IF anyone can help me with this involved question it would be greatly appreaciat
ID: 810295 • Letter: I
Question
IF anyone can help me with this involved question it would be greatly appreaciated.
the following data were collected from two separate experiments for the following reaction in aqueous, by measuring the number of moles of Hg2Cl2 that precipated per liter per minute
2HgCl2(aq) + C2O4^2-(aq) ---> 2Cl-(aq) + 2CO2(g) Hg2Cl2 (s)
Experiment 1 Experiment 2
time HgCl2 mol/L excess C2O4^2- C2O4^2- mol/L excess HgCl2
0 0.150 0.300
3.33 0.0843 0.283
5.00 0.0755 0.275
6.67 0.0676 0.268
8.33 0.0606 0.261
10.00 0.0543 0.254
16.67 0.0340 0.229
23.33 ? ?
d.) In a separate experiment when HgCl2 = 0.105 M and C2O4^2- = 0.300 M the initial rate was 7.1 * 10^-5 mol L^-1min-1. Determine the overall rate law constant (K). What would be the initial rate of the reaction if HgCl2 = 0.02 M and C2O4^2- = 0.22 M under this condition?
Explanation / Answer
Given equation is :
2HgCl2+C2O42- ----> 2Cl- + 2CO2 +Hg2Cl2
The rate of appearance of product and disappearance of reactant will be as such:
-1/2d/dt [HgCl2] = -d/dt[C2O42-] =+I/2d/dt[Cl-] = +d/dt [Hg2Cl2]
The overall rate rate law expression would be
-d/dt[HgCl2] = - 2d/dt [C2O42-] = +d/dt[Cl-]
This means that 1 mole of HgCl2 reacts with 2 moles of C2O42- to give 1 mole of Cl- ion
As the rate law would be given as
r = K [HgCl2] [C2O42-]2
From the d) part when the HgCl2 = 0.105M and C2O42- = 0.22M and the initial rate (r) =7.1 X 10-5mol L-1 min-1.
for the above rate law expression
K = 7.1 X 10-5 mol L-1 min-1 / [0.105 M][0.3M]2 = 7.5 X 10-3
Again as we know value of K, from the rate law expression r is equal to as follows for HgCl2 = 0.02M, C2O42- = 0.22M
r = 7.5 X 10-3 [0.02] [0.22]2 = 7.2 X 10-6.
now calculate initial rate of each reaction with K= 7.5 X10-3
the overall rate of the reaction is 3rd order.
the conc of HgCl2 and C2O42- at 23.33 from the intergarte law expression for the 3rd order reaction.
time HgCl2 C2O42- initial rate 0 0.15 0.3 7.1X10-5 3.33 0.0843 0.283 5.06 X 10-5 5 0.0755 0.275 4.2 X10 -5 6.67 0.0676 0.268 3.64X10-5