[A], →, [B]; Reactant, Product
If formation of compound [B] follows the first order of kinetics and after 70 minutes the concentration of [A] was found to be half of its initial concentration. Then the rate constant of the reaction is x × 10⁻⁶ s⁻¹. The value of x is ______________. (Nearest Integer)

Solution

Chemistry ma’am · 3 steps

Half the reactant is gone in 70 minutes, so that is the half-life; convert it to seconds first.

1. Spot the half-life. The concentration of [A] falls to half its starting value in 70 minutes. By definition, that time is the half-life, so t_(1/2) = 70 min.
2. Match the units. The answer is asked in s⁻¹, so change the half-life to seconds: t_(1/2) = 70 × 60 = 4200 s.

   k = 0.693/t_(1/2) = 0.693/(4200 s) = 1.65 × 10⁻⁴ s⁻¹

3. So. k = 165 × 10⁻⁶ s⁻¹, which gives x = 165.

The common slip: 9900. Keeping the time in minutes gives k = 0.0099 min⁻¹, which reads as 9900 once written as x × 10⁻⁶. Convert 70 minutes to 4200 s first.

The idea to keep. For a first-order reaction, t_(1/2) = 0.693/k and does not depend on the starting concentration. Always check the time unit the answer wants before you divide.

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