For a zero order reaction of the type `A rarr` products, the integrated rate equation may be expressed asA. `k = ([A]_(0)-[A])/(2).t`B. `k = ([A]_(0)-[A])/(2t)`C. `k = ([A]-[A]_(0))/(t)`D. `k = ([A]_(0)-[A])/(t)`

For a zero order reaction of the type `A rarr` products, the integrate

1.	For a zero order reaction of the type `A rarr` products, the integrated rate equation may be expressed asA. `k = ([A]_(0)-[A])/(2).t`B. `k = ([A]_(0)-[A])/(2t)`C. `k = ([A]-[A]_(0))/(t)`D. `k = ([A]_(0)-[A])/(t)`
Answer» Correct Answer - D Consider the reaction `Ararrprocts` Rate `=-(d[A])/(dt)=K[A]^(@)` As any equantity raised to power zero is unity Rate `=-(d[A])/(dt)=Kxx1` or `d[A]=-Kdt` Intergrating both sides `[A]=-KtI` where `I` is the constant of integration. At `t=o` , the concentration of the reaction `A=[A]_(0)` , where `[A]_(0)` is intial concentration of the reaction. Substitution in previous equation, we get `[A]_(0)=-K(0)+I` or `[A]_(0)=I` Substituting the value of `I` in the equation `(1)` , we get `[A]=-Kt+[A]_(0)` Further simplifying equation `(2)` , we get the rate constant, `K` as `K=([A]_(0)-[A])/(t)`

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For a zero order reaction of the type `A rarr` products, the integrated rate equation may be expressed asA. `k = ([A]_(0)-[A])/(2).t`B. `k = ([A]_(0)-[A])/(2t)`C. `k = ([A]-[A]_(0))/(t)`D. `k = ([A]_(0)-[A])/(t)`

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