A certain first-order reaction (A→products) has a rate constant of 6.30×10^−3 s^−1 at 45 ∘C. ?

[匿名]

2021-03-10 11:16 am

How many minutes does it take for the concentration of the reactant, [A], to drop to 6.25% of the original concentration?

A certain second-order reaction (B→products) has a rate constant of 1.55×10^−3 M−1⋅s−1 at 27 ∘C and an initial half-life of 210 s . What is the concentration of the reactant B after one half-life?

更新1:

For a first-order reaction, the half-life is constant. It depends only on the rate constant k and not on the reactant concentration. It is expressed as t 1/2=0.693/k For a second-order reaction, the half-life depends on the rate constant and the concentration of the reactant and so is expressed as t 1/2=1/k[A]0

回答 (1)

Uncle Michael

2021-03-10 12:13 pm

✔ 最佳答案

1.
Let n be the number of half-lives.
(1/2)ⁿ = 6.25%
(0.5)ⁿ = (0.5)⁴
n = 4

For a first-order reaction: t½ = 0.693/k
Half-life, t½ = 0.693/(6.30 × 10⁻³) s = 110 s

Time taken = 110 × 4 s = 440 s = 7.33 min

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2.
For a second-order reaction: t½ = 1/(k[B]ₒ)
Then, [B]ₒ = 1/(kt½)
Initial concentration, [B]ₒ = 1/(1.55 × 10⁻³ × 210) M = 3.072 M

Concentration after one half-life, [B] = 3.072 × (1/2) M = 1.54 M

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