Question

(a) Derive integrated rate equation for $1^{st}$ order rate equation

(b) A first order reaction is found to have a rate constant, $k=5.5\times {10}^{-14}{S}^{-1}$ Find the half-life of the reaction.

Answer 1

(a) Integrated equation of I order rate equation

$R\to P$

$Rate=\frac{d\left[R\right]}{dt}=K\left[R\right]$

$\left(Or\right)=\frac{d\left[R\right]}{\left[R\right]}=-Kdt$

Integrating this equation we get

ln [R] = – Kt + I ___ (1)

I is the constant of integration and its value can be determined.

When $t=0,R=[R{]}_0$

$In[R{]}_0=-K\times 0+I$

$In[R{]}_0=l$

Substitution the value of I in eq (1) we get

In [R] = -Kt + In[R] ___(1a)

$In\frac{\left[R\right]}{[R{]}_0}=-Kt$

$OrK=\frac{1}{t}In\frac{\left[R\right]}{\left[R\right]}$____(1b)

At time $t_1$ from equation 1

$In[R{]}_1=K{t}_1+ln[R{]}_O$ ____ (2)

At time $t_2$

$In[R{]}_2=K{t}_2+ln[R{]}_O$ ____ (3)

Where [R], and $[R{]}_2$ are concentration of the reactions at time $t_{1}and{t}_2$ respectively.

Subtracting 3 from 2

$Ln\left[R\right],-ln[R{]}_2=-K{t}_1-(-K{t}_2)$

$\frac{In[R{]}_1}{[R{]}_2}=K(t_2-t_1)$

$K=\frac{1}{(t_2-t_1)}In\frac{[R{]}_1}{[R{]}_2}$

Equation 1a can also be written as

$\frac{In\left[R\right]}{[R{]}_0}=-Kt$

Taking antilog of both sides

$\left[R\right]=[R{]}_0{e}^{-Kt}$

Equation 1b, the first order rate equation can also be written in the form.

$K=\frac{2.303}{t}log\frac{[R{]}_0}{\left[R\right]}$

(b) $K=\frac{0.693}{t\frac{1}{2}}K=5.5\times {10}^{-14}{s}^{-1}$
$5.5\times {10}^{-14}=\frac{0.693}{t\frac{1}{2}}t\frac{1}{2}=\frac{0.693}{5.5\times {10}^{-14}}$
$=0.126\times {10}^{14}s$