IIT JEE past question Kinetics 3

The rate of first order reaction is 0.04 mol L^-1 s^-1 at 10 minutes and 0.03 mol L^-1 s^-1 at 20 minutes after initiation.

Find the half life of the reaction.

5 Answers

n1/n2 = 4/3

n1 = n₀e^-Î»10

n2 = n₀e^-Î»20

dividing we get

4/3 = e^10Î»

10Î» = ln (4/3)

t_1/2 = 10 ln 2/ln(4/3)

something like this [7]

tapan see carefully it has to be more than 10 minutes!

tapan is it 2 half lives?!! :O

Got my mist.....

I had taken rate as the CONcentration [16] [17]

24min perhaps [7] oh i am messing up with the final calculation
give me a break !!

ya now 24 looks more close to the answer

√ √ || {} [] () → ~ ^ x x x → → ln log lg lim lim cos sin tan cot arcsin arccos arctan

∫ ∫ ∫ ∫ ∫ ∬ ∬ ∬ ∬ ∬ ∭ ∭ ∭ ∭ ∭ ⨌ ⨌ ⨌ ⨌ ⨌

∑ ∑ ∑ ∑ ∑ ∮ ∮ ∮ ∮ ∮ ∏ ∏ ∏ ∏ ∏ ∐ ∐ ∐ ∐ ∐

+ - × ÷ ± ∓ = ∪ ∩ ≠ ∼ ≈ ∅ ∀ ∃ ∈ ∋ ∝ ≤ ≥ ≃ ≅ ≡ < > ≪ ≫ ⊂ ∉ ⊃ ⊆ ⊇ ⇒ ⇐ ⇔ ⇒ ⇐ ⇔ ⇑ ⇓ ⇕ → ← ↔ → ← ↔ ↑ ↓ ↕ ↖ ↗ ↘ ↙ ↪ ↩ ⇀ ↼ ⇁ ↽

° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω