Q17 / 90 · JEE Main 2019 (08 Apr Shift 1) · Chemistry

For the reaction 2A+B→C , the values of initial rate at different reactant concentrations are given in the table below. The rate law for the reactions is:
[A](mol L-1) [B](mol L-1)
Initial Rate
(mol L-1s-1)
0.05 0.05 0.045
0.10 0.05 0.090
0.20 0.10 0.72

  • A
    Rate=k[A]2[B]
  • B
    Rate=k[A]2[B]2
  • C
    Rate=k[A][B]
  • Correct answer
    D
    Rate=k[A][B]2

Solution

From the experiment (1) and (2)
0.050.10x0.050.05y=0.0450.090
12x=12⇒x=1

From the experiment (2) and (3)
0.10.2x0.050.1y=0.090.72
1212y=0.090.72
12y=0.09×20.72
12y=14
y=2
Order of reaction with respect to A is 2 and B=2
∴R∝AB2
Or R=KAB2

Hence, the correct answer is option: 'D'.

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