For a 2nd-order reaction with [A]₀ = 0.5 M and k = 0.1 L mol⁻¹ s⁻¹, half-life:
A10 s
B20 s (t₁/₂ = 1/(k[A]₀) = 1/(0.1 × 0.5))
C5 s
D100 s
Answer & Solution
Correct answer: B. 20 s (t₁/₂ = 1/(k[A]₀) = 1/(0.1 × 0.5))
We have t₁/₂ = 1/(k [A]₀).
That is t₁/₂ = 1/(0.1 × 0.5).
So t₁/₂ = 20 s.
Note dependence on [A]₀ (unlike 1st order).
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