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Firdavs [7]
3 years ago
10

What is the value of the rate constant for a second order reaction if the reactant conceretation drop sform 0.657 M to 0.0981 M

in 17.0 s?
Chemistry
1 answer:
Ainat [17]3 years ago
6 0

Answer:

The rate constant for a second order reaction is

k = 0.51 dm-3 s-1.

Explanation:

In a regular second-order reaction the rate equation is given by v = k[A][B], if the reactant B concentration is constant then v = k[A][B] = k'[A], where k' the pseudo–first-order rate constant = k[B].

Also 1/|A| = Kt + 1/|Ao|

But Ao = 0.657 M and

A = 0.0981 M also

t = 17.0 s

Therefore

1/| 0.0981 M| =K × 17.0 s + 1/| 0.657 M|

→ 10.19/M = 17K + 1.52/M

10.19/M - 1.52/M = 8.67/M = 17K

K = 8.67M/17s = 0.51 dm-3 s-1.

k = 0.51 dm-3 s-1.

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