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BITSAT 2011
Exam
BITSAT
Year
2011
Subject
Chemistry
Chapter
Chemical Kinetics

BITSAT 2011 · Chemistry · Chemical Kinetics

During the kinetic study of the reaction, 2 A+B arrow C+D ,...

During the kinetic study of the reaction, 2A+B→C+D2 \mathrm{ A}+\mathrm{B} \rightarrow \mathrm{C}+\mathrm{D}, following results were obtained:
Run{[A]/ mol L^{-1}}{[ B]/ mol L^{-1}}Initial rate of
formation of
D/ mol L^{-1} min^{-1}
I0.10.16.0 \times10^{-1}
II0.30.27.2 \times10^{-1}
III0.30.428.8 \times10^{-1}
IV0.40.124.0 \times10^{-1}
Based on the above data which one of the following is correct?

Options

  1. Arate =k[A]2[B]=k[\mathrm{ A}]^{2}[\mathrm{ B}]
  2. Brate =k[A][B]=k[\mathrm{ A}][\mathrm{B}]
  3. Crate =k[A]2[B]2=k[\mathrm{ A}]^{2}[\mathrm{ B}]^{2}
  4. Drate =k[A][B]2=k[\mathrm{ A}][\mathrm{B}]^{2}

Correct answer

D. rate =k[A][B]2=k[\mathrm{ A}][\mathrm{B}]^{2}

Step-by-step solution

In case of (II) and (III) Keeping concentration of [A] constant, when the concentration of [B][\mathrm{B}] is doubled, the rate quadruples. Hence it is second order with respect to B\mathrm{B}. In case of \mathrm{I} \ IV Keeping the concentration of [B] constant, when the concentration of [A][\mathrm{A}] is increased four times, rate also increases four times. Hence, the order with respect to A\mathrm{A} is one. Hence Rate =k[A][B]2=k[\mathrm{ A}][\mathrm{B}]^{2}
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