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Chemical Kinetics — Chemistry STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceChemistryChemical Kinetics3 Marks
Question
A first order reaction takes 40 min for 30% decomposition. Calculate $t_{1 / 2}$.
✓

Answer

The reaction is progressing at 30%
Hence, assuming [R]0 = 1
[R] = 1 - 0.3 = 0.7 and t = 40 minute
Hence, rate constant,
$k=\frac{2.303}{t} \log \frac{[R]_0}{[R]}$
$\begin{array}{l}k=\frac{2.303}{40} \log \frac{1}{0.7} \\ k=0.05757 \log \frac{10}{7} \\ k=0.05757(\log 10-\log 7)\end{array}$
$\begin{array}{l} k =0.05757(1-0.8451) \\ k =0.05757 \times(0.1549)\end{array}$
$\begin{array}{l} k =8.917 \times 10^{-3} \text { minute }^{-1} \\ k =8.92 \times 10^{-3} \text { minute }^{-1}\end{array}$
$t_{1 / 2}=\frac{0.693}{k}=\frac{0.693}{8.92 \times 10^{-3}}$
$\begin{array}{l} t _{1 / 2}=0.07769 \times 10^3 \text { minute } \\ t _{1 / 2}=77.7 \text { minute }\end{array}$

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