For a given reaction R → P, t_(1 / 2) is related to [A]₀ as given in table.
Given: log 2=0.30
Which of the following is true?
A. The order of the reaction is 1 / 2.
B. If [A]₀ is 1 M , then t_(1 / 2) is 200 √10 min
C. The order of the reaction changes to 1 if the concentration of reactant changes from 0.100 M to 0.500 M.
D. t_(1 / 2) is 800 min for [A]₀=1.6 M
Choose the correct answer from the options given below:

Solution

Chemistry ma’am · 3 steps

Concentration falls 4 times but t_(1/2) falls only 2 times: that square root gives you the order.

1. Half-life rule. For a reaction of order n, t_(1/2) ∝ [A]₀¹⁻ⁿ. Take the ratio of the two rows in the table so the constant cancels.

   200/100 = (0.100/0.025)¹⁻ⁿ ⇒ 2 = 4¹⁻ⁿ

2. Find the order. Take logs: 0.30 = (1-n)(0.60), so 1-n = 1/2 and n = 1/2. This means t_(1/2) ∝ √[A]₀, so statement A is true.

   Statement	Check	Verdict
   A	n = 1/2 from the ratio above	true
   B	1 M is 10 times 0.100 M, so t_(1/2) = 200√10 min	true
   C	Order is a fixed property of this reaction, it does not depend on the starting concentration	false
   D	1.6 M is 16 times 0.100 M, so t_(1/2) = 200 × √16 = 800 min	true

3. So. A, B and D are true and C is false, which is option (b).

The idea to keep. For order n, t_(1/2) ∝ [A]₀¹⁻ⁿ. Compare two data rows to get n, then scale t_(1/2) by that power. Changing the starting concentration does not change the order.

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