The chemical reaction between $K_{2} C_{2} O_{4}$ and $H g C l_{2}$ is: $2 H g C l_{2} + K_{2} C_{2} O_{4} \to 2 K C l + 2 C O_{2} + H g_{2} C l_{2}$ .
The masses of $H g_{2} C l_{2}$ are precipitated from different solutions in given time are as follows:
Time
$H g C l_{2}$
(mol $l i t r e^{1})$
$K_{2} C_{2} O_{4}$
(mol $l i t r e^{1})$ $H g_{2} C l_{2}$ formed
(mol $l i t r e^{1})$
60
0.0418
0.404
0.0032
65
0.0836
0.404
0.0068
120
0.0836
0.202
0.0031
What is the overall order of the reaction?

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Solution

The rate law expression is as given below:

$R a t e = K [H g C l_{2}]^{m} [K_{2} C_{2} O_{4}]^{n}$

$∴ \frac{0.0032}{60} = K [0.0418]^{m} [0.404]^{n}$ .....(i)

$\frac{0.0068}{65} = K [0.0836]^{m} [0.404]^{n}$ .....(ii)

$\frac{0.0031}{120} = K [0.0836]^{m} [0.202]^{n}$ .....(iii)

Solving Eqs. (i), (ii) and (iii), we get-

$m = 1; n = 2$

Overall order $= m + n = 1 + 2 = 3$ .

Hence, the reaction is of first order w.r.t. $[H g C l_{2}]$ , second order w.r.t $[K_{2} C_{2} O_{4}]$ and the overall order is 3.

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Time	$H g C l_{2}$ (mol $l i t r e^{1})$	$K_{2} C_{2} O_{4}$ (mol $l i t r e^{1})$	$H g_{2} C l_{2}$ formed (mol $l i t r e^{1})$
60	0.0418	0.404	0.0032
65	0.0836	0.404	0.0068
120	0.0836	0.202	0.0031