Question

S. No	$\left[A\right]$	$\left[B\right]$	$\left[C\right]$	Rate of reaction
$1$	$0.2$	$0.2$	$0.2$	$2\times {10}^{-4}$
$2$	$0.2$	$0.2$	$0.4$	$4\times {10}^{-4}$
$3$	$0.2$	$0.4$	$0.2$	$2\times {10}^{-4}$
$4$	$0.4$	$0.2$	$0.2$	$8\times {10}^{-4}$

$\left(i\right)$ Find the order of the reaction.
$\left(ii\right)$ Find the rate constant.
$\left(iii\right)$ Find the rate of appearance of $\left(i\right)$ when $\left[A\right]=0.1,\left[B\right]=0.1,\left[C\right]=0.1$

Answer 1

$r=\left[A{]}^m\right[B{]}^n[C{]}^p$ $m,n,p$ are the order of $\left[a\right],\left[b\right],\left[c\right]$ respectively.

From equation $\left(1\right)$ and $\left(2\right)$, $\frac{r_2}{r_1}=\frac{{\left[A\right]}^m{\left[B\right]}^n{\left[2C\right]}^p}{{\left[A\right]}^m{\left[B\right]}^n{\left[C\right]}^p}\Rightarrow {\left(2\right)}^1={\left(2\right)}^p$

So, $p=1$ (first order w.r.t $\left[C\right]$)

From equation $\left(1\right)$ and $\left(3\right)$, $\frac{r_3}{r_1}={\left(2\right)}^m\Rightarrow 1=2^n\Rightarrow 2^0=2^n$

So, $n=0$ (zero order w.r.t $\left[B\right]$)

From equation $\left(1\right)$ and $\left(4\right)$, $\frac{r_4}{r_1}={\left(2\right)}^m\Rightarrow 4=2^n\Rightarrow 2^2=2^n$

So, $m=2$, (second order w.r.t $\left[A\right]$)
$\left(i\right)$ Order of the reaction is $2+0+1=3$

$\left(ii\right)$ Rate constant, $K=\frac{r}{{\left[A\right]}^2{\left[B\right]}^0{\left[C\right]}^1}=\frac{2\times {10}^{-4}}{{\left(0.2\right)}^2\times 1\times 0.2}=\frac{1}{40}=0.025$

$\left(iii\right)$ $Rate=K\left[A{]}^2\right[B{]}^0[C{]}^1$

$=0.025\times (0.1{)}^2\times (0.1{)}^0\times (0.1{)}^1$

$=2.5\times {10}^{-5}$