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Chemical Kinetics
Integrated Rate Equation for a First Order Reaction
3 questions from 3 papers, each with its answer.
- questions
- 3
- papers
- 3
- years asked, 2017–2026
- 3 of 10
- of Chemical Kinetics
- 3.75%
Appears in
Concepts tested
- rate constant of a first order reaction
- time for 99% completion
- time for a first order reaction to reach 10%
All Questions
2022 1 question
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For a first order reaction $A\rightarrow$ Products, initial concentration of A is 0.1 M, which becomes 0.001 M after 5 minutes. Rate constant for the reaction in $min^{-1}$ isFor a first order reaction: $k = \frac{2.303}{t}\log\frac{[A]_0}{[A]_t}$ $k = \frac{2.303}{5}\log\frac{0.1}{0.001} = \frac{2.303}{5} \times 2$ $k = 0.9212\ min^{-1}$
2020 1 question
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The rate constant for a first order reaction is $4.606\times10^{-3}\ s^{-1}$. The time required to reduce 2.0 g of the reactant to 0.2 g is:For a first order reaction: $k = \frac{2.303}{t}\log\frac{A_0}{A}$ $4.606 \times 10^{-3} = \frac{2.303}{t}\log\frac{2}{0.2}$ $t = \frac{2.303}{4.606 \times 10^{-3}} \times \log 10 = \frac{10^3}{2} = 500$ s
2019 1 question
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If the rate constant for a first order reaction is $k$, the time ($t$) required for the completion of 99% of the reaction is given by:For a first order reaction: $k=\frac{2.303}{t}\log\frac{[A]_0}{[A]_t}$ For 99% completion, $\frac{[A]_0}{[A]_t}=\frac{100}{1}$ $k=\frac{2.303}{t}\log10^2=\frac{2.303\times2}{t}$ $t=\frac{4.606}{k}$