Question bank/JEE Main/Chemistry

JEE MainChemistryModerateChemical Kinetics

Compound A and B react to form C and D in a reaction that was
found to be second-order over all and second-order in A. The rate
constant -at 30ºC is 0.622 L mol-1 min-1. What is the half-life of A
when 4.10 × 10-2 M of A is mixed with excess B?

Show answer and solution
A + B → C + D
Second order with respect to A
then,
t1/2=1Ka0⇒t12=10.622×4.10×10−2=39.2mint_{1/2} = \frac{1}{{Ka}_{0}} \Rightarrow t_{12} = \frac{1}{0.622 \times 4.10 \times 10^{- 2}} = 39.2min

Source: Grass-G / CH-15 Chemistry Paper 11 Nov..docx

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