Derive the integrated rate equation for the rate constant for a first order reaction. What would be units of the first order rate constant, if the concentration is expressed in moles per litre and time to seconds ? Also give graphical representation of integrated rate law equation.

Explanation

Let us consider the following hypothetical reaction of first order,
`nR rarr "Products"`
If the concentration of reactant, R is [R] at a particular instant of time, then instantaneous rate of reaction at that instant of time is given by the expression.
`-(d[R])/(dt)=k[R]`
where k is the rate constant.
Rearranging the equation, we get
`-(d[R])/(dt)=kdt`
On integrating the above equation.
`-(d[R])/(dt)=kdt`
- In [R]-kt+I
where I is the constant of integration. The value of this constant, I can be calculated from the initial conditions. When I = 0, [R] = `[R]_0` i.e. initial concentration of R. Thus becomes,
`-In[R]_0=I` .....(i)
Substituting the value of I in ean. (i), we get
`-In[R]=kt-In[R]_0`
`In [R]_0-In [R]=kt`
or `In ([R]_0)/([R])=kt`
or `k=1/t log ([R]_0)/([R])` ... (iii)
Converting the above expression to log to the 10, we get or `k=(2.303)/(t)log([R]_0)/([R])`
The equation (iv) is integrated .rate equation. This equation can be used to check if a given reaction is of 1st order with respect to a reactant or not.
Unit of rate constant for first order reaction.
For first order reaction
Reaction rate = k[R]
`mol L^(-1)s^(-1)=kxx"mol L"^(-1)`
Unit of `k=s^(-1)`
Graphical representation. Equation (iv) can be written as,
`(k)/(2.303)t=log([R]_0)/([R])`

or `(k)/(2.303)t=log[R]_0-log[R]`
or `log[R]=(k)/(2.303)t log [R]_0`
When a graph is plotted between log [A] (along Y-axis) and t (along X-axis) a straight line is obtained.
slope=`(k)/(2.303)t` and Y-intercept=log `[R]_0`
Rate constant k can be also calculated from the slope of the line so obtained.