1.

Calculate the rate of reaction from the following rate law: -(d[A])/(dt) = k[A]^(1)[B]^(2) when the concentrations of A and B are 0.01 M and 0.02 M respectively and k=5.1xx10^(-3)L^(2)"mol"^(-2)s^(-1).

Answer»

Solution :`-(d[A])/(dt)=k[A]^(1)[B]^(2)`
SUBSTITUTING the values of k, [A] and [B], we have
RATE of reaction `=5.1xx10^(-3) xx (0.01)(0.02)^(2)=5.1xx10^(-3)xx10^(-2)xx4xx10^(-4)`
`=20.4 xx 10^(-9) =2.04xx10^(-8)"MOL L"^(-1)s^(-1)`


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