1.

Rae law for the reaction A+2BtoC is found to be Rate =K[A][B] concentration of reactant .B. is doubled ,keeping the concentration of .A. constant ,the value of rate constant will be……

Answer»

the same
doubled
quadrupled
halved

Solution :Initial RATE `r_(1)`=K[A][B]
If the concentration of B is double than new concentration [2B]
So `r_(2)`=k[A][B]=2K [A][B]
`therefore (r_(2))/(r_(1))=2` so the rate is double


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