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For the following rate law determine the unit of rate constant. Rate k[A]^((1)/(2))[B]^(2)=[R]^((5)/(2)) |
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Answer» Solution :The total order of reaction `n = 1/2 + 2 = 5/2 = 2.5 ` Rate `K[A]^(1/2) [B]^(2)= [R]^(5/2)` `therefore k=(Rate)/([5]^((5)/(2)))` `therefore` UNIT of k=`("unit of rate")/(("unit of concentration")^((5)/(2)))` `=((mol L^(-1))^(1)s^(-1))/((mol L^(-1))^((5)/(2)))` `(mol L^(-1))^(-(3)/(2))s^(-1)` `=(mol)^(-(3)/(2))(L^(-1))^(-(3)/(2))S^(-1)` `=(mol)^(+(3)/(2))(L)^(-(3)/(2))S^(-1)` If the order of reaction =`(5)/(2)` then unit of rate CONSTANT k is `L^((+3)/(2)) mol^((-3)/(2))s^(-1)` |
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