MHT CET · Chemistry · Chemical Kinetics
A first order reaction has rate constant \(1 \times 10^{-2} \mathrm{~s}^{-1}\). What time will it take for \(20 \mathrm{~g}\) of reactant to reduce to \(5 \mathrm{~g}\) ?
- A \(346.5 \mathrm{~S}\)
- B \(238.6 \mathrm{~S}\)
- C \(138.6 \mathrm{~S}\)
- D \(693.0 \mathrm{~S}\)
Answer & Solution
Correct Answer
(C) \(138.6 \mathrm{~S}\)
Step-by-step Solution
Detailed explanation
(A)
\(\mathrm{k}=1 \times 10^{-2} \mathrm{~s}^{-1}, \quad[\mathrm{~A}]_{0}=20 \mathrm{~g},[\mathrm{~A}]_{\mathrm{t}}=5 \mathrm{~g}\)
For first order reaction,
\(t=\frac{2.303}{k} \log _{10} \frac{[A]_{0}}{[A]_{t}}\)
\(\therefore t=\frac{2.303}{1 \times 10^{-2}} \log _{10} \frac{20}{5}\)
\(\therefore \mathrm{t}=2.303 \times 0.602 \times 10^{2}=138.6 \mathrm{~s}\)
\(\mathrm{k}=1 \times 10^{-2} \mathrm{~s}^{-1}, \quad[\mathrm{~A}]_{0}=20 \mathrm{~g},[\mathrm{~A}]_{\mathrm{t}}=5 \mathrm{~g}\)
For first order reaction,
\(t=\frac{2.303}{k} \log _{10} \frac{[A]_{0}}{[A]_{t}}\)
\(\therefore t=\frac{2.303}{1 \times 10^{-2}} \log _{10} \frac{20}{5}\)
\(\therefore \mathrm{t}=2.303 \times 0.602 \times 10^{2}=138.6 \mathrm{~s}\)
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