JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Chemical Kinetics

Under the same reaction conditions, initial concentration of

1.386 mol dm^{- 3}

of a substance becomes half in 40 seconds and 20 seconds through first order and zero order kinetics, respectively. Ratio

(\frac{k _{1}}{k _{0}})

of the rate constants for first order

(k_{1})

and zero order

(k_{0})

of the reactions is

Answer: (A)

Solution

For first-order kinetics, the rate constant is:

k_{1} = \frac{0.693}{t _{1/2}} = \frac{0.693}{40}

For zero-order kinetics, the rate constant is:

k_{0} = \frac{A _{0}}{2 t _{1/2}} = \frac{1.386}{2 \times 20} = \frac{1.386}{40}

Therefore, the ratio is:

\frac{k _{1}}{k _{0}} = \frac{0.693/40}{1.386/40} = \frac{0.693}{1.386} = 0.5 mol^{- 1} dm^{3}

JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Single correct option