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?

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.

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abhishekprakash2625 abhishekprakash2625 · Answer 2 · 2021-11-22T13:42:37+01:00

A chemical reaction in which the rate of reaction is directly proportional to the concentration of the reacting substance.

The example of the first-order reaction is as follows:-

The hydrolysis of aspirin
The reaction of t-butyl bromide with water

To solve this question, the formula we are using this as follows:-

The integrated first order rate law is:

[tex]IN\frac{C_{2}}{C_{1}} = -k\*\ t[/tex]

[tex]ln(\frac{0.34}{0.42} ) = -k* 2520 s[/tex]

[tex]k = \frac{0.34}{2520}[/tex]

[tex]k = 0,0001351/s.[/tex]

For 90% :[tex]ln(\frac{0.045}{0.45} ) = -0,000135 1/s[/tex]

[tex]t= \frac{2.30}{0.000135} .[/tex]

t =284 mins

Hence, the time required to complete 90% of the reaction is 284 mins.

For more information, refer to the link:-

https://brainly.com/question/13179326