For a first order reaction the Rate = k[R] solving it

At t=0 value of R =R₀ and at t=t value of R=R …(let),

On integrating w.r.t t on both sides we get,

……………………. Equation 1

Here, R is the molar concentration of a reactant at time t.

Taking ln on both the side we get,

………………… Equation 2

ln =2.303log

for t_1/2 at t=t_1/2 [R]=[R₀] for first-order reaction

so k=

t_1/2 =(2.303 log2)/k t_1/2 is independent of [R₀]………. Equation 3

t_1/2=.693/k

so from Equation 1,2,3, the above graph can be plotted t_1/2 independent of [R₀] so the line is constant and molar concentration is exponentially related with time.

The Equation 2 gives an equation of a line i.e. y=mx +c, with m=k/2.303 and c=0.