If the rate constant of a reaction is 0.03 s–1, howmuch time does it take for 7.2 mol L–1 concentrationof the reactant to get reduced to 0.9 mol L–1?(Given : log 2 = 0.301)

Answer

First-Order Kinetics Calculation

For a first-order reaction, the time t required for the concentration of a reactant to change from [R]₀ to [R] is given by the integrated rate law:

t = (2.303 / k) × log([R]₀ / [R])

Given:

• Rate constant, k = 0.03 s⁻¹
• Initial concentration, [R]₀ = 7.2 mol·L⁻¹
• Final concentration, [R] = 0.9 mol·L⁻¹

Step 1: Calculate the ratio

[R]₀ / [R] = 7.2 / 0.9 = 8

Step 2: Calculate the logarithm

log(8) = log(2³) = 3 × log(2) = 3 × 0.301 ≈ 0.903

Step 3: Plug into the equation

t = (2.303 / 0.03) × 0.903 ≈ 76.77 × 0.903 ≈ 69.3 s

Conclusion: The time required for the concentration to drop from 7.2 mol·L⁻¹ to 0.9 mol·L⁻¹ is approximately 69.3 seconds.