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# Can you please help me with this chemistry question?

Consider a certain type of reaction has a rate constant of 2.95 × 10–2 min–1. Calculate the time required for the sample to decay to one-fourth of its initial value.

a) 2.95 min

b) 0.0590 min

c) 23.5 min

d) 29.4 min

e) 47.0 min

### 2 Answers

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- ChemTeamLv 72 months ago
Integrated form of the first order rate law:

ln [A] = –kt + ln [A]_o

ln 0.25 = –(2.95 × 10–2 min–1) (t) + ln 1

-1.3865 = –(2.95 × 10–2 min–1) (t)

t = -1.3865 / –2.95 × 10–2 min–1

t = 47.0 min

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- Roger the MoleLv 72 months ago
t½ = (ln 2) / k = 0.693147 / (2.95 × 10^–2 min^–1) = 23.5 min

It takes two half-lives to reach one-fourth of its initial value.

2 × 23.5 min = 47.0 min

So answer e)

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