Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half-life of 3.33 h at 25°C. After 9 h, the fraction of sucrose remaining is f. The value of log ₁₀( 1/f ) is ________ × 10⁻². (Rounded off to the nearest integer)
[Assume : ln 10 = 2.303, ln 2 = 0.693]

Solution

Chemistry ma’am · 4 steps

For a first-order reaction, ln(1/f) is just kt; convert to log₁₀ at the very end.

1. Rate constant. For first order, k = (ln 2)/t_(1/2) = 0.693/(3.33 h) = 0.2081 h⁻¹.
2. First-order law. The fraction left is f = [A]/[A]₀, so the integrated rate law gives ln(1/f) directly. Dividing by ln 10 turns it into a base-10 log.

   log₁₀ 1/f = kt/(ln 10) = (0.2081 × 9)/2.303

3. Put in numbers. kt = 0.2081 × 9 = 1.873, and 1.873/2.303 = 0.8133.
4. So. log₁₀(1/f) = 0.8133 = 81.33 × 10⁻², which rounds to 81.

The common slip: 187. Stopping at ln(1/f) = 1.87 and not dividing by 2.303 gives 187, not 81.

The idea to keep. In first-order kinetics, ln([A]₀/[A]) = kt with k = 0.693/t_(1/2). If the question asks for log₁₀, divide by 2.303 at the end.

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