Question

Experiment No.	Initial $\left[NO\right]\left(M\right)$	Initial $\left[C{l}_2\right]\left(M\right)$	Initial rate of disappearance of $C{l}_2\left(M/min\right)$
1	0.15	0.15	0.60
2	0.15	0.30	1.20
3	0.30	0.15	2.40
4	0.25	0.25	?

For the reaction, $2N{O}_{\left(g\right)}+C{l}_{2\left(g\right)}\to 2NOC{l}_{\left(g\right)}$
The following data were collected. All the measurements were taken at $263K$:

(a) Write the expression for rate law.
(b) Calculate the value of rate constant and specify its units.
(c) What is the initial rate of disappearance of $Cl$, in experiment no. 4?

Answer 1

$2NO\left(g\right)+{Cl}_2\left(g\right)⟶2NaCl\left(g\right)$

Let the rate be, $R=K{\left[NO\right]}^x{\left[{Cl}_2\right]}^y$

$\therefore R=\frac{-d\left[{Cl}_2\right]}{dt}$=Rate of disappearance of $C{l}_2$

$\therefore 0.60=K{\left[0.15\right]}^x{\left[0.15\right]}^y⟶\left(i\right)\therefore 1.20=K{\left[0.15\right]}^x{\left[0.30\right]}^y⟶\left(ii\right)\therefore 0.40=K{\left[0.30\right]}^x{\left[0.15\right]}^y⟶\left(iii\right)$

Dividing (i) & (ii),

$\frac{1.6}{1.2}={\left[\frac{0.15}{0.30}\right]}^{y}y=1$

Dividing (i) & (ii),

$\frac{0.6}{2.4}={\left[\frac{0.15}{0.30}\right]}^{x}x=2$

$\left(a\right)\therefore$ Rate$=K{\left[NO\right]}^2\left[{Cl}_2\right]⟹$ Expression for rate law

(b) From (i),

$0.60={K\left(0.15\right)}^2\left(0.15\right)K=\frac{0.60}{{\left(0.15\right)}^3}=177.78M^{-2}.{min}^{-1}$

$K=177.78{mol}^{-2}{L}^{+2}{min}^{-1}\Rightarrow$ Rate constant.

(C) Rate of disappearance$=K{\left[NO\right]}^2\left[{Cl}_2\right]=177.78\times {\left(0.25\right)}^2\times 0.25=2.78M/min$