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Question

For the reaction, 2A+BA2B the rate =k[A][B]2 with k=2.0×106 mol2 L2 s1. Calculate the initial rate of reaction when [A]=0.1 mol L1,[B]=0.2 mol L1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L1.

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Solution

(i) Case I: Rate =k[A][B]2
=(2.0×106 mol2 L2 s1)×(0.1 mol L1)×(0.2 mol L1)2
=8.0×109 mol L1 s1

(ii) Case II: Concentration of A at a particular time =0.06 mol L1
Amount of A reacted =(0.10.06)=0.04 mol L1
Amount of B reacted =12×0.04 mol L1=0.02 mol L1
Concentration of B at a particular time
=(0.20.02) mol L1=0.18 mol L1
Rate =k[A][B]2
=(2.0×106 mol2 L2 s1)×(0.06 mol L1)×(0.18 mol L1)2
=3.89×109 mol L1 s1


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