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The rate constant of the first order reaction is 1.386 min–1. Calculate the time required for 80% reactant to decompose?

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Question

The rate constant of the first order reaction is 1.386 min–1. Calculate the time required for 80% reactant to decompose?

Numerical
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Solution

Given: [A]0 = 100%, k = 1.386 min–1

To find: Time required for 80% reactant to decompose

Formula: t = `2.303/"k" log_10  ["A"]_"o"/["A"]_"t"`

Calculations: The reactant is 80% decomposed. Hence, [A]t = 20%.

t = `2.303/"k" log_10  ["A"]_"o"/["A"]_"t"`

Substituting the values,

t = `2.303/(1.386  "min"^-1) log_10  100/20`

= `2.303/(1.386  "min"^-1) xx 0.69897`

= 1.16 min × `(60  "s")/(1  "min")`

= 69.6 s

Time required for reducing concentration of reactant to 20 % is 1.16 min or 69.6 s.

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Chapter 6: Chemical Kinetics - Short answer questions (Type- II)

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SCERT Maharashtra Chemistry [English] 12 Standard HSC
Chapter 6 Chemical Kinetics
Short answer questions (Type- II) | Q 2.2

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