Consider the following reaction
A+B → C
The time taken for A to become 1 / 4ᵗʰ of its initial concentration is twice the time taken to become 1 / 2 of the same. Also, when the change of concentration of B is plotted against time, the resulting graph gives a straight line with a negative slope and a positive intercept on the concentration axis.
The overall order of the reaction is ________.

Solution

Chemistry ma’am · 3 steps

Read each clue on its own: one tells you the order in A, the other the order in B, then add them.

1. Order in A. Going from [A]₀ to [A]₀/2 takes one half-life, and going on to [A]₀/4 takes a second one. If the time to reach 1/4 is exactly twice the time to reach 1/2, both half-lives are equal. A half-life that does not depend on concentration is the mark of a first order reaction, so the order in A is 1.
2. Order in B. A plot of [B] itself against time is a straight line with a negative slope and an intercept of [B]₀. That is the zero order law, so the order in B is 0.

   [B] = [B]₀ - kt

3. So. Rate = k[A]¹[B]⁰, so the overall order is 1 + 0 = 1.

The common slip: 2. Taking the straight line for B as first order gives 2, not 1. A straight line in [B] itself, not in ln[B], means zero order.

The idea to keep. Constant half-life means first order. A straight line for concentration against time means zero order, while first order needs ln[conc] against time. Overall order is the sum.

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