps

That half-life of 0.3010 min is log 2 in disguise, so k comes out as exactly 2.303 min⁻¹.

1. Find k. For a first order reaction, k = 0.693/t_(1/2). Since 0.693 = 2.303 × 0.3010, this gives k = (2.303 × 0.3010)/0.3010 = 2.303 min⁻¹.

   k = 2.303/t log[A]₀/[A]

2. Put in numbers. With t = 2.0 min: 2.303 = 2.303/2 log[A]₀/[A]. The 2.303 cancels on both sides, leaving log[A]₀/[A] = 2.
3. Cross-check. You can also count half-lives: n = 2.0/0.3010, and the ratio is 2ⁿ. Then log 2ⁿ = n × 0.3010 = 2.0, the same result.
4. So. [A]₀/[A] = 10² = 100.

The idea to keep. For first order kinetics, the concentration ratio is 2^(t/t_(1/2)). When the numbers look like log 2 = 0.3010, take the log of the ratio and the answer drops out as a clean power of 10.

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