In a reaction involving one single reactant, the fraction of the reactant consumed may be defined as f = (1 - C /Co ) where
Co and C are the concentrations of the reactants at the start and after time ' t ' . For a first order reaction :
(A) df / dt = k (1 - f )
(B) - df / dt = kf
(C) - df / dt = k( 1 - f )
(D) df /dt = kf
The correct answer is (A) df / dt = k (1 - f ).
We know that the integrated rate equation for first order reaction is
where k is the rate constant, t is the time, [A]1 is the initial concentration of reactant and [A]2 is the concentration of A after time t. Now we are given that the initial concentration is [C]0 and the concentration after time t is [C]. So on integrating (A), we will get
ln (1-f) = -kt (1)
As f = (1 - [C ]/[C]0 ), therefore substituting the value of 'f' in equation (1) will give us
ln ([C]0/[C]) = kt
or k = (2.303/t) log ([C]0/[C])