JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
For a certain first order reaction \(32 \%\) of the reactant is left after \(570 \,s\). The rate constant of this reaction is ........... \(\times 10^{-3} s ^{-1}\). (Round off to the Nearest Integer). \(\left[\right.\) Given \(\left.: \log _{10} 2=0.301, \ln 10=2.303\right]\)
- A \(2\)
- B \(4\)
- C \(6\)
- D \(8\)
Answer & Solution
Correct Answer
(A) \(2\)
Step-by-step Solution
Detailed explanation
For \(1^{\text {st }}\) order reaction, \(K =\frac{2.303}{ t } \cdot \log \frac{\left[ A _{0}\right]}{\left[ A _{ t }\right]}=\frac{2.303}{570 sec } \cdot \log \left(\frac{100}{32}\right)\) \(=1.999 \times 10^{-3} sec ^{-1} \approx 2 \times 10^{-3} sec ^{-1}\)
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