Consider the two different first order reactions given below
A+B → C (Reaction 1) ; P → Q (Reaction 2)
The ratio of the half life of Reaction 1 : Reaction 2 is 5: 2 If t₁ and t₂ represent the time taken to complete 2 / 3ʳᵈ and 4 / 5ᵗʰ of Reaction 1 and Reaction 2 , respectively, then the value of the ratio t₁: t₂ is _________ × 10⁻¹ (nearest integer). [Given : log ₁₀(3)=0.477 and log ₁₀(5)=0.699]

Solution

Chemistry ma’am · 4 steps

For first order, the time for any fraction scales with half-life, so you only need a ratio of two logs.

1. Time for any fraction. For a first order reaction, k = 0.693/t_(1/2), and the time to fall from [A]₀ to [A] is t = 2.303/k log[A]₀/[A]. So t is proportional to t_(1/2) × log[A]₀/[A]. Reaction 1 is first order overall as given, even though it has two reactants.
2. Fraction left. When 2/3 of Reaction 1 is complete, 1/3 is left, so [A]₀/[A] = 3. When 4/5 of Reaction 2 is complete, 1/5 is left, so [P]₀/[P] = 5.

   t₁/t₂ = (t_(1/2))₁/(t_(1/2))₂×(log 3)/(log 5) = 5/2×0.477/0.699

3. Put in numbers. 0.477/0.699 = 0.6824, and 2.5 × 0.6824 = 1.706.
4. So. t₁ : t₂ = 1.706 = 17.06 × 10⁻¹, which rounds to 17.

The common slip: 3. Taking k₁ : k₂ = 5 : 2 instead of t_(1/2) ratio flips the factor to 2/5 and gives 2.73 × 10⁻¹, which rounds to 3, not 17.

The idea to keep. In first order kinetics, t = t_(1/2)/0.693 × 2.303 log[A]₀/[A]. Use the fraction that is left, not the fraction that has reacted, and a longer half-life means a longer time.

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