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InterviewSolution
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A reaction is of second order with respect to its reactant. How will its reaction rate be affected if the concentration of the reactant is (i) doubled (ii) reduced to half ? |
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Answer» Let the concetration of the reactant , [A] =a Order of reaction =2 so that Rate fo reactioin will Rate =`k[A]^(2)" "…(i)` Plug the values we get Rate `=ka^(2)` (i) Given that concentration of the reactant of the reactant is double So that `[A]=2a,` Plug the value in equation (1) we get New Rate of reaction, `R_(1)=k(2a)^(2)=4ka^(2)` We have already calculate that initial rate `R=Ka^(2)` Plud the value we get `R_(1)=4R` Hence rate of reaction will increased to 4 times (ii) Given that concentration of the reactant is reduced to half So that [A]=(1/2)a Plug the values in equation (1) we get New Rate of reaction , `R_(2)=k((1//2)a)^(2)` `=(1//4)ka^(2)` We have already calculate that initial rate `R=ka^(2)` Plug the value we get `R_(1)=(1//4)R` Hence rate of reaction will reduced to 1/4 |
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