Saved Bookmarks
| 1. |
What is first order reaction?Determine the integrated rate equation for first order reaction RtoP |
|
Answer» <P> Solution :First order reaction :The rate of the reaction is proportional to the first power of the cocentration of the reactant R is called first order reaction .The rate of first order reaction. The rate of first order reaction `prop[R]^(1)` Where ,reaction R `to` P is first order reaction. The differential rate expression of first order reaction is as under Rate =`-(d[R])/(dt)=k[R]` ..........(i) `therefore(d[R])/([R])=-k dt`.......(ii) Intergrating this equation both side , In [R] =-kt+1 .......(iii) Here,I=Constant of integration When ,t= [R]=`[R]_(0)` .Where `[R]_(0)` is the initial concentration of the reactant.Put this value in eq. (iii) In `[R]_(0)=-kxx(0)+1` `therefore In [R]_(0)=I` .......(iv) If we put I =In `[R]_(0)` in equation (iii),we get In [R]=-kt+In `[R]_(0)`......(VA) `therefore` kt =In `[R]_(0)`-In [R] `therefore kt=In ([R]_(0))/([R])` `therefore k=(1)/(t)xxIn ([R]_(0))/([R])` .......(VB) Equation V(A) and V(B) both are integrated rate equation (VA) and (VB) can written in the form log [R]=`-(kt)/(2.303)+log [R]_(0)` ......(ViA) `K=(2.303)/(t)` log `([R]_(0))/([R])` .....(VIB) Note:Equation VA and VIA are stright equations .So,if we plot in `[R]tot` and log [R]`to` t we get a straight LINE with slope=-k and intercept equal to In `[R]_(0)` and log `[R]_(0)` Taking ANTILOG of equation (VA) at both ide ,we get [R]=`[R]_(0)E^(-kt)` .........(vii) |
|