For the reaction A → B the following graph was obtained. The time required (in seconds) for the concentration of A to reduce to 2.5 g L⁻¹ (if the initial concentration of A was 50 g L⁻¹ ) is . (Nearest integer)
Given : log 2=0.3010

Solution

Chemistry ma’am · 4 steps

If the half-life stays the same on the graph, it is first order, and the rest is one log.

1. Read the graph. The concentration falls from 40 to 20 g L⁻¹ between 5 s and 15 s. It then falls from 20 to 10 g L⁻¹ between 15 s and 25 s. Each halving takes 10 s.
2. Identify the order. The half-life does not depend on the starting concentration, so the reaction is first order with t_(1/2) = 10 s. That gives k = 0.693/10 s⁻¹.

   t = 2.303/k log[A]₀/[A] = t_(1/2) × log(50/2.5)/(log 2)

3. Put in numbers. 50/2.5 = 20 and log 20 = 1 + 0.3010 = 1.3010. So t = 10 × 1.3010/0.3010 = 43.2 s.
4. So. The time needed is about 43.2 s, which rounds to 43.

The common slip: 65. Taking the half-life as 15 s, the time at which the curve reaches 20, gives about 65 s. The half-life is the time between two readings where the value halves, and here that is 10 s.

The idea to keep. A constant half-life on a concentration–time graph means first order. Then t = t_(1/2) × log(fold drop) / log 2, and you do not need to find k separately.

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