Solution
Related Formula
Rate = k[A]
[A] = [A]₀ e-kt t1/2 = (ln 2)/(k)Core Logic
For a first order reaction, rate is directly proportional to concentration.
R₁ = k[A]₁₀ = k[A]₀ e-k(10 × 60) R₂ = k[A]₂₀ = k[A]₀ e-k(20 × 60)Step 1: Setting up equations
0.04 = k[A]₀ e-600k (1) 0.03 = k[A]₀ e-1200k (2)Step 2: Solving for k
Dividing equation (1) by (2):
(0.04)/(0.03) = e-600ke-1200k (4)/(3) = e600kTake natural log on both sides:
ln((4)/(3)) = 600k k = (ln(4/3))/(600) s⁻¹Step 3: Calculating half life
t1/2 = (ln 2)/(k) = (ln 2)/((ln(4/3))/(600)) = (600 ln 2)/(ln 4 - ln 3) secondsConvert to minutes by dividing by 60:
t1/2 = (10 ln 2)/(ln 4 - ln 3) minutesSubstitute log values (since ln x = 2.303 x, the 2.303 cancels out):
t1/2 = 10 × ( 2)/( 4 - 3) minutes t1/2 = 10 × (0.3010)/(2(0.3010) - 0.4771) t1/2 = 10 × (0.3010)/(0.6020 - 0.4771) = 10 × (0.3010)/(0.1249) t1/2 = 24.099 ≈ 24 minutesChapter Mix
Class 12 Chemistry: Chemical Kinetics