For a first order reaction. When the initial concentration, a is not know, the integrated rate expression becomes

Options

A. `k = (2.303)/(t_(2)-t_(1))log ((a-x_(1))/(a-x_(2)))`
B. `k = n(2.303)/(t.a)log ((a)/(a-x))`
C. `k = (2.303)/(t_(2)-t_(1))log((a)/(a-x))`
D. `k = (2.303)/(t)log (a)/(a-x)`

Answer

A

Explanation

For a first order reaction, the rate of the reaction Is proportional to the first power of the concentration of the reactant `R` . For example
`RrarrP` lt brgt Initial concentration `ao`
Concentration at time `t_(1)a-x_(1)x_(1)`
Concentration `t_(2)a-x_(2)x_(2)`
Differential rate law is
Rate `=-(d[R])/(dt)=K[R]`
or `(d[R])/([R])=-Kdt`
Integrating this equation, we get
In `[R]=-Kt+I`
where `I` is the constant of integration. When `t=0,[R]_(0)` ,
where `[R]_(0)` is the initial concentration of the reactant. Therefore, equation `(1)` can be written as
In `[R]_(0)=-K(0)+I`
In `[R]_(0)=I`
Substituting the value of `I` in equation `(1)` , we get
In n`[R]=-Kt+In[R]_(0)`
At time `t_(1)`
In `[R]_(t)=-Kt_(1)+In[R]_(0)`
At time `T_(2)`
In `[R]_(2)=-Kt_(2)+In[R]_(0)`
where `[R]_(1)` and `[R]_(2)` are the concentrations of the reactants at time `t_(1)` and `t_(2)` respectively.
Subtracting `(4)` from `(3)` , we get
In `[R]_(1)-In[R]_(2)=-Kt_(1)-(-Kt_(2))` lt.brgt `In([R]_(1))/([R]_(2))=K(t_(2)-t_(1))`
`K=(1)/((t_(2)-t_(1))`in([R]_(1))/([R]_(2))` or `K=(2.303)/((t_(2)-t_(2)))log((a-x_(1))/(a-x_(2)))`