II. 0.050 mol of H 2 O and 0.040 mol of Cl 2 O were mixed in a 0.500 L containe
ID: 783308 • Letter: I
Question
II. 0.050 mol of H2O and 0.040 mol of Cl2O were mixed in a 0.500 L container and allowed to react:
H2O (g) + Cl2O (g) 2 HOCl (g) K = 0.090
Calculate the equilibrium concentraions
[H2O] = _______________ [Cl2O] = _______________ [HOCl] = _______________
Repeat your calculation using 0.40 mol of H2O (8 times as much) and the same amount of Cl2O as above. Are the results consistent with Le Chatelier%u2019s Principle?
[H2O] = _______________ [Cl2O] = _______________ [HOCl] = _______________
Explanation / Answer
H2O (g) + Cl2O (g) --------> 2 HOCl (g)
0.05-x 0.04-x 2x
k = [HOCl]^2/[H2O][Cl2O] = (2x)^2/(0.05-x)(0.04-x) = 0.09
4x^2 = 0.09(x^2 + 0.002 - 0.09x)
3.91x^2 - 0.00018 + 0.0081x = 0
x = 0.0058
[H2O] = (0.05-0.0058)/0.5 = 0.0442/0.5 = 0.0884M
[Cl2O] = 0.0342/0.5 = 0.0684 M
[HOCl] = 2x/0.5 = 0.0116/0.5 = 0.0232 M
H2O (g) + Cl2O (g) --------> 2 HOCl (g)
0.04-x 0.04-x 2x
K = (2x)^2/(0.04-x)^2 = 0.09
2x/(0.04-x) = sqrt(0.09) = 0.3
x = 0.0052
[H2O] = 0.04-0.0052)/0.5 = 0.0696 M
[CL2O] = 0.0696 M
[HOCl] = 0.0052*2/0.5 = 0.0208 M
yes the results are consistent