JEE Advanced · Chemistry · 18. Chemical Kinetics
Under the same reaction conditions, initial concentration of \(1.386 \mathrm{~mol} \mathrm{dm}^{-3}\) of a substance becomes half in 40 s and 20 s through first-order and zero-order kinetics, respectively. Ratio \(\left(\frac{k_1}{k_0}\right)\) of the rate constants for first order \(\left(k_1\right)\) and zero \(\operatorname{order}\left(k_0\right)\) of the reaction is
- A \(0.5 \mathrm{~mol}^{-1} \mathrm{dm}^3\)
- B \(1.0 \mathrm{~mol} \mathrm{dm}^3\)
- C \(1.5 \mathrm{~mol} \mathrm{dm}^3\)
- D \(2.0 \mathrm{~mol}^{-1} \mathrm{dm}^3\)
Answer & Solution
Correct Answer
(A) \(0.5 \mathrm{~mol}^{-1} \mathrm{dm}^3\)
Step-by-step Solution
Detailed explanation
First order kinetics, \(k_1=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{40} \mathrm{~s}^{-1}\)
Zero order kinetics, \(k_1=\frac{C_0}{2 t_{1 / 2}}=\frac{1.386}{2 \times 20}\)
Hence, \(\frac{k_1}{k_0}=\frac{0.693}{1.386}=0.5\)
Zero order kinetics, \(k_1=\frac{C_0}{2 t_{1 / 2}}=\frac{1.386}{2 \times 20}\)
Hence, \(\frac{k_1}{k_0}=\frac{0.693}{1.386}=0.5\)
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