Question

The initial rate of reaction
A + 5B + 6C = 3L + 3M
has been determined by measuring the rate of disappearance of A under the following conditions:

Experiment No.	$[A{]}_0$ (M)	$[B{]}_0$ (M)	$[C{]}_0$ (M)	Initial rate (M $mi{n}^1$)
1.	0.02	0.02	0.02	$2.08\times {10}^3$
2.	0.01	0.02	0.02	$1.04\times {10}^3$
3.	0.02	0.04	0.02	$4.16\times 103$
4.	0.02	0.02	0.04	$8.32\times {10}^3$

Determine the order of reaction with respect to each reaction and overall order of reaction:

Answer 1

(I.) In experiments 1 and 2, initial concentrations of B and C remains unchanged and hence from the rate expression.
r = $k\left[A_0{]}^a\right[B_0{]}^b[C_0{]}^c=k_0[A_0{]}^a$, we get
$\frac{(r_0{)}_1}{(r_0{)}_2}=\frac{2.08\times {10}^{-3}}{1.04\times {10}^{-3}}=\frac{[A_0{]}_1^a}{[A_0{]}_2^a}=\frac{{0.02}^a}{{0.01}^a}$
or $(2{)}^1=(2{)}^a\Rightarrow a=1;ORw.r.t.A=1$
(ii)
Similarly, $\frac{(r_0{)}_3}{(r_0{)}_1}=\frac{4.16\times {10}^{-3}}{1.04\times {10}^{-3}}=\frac{[B_0{]}_3^b}{[B_0{]}_1^b}=\frac{{0.04}^b}{{0.02}^b}$
or $(2{)}^t=(2{)}^b\Rightarrow b=1;ORw.r.t.B=1$
(iii)
and $\frac{(r_0{)}_4}{(r_0{)}_1}=\frac{8.32\times {10}^{-3}}{2.08\times {10}^{-3}}=\frac{[C_0{]}_4^b}{[C_0{]}_1^c}=\frac{{0.04}^c}{{0.02}^c}$
or 4 = $(2{)}^c\Rightarrow (2{)}^2=(2{)}^c;ORw.r.t.C=2$
Therefore, orders of reaction with respect to A, B, and C are 1, 1, and 2, respectively.
(iv) The overall order of the reaction = 1 + 1 + 2 = 4
k = $\frac{a}{\left[A_0{]}^a\right[B_0{]}^b[C_0{]}^c}$ = $\frac{r}{\left[A_0\right]\left[B_0\right][C_0{]}^2}$
We can put the values from the set 1.
(v) k = $\frac{2.08\times {10}^{-3}Mmi{n}^{-1}}{16\times {10}^{-8}{M}^3}=1.3\times M^{-3}mi{n}^{-1}$
(b) (i) Rate of the reaction when,
$\left[A_0\right]=\left[C_0\right]=\left[C_0\right]=0.01M$
= $1.3\times {10}^4{M}^{-3}mi{n}^{-1}$
= [0.01 M] [0.01 M] $[0.01M{]}^2$
= $1.3\times {10}^{-4}Mmi{n}^{-1}$
(ii)
$\frac{-d\left[A\right]}{dt}=r\frac{-1}{5}\frac{d\left[B\right]}{dt}=\frac{1}{3}\frac{d\left[L\right]}{dt}=1.3\times {10}^{-4}Mmin-1$
(iii) $\frac{-d\left[B\right]}{dt}=5\times 1.3{10}^{-4}Mmi{n}^{-1}=6.5\times {10}^{-4}Mmi{n}^{-1}$
(iv) $\frac{-d\left[B\right]}{dt}=3\times 1.3{10}^{-4}Mmi{n}^{-1}=3.9\times {10}^{-4}Mmi{n}^{-1}$