It takes 42.0 min for the concentration of a reactant in a first-order reaction to drop from 0.45 m to 0.32 m at 25°c. how long will it take for the reaction to be 90% complete?

Question

Kaya Blackwell

Chemistry

19 February, 06:25

It takes 42.0 min for the concentration of a reactant in a first-order reaction to drop from 0.45 m to 0.32 m at 25°c. how long will it take for the reaction to be 90% complete?

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Lola · Answer 1

Answer is: 284 min.

t₁ = 42 min = 2520 s.

t₂ = ?

c₁ = 0,45 mol/L.

c₂ = 0,32 mol/L.

The integrated first order rate law is:

ln (c₂/c₁) = - k·t.

ln (0,32 mol/L : 0,45 mol/L) = - k·2520 s.

k = 0,34 : 2520 s.

k = 0,000135 1/s.

For 90%: ln (0,045 mol/L : 0,45 mol/L) = - 0,000135 1/s · t.

t = 2,30 : 0,000135 1/s.

t = 17056 s = 284 min.