A first order reaction has the rate constant, k=4.6×10⁻³ s⁻¹. The number of correct statement/s from the following is/are __________
Given : log 3=0.48
A. Reaction completes in 1000 s.
B. The reaction has a half-life of 500 s.
C. The time required for 10% completion is 25 times the time required for 90% completion.
D. The degree of dissociation is equal to (1-e⁻ᵏᵗ)
E. The rate and the rate constant have the same unit.

Solution

Chemistry ma’am · 3 steps

Test every statement against the first-order law. Only the degree of dissociation one holds up.

1. First-order tools. For a first-order reaction, [A] = [A]₀ e⁻ᵏᵗ. The time to reach any extent of reaction comes from the equation below, where x is the amount reacted out of a.

   t = 2.303/k log a/(a-x), 2.303/k = 2.303/(4.6 × 10⁻³) ≈ 500.7 s

2. Log values needed. For 10% completion, log 10/9 = 1 - 2 log 3 = 1 - 0.96 = 0.04. For 90% completion, log 10/1 = 1.

   Statement	Check	Verdict
   A. Completes in 1000 s	[A] falls exponentially and never reaches zero. At 1000 s, kt = 4.6, so about 1% is still left.	False
   B. Half-life is 500 s	t_(1/2) = 0.693/(4.6 × 10⁻³) ≈ 151 s. The 500 s figure is the 90% time.	False
   C. t_(10%) = 25 × t_(90%)	t_(10%) = 500.7 × 0.04 ≈ 20 s and t_(90%) ≈ 501 s, so t_(90%) = 25 t_(10%): the reverse.	False
   D. α = 1 - e⁻ᵏᵗ	α = ([A]₀ - [A])/[A]₀ = 1 - [A]/[A]₀ = 1 - e⁻ᵏᵗ	True
   E. Same units	Rate is in mol L⁻¹ s⁻¹, while k is in s⁻¹.	False

3. So. Only statement D is correct, so the number of correct statements is 1.

The common slip: 2. Accepting C as true gives 2. With log 10/9 = 0.04, the 90% time is 25 times the 10% time, not the other way round.

The idea to keep. A first-order reaction never truly completes. To compare times for two extents of reaction, compare their log a/(a-x) values, and check which way round the ratio goes.

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