Question

Derive the integrated rate equation for expressing the rate constant of a first order :
$R\to P$

Answer 1

In this type of reaction, the sum of the powers of concentrations of reactants in rate law is equal to 1, that is the rate of the reaction is proportional to the first power of the concentration of the reactant. Consider the reaction $R\to P$ again. Therefore, the rate law for this reaction is,

Rate is directly proportional to [R]

We know that $\left[R\right]=-kt+[R{]}_0$ ( from equation II). Taking log of both sides, we get

$ln\left[R\right]=-kt+ln[R{]}_0$ ………………………….(IV)

$ln\left[R\right]/[R{]}_0=-kt$ …………………………….(V)

$k=\left(1/t\right)ln\left[R{]}_0/\right[R]$………………………(VI)

Now, consider equation II again. At time t1 and time t2, the equation II will be [R]1 = -kt1 + [R]0 and [R]2 = -kt2 + [R]0 respectively, where [R]1 and [R]2 are concentrations of the reactants at time t1 and t2 respectively. Subtracting second equation from first one, we get

$ln[R{]}_1–ln[R{]}_2=-k{t}_1–(-k{t}_2)$

$ln\left[R{]}_1/\right[R{]}_2=k(t_2–t_1)$

$k=\left[1/\right(t_2–t_1\left)\right]ln\left[R{]}_1/\right[R{]}_2$

Now, taking antilog of both sides of equation V, we get $\left[R\right]=[R{]}_0{e}^{-kt}$

Comparing this equation with equation of a straight line y = mx + c, if we plot ln [R] against t, we get a straight line with slope = -k and intercept $=ln[R{]}_0$

On removing natural logarithm from equation VI, the first-order reaction can also be written as,

$k=2.303/tlog\left[R{]}_0/\right[R]$ …………..(VII)

If we plot a graph of $log\left[R{]}_0/\right[R]$ against t, we get slope = k/2.303