For the reaction : 2A+B+CtoA_(2)B+C, the rate law has been determined to be Rate =k[A][B]^(2)" with "k=2.0xx10^(-6)" mol"^(-2)L^(2)s^(1) For this reaction determine the initial rate of the reaction with [A]=0.1" mol L"^(-1),[B]=0.2" mol L"^(-1),[C]=0.8" mol L"^(-1). Determine the rate after 0.04" mol L"^(-1). Determine the rate after 0.04 "mol L"^(-1) of A has reacted.

For the reaction : 2A+B+CtoA_(2)B+C, the rate law has been determined

1.	For the reaction : 2A+B+CtoA_(2)B+C, the rate law has been determined to be Rate =k[A][B]^(2)" with "k=2.0xx10^(-6)" mol"^(-2)L^(2)s^(1) For this reaction determine the initial rate of the reaction with [A]=0.1" mol L"^(-1),[B]=0.2" mol L"^(-1),[C]=0.8" mol L"^(-1). Determine the rate after 0.04" mol L"^(-1). Determine the rate after 0.04 "mol L"^(-1) of A has reacted.
Answer» Solution :Initial rate `=(2.0xx10^(-6)" MOL"^(-2)L^(2)s^(-1))xx0.1" mol L"^(-1)xx(0.2" mol L"^(-1))^(2)=8xx10^(-9)" mol L"^(-1)s^(-1)` After `0.04" mol L"^(-1)" of A has REACTED. "[A]=0.1-0.04=0.06" mol L"^(-1)` `[B]=0.2-0.02=0.18" mol L"^(-1)` Then rate `=(2XX10^(-6))xx(0.06)xx(0.18)^(2)=3.9xx10^(-9)" mol L"^(-1)s^(-1).`