1.

For the reaction 2" A"+" B"toA_(2)B," rate "=k[A][B]^(2) with k=2.0xx10^(-6)mol^(-2)L^(2)s^(-1). Calculate the initial rate of the reaction when [A]=0.1molL^(-1)" and "[B]=0.2molL^(-1). Calculate the rate of reaction after [A] is reduced to 0.06molL^(-1).

Answer»

Solution :Initial rate `=K[A][B]^(2)=(2.0xx10^(-6)mol^(-2)L^(2)s^(-1))(0.1molL^(-1))(0.2molL^(-1))^(2)=8xx10^(-9)MOLL^(-1)s^(-1)` When [A] is reduced from `0.10molL^(-1)" to "0.06molL^(-1)`, i.e., `0.04molL^(-1)` of A has reacted, B reacted
`=(1)/(2)xx0.04molL^(-1)=0.02molL^(-1).`
Hence, new `[B]=0.2-0.02=0.18molL^(-1).`
Now,rate `=(2.0xx10^(-6)mol^(-2)L^(2)s^(-1))(0.06molL^(-1))(0.18molL^(-1))^(2)=3.89xx10^(-9)molL^(-1)s^(-1).`


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