JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
\([ A ] Reactant \quad \rightarrow\) \([ B ]Product\) If formation of compound \([B]\) follows the first order of kinetics and after 70 minutes the concentration of \([ A ]\) was found to be half of its initial concentration. Then the rate constant of the reaction is \(x \times 10^{-6} s ^{-1}\). The value of \(x\) is \(......\) (Nearest Integer)
- A \(166\)
- B \(165\)
- C \(167\)
- D \(186\)
Answer & Solution
Correct Answer
(B) \(165\)
Step-by-step Solution
Detailed explanation
\(K =\frac{0.693}{ t _{1 / 2}}=\frac{0.693}{70 \times 60}\) \(=\frac{6930}{7 \times 6} \times 10^{-6}\) \(=165 \times 10^{-6}\,s ^{-1}\)
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