Question

Part A : A certain first-order reaction (A→products) has a rate constant of 9.30×10−3 s−1 at 45 ∘C. How many minutes does it take for the concentration of the reactant, [A], to drop to 6.25% of the original concentration?

Part B : A certain second-order reaction (B→products) has a rate constant of 1.10×10⁻³M−1⋅s−1 at 27 ∘C and an initial half-life of 278 s . What is the concentration of the reactant B after one half-life?

Answer 1

Part-A: Given rate constant, k = 9.30x10^-3 s^-1 at 45 DegC

Let's consider the initial concentration of A, [A]₀ = 100

Final concentration after time t, [A]_t = 6.25

For first order reaction

t = (1/k) xln [A]₀ / [A]_t = (1 / 9.30x10^-3 s^-1) x ln[100/6.25] = 298 s

=> t = 298 s = 298 s x (1 mimute / 60 s) = 4.97 minutes (answer)

Part-B: Given rate constant, k = 1.10x10^-3 M^-1s^-1 at 27 DegC

Let's consider the initial concentration of B = [B]₀

concentration of B after first half life, [B]_t1/2 =  [B]₀ / 2

t_1/2 = 278 s

For second order reaction involving a single reactant

t = (1/k) x{1/[B]_t1/2 - 1/[B]₀}

=> t_1/2 = 278 s = (1 / 1.10x10^-3 M^-1s^-1) x {2/[B]₀ - 1/[B]₀}

=> 278 s = (1 / 1.10x10^-3 M^-1s^-1) x 1/[B]₀

=> [B]₀ = 1 / 1.10x10^-3 M^-1s^-1x278 s = 3.27 M

Hence concentration after one half-life = /[B]_t1/2 = [B]₀ / 2 = 3.27 M / 2 = 1.635 M (answer)