IIT-JEE 2008 · Paper 1 · Chemistry Q51

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ChemistrySingle correct+3 / -1Question 51 of 69

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 $\left(\dfrac{k_1}{k_0}\right)$ of the rate constants for first order $(k_1)$ and zero order $(k_0)$ of the reactions is

0.5 mol⁻¹ dm³

1.0 mol dm⁻³

1.5 mol dm⁻³

2.0 mol⁻¹ dm³

Answer and worked solution

Official answer: A

First order. $t_{1/2}=\dfrac{\ln2}{k_1}=40$ s, so $k_1=\dfrac{0.693}{40}$ s⁻¹.

Zero order. $t_{1/2}=\dfrac{[A]_0}{2k_0}=20$ s, so $k_0=\dfrac{1.386}{40}=\dfrac{0.693}{20}$ mol dm⁻³ s⁻¹.

Ratio. $\dfrac{k_1}{k_0}=\dfrac{20}{40}=0.5$ mol⁻¹ dm³.

(A)

Topics:Half-life

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IIT-JEE 2008 · Paper 1, Chemistry question 51