Consider the following reaction, the rate expression of which is given below
A+B → C ; rate =k[A]^(1 / 2)[ B]^(1 / 2)
The reaction is initiated by taking 1 M concentration of A and B each. If the rate constant (k) is 4.6 × 10⁻² s⁻¹, then the time taken for A to become 0.1 M is _________ sec. (nearest integer)

Solution

Chemistry ma’am · 4 steps

A and B start equal and fall one-for-one, so they stay equal and the rate law collapses to first order.

1. Equal at all times. Both start at 1 M, and each reaction event uses one A and one B. So [A] = [B] at every moment.
2. Simplify the rate. Put [B] = [A] into the rate law: k[A]^(1/2)[A]^(1/2) = k[A]. This is a first-order decay of A, which matches the unit of k, s⁻¹.

   t = 2.303/k log[A]₀/[A] = 2.303/(4.6 × 10⁻²)log 1/0.1

3. Put in numbers. log 10 = 1, so t = 2.303/0.046 = 50.07 s.
4. So. The time taken for A to fall to 0.1 M is about 50 s.

The common slip: 22. Using log but leaving out the factor 2.303 gives 1/0.046 ≈ 22 s. Use either ln 10 or 2.303 log 10, not plain log 10.

The idea to keep. When reactants start equal and react 1:1, their concentrations stay equal. Merge them into one term, and the overall order tells you which integrated rate law to use.

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