JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
If compound A reacts with \(B\) following first order kinetics with rate constant \(2.011 \times 10^{-3} s ^{-1}\). The time taken by \(A\) (in seconds) to reduce from \(7\,g\) to \(2\,g\) will be \(.........\) (Nearest Integer) \([\log 5=0.698, \log 7=0.845, \log 2=0.301]\)
- A \(620\)
- B \(623\)
- C \(622\)
- D \(625\)
Answer & Solution
Correct Answer
(B) \(623\)
Step-by-step Solution
Detailed explanation
\(A + B \rightarrow P\) \(\begin{array}{ll} t =0 & 7\,g \\ t = t & 2\, g \end{array}\) at constant volume \(t =\frac{2.303}{ K } \log \frac{[ A ]_0}{[ A ]_{ t }}\) \(=\frac{2 \cdot 303}{2 \cdot 011 \times 10^{-3}} \log \frac{7}{2}\)…
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