For a certain first order reaction 32% of the reactant is left after 570s. The rate constant of this reaction is _________ × 10⁻³ s⁻¹. (Round off to the Nearest Integer). [Given : log₁₀2 = 0.301, ln10 = 2.303]

Solution

Chemistry ma’am · 4 steps

First order means use the log formula with [A]₀/[A] = 100/32, and the numbers come out exact.

1. Set up the ratio. Take the starting amount as 100. After 570 s, 32 is left, so [A]₀/[A] = 100/32.

   k = 2.303/t log₁₀[A]₀/[A]

2. Find the log. log₁₀ 100/32 = log₁₀ 100 - log₁₀ 2⁵ = 2 - 5 × 0.301 = 0.495.
3. Put in numbers. k = (2.303 × 0.495)/570 = 1.14/570 = 2.0 × 10⁻³ s⁻¹.
4. So. The rate constant is 2 × 10⁻³ s⁻¹, so the answer is 2.

The common slip: 1. Using log₁₀ but dropping the 2.303 factor gives k = 0.495/570 ≈ 0.87 × 10⁻³ s⁻¹, which rounds to 1, not 2.

The idea to keep. For first order, write 32 as 2⁵ and 100 as 10² so a single log 2 value does all the work. The factor 2.303 belongs with log₁₀, never with ln.

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