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Derive the expression for integrated rate law(equation) for the first order reaction. |
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Answer» Solution :CONSIDER the FOLLOWING first order reaction, A`to` B. The rate of the chemical reaction is given by the rate lawexpression as, Rate R = K [A] where [A] is the concentration of the reactant A and k is the velocity constant or specific rate of the reaction. The instantaneous rate is given by, `R = k[A] = (-d[A])/(dt)` ` (-d[A])/([A]) = k.dt` If `[A_(0)]` is the INITIAL concentration of the reactant and `[A]_(t)` at TIME t. then by integrating the above equation, `-underset([A_(0)])overset([A]_(t))int (d[A])/([A]) = k underset(0)overset(t)int dt` ` therefore- log_(e).([A_(0)])/([A]_(t)) = kt ""thereforelog_(e).([A_(0)])/([A]_(t)) = kt` `therefore k = (1)/(t) log_(e).([A_(0)])/([A]_(t)) ""thereforek = (2.303)/(t) log_(10).([A_(0)])/([A]_(t))` Thisis theintegranted rateequationfor the firstorderreaction . Thisis also calledintegrated rate law. |
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