Question

In a fluorine molecule (F_2) the considerable repulsion that exisys between lone pairs of electrons makes the F-F bond relatively weak since lone paris on one F atom repels the lone pairs on the other F atom. The situation can be represented as follows: F_2(g) K_p (g) = 7.83 at 1500K (i) If the equilibrium partial pressure of F2 molecules at 1500K is 0.200 atm, what is the equilibrium partial pressure of F atoms in atm? (ii) Calculate the fraction of the F_2 molecules that will dissociate at 1500K. (b) Carbon disulfide, CS_2, has P_Vap = 100mm at -5.1 degreeC and a normal boiling point of 46.5 degreeC. What is DeltaH_vap for CS_2 in kJ/mol? (c) An aqueous solution containing 250 mg of sodium dodecylsulfate (SDS, NaC_12H_25SO-4, a common ingredient in soap) was prepared in a 50.0 mL volumetric flask and was found to have an osmotic pressure of 1.60 mmHg at 298K (1 atm = 760 mmHg). (1) Calculate the nominal molecular weight of the particles in the solution (g/mol) and (2) explain its difference from the molecular weight of SDS.

Answer 1

The reaction is:

The reaction is: F_2(g) rightarrow 2 F(g).....K_p = 7.83 The Kp expression is: K_p = P_F^2/P_F_2 = 7.83 Using the data you given: P_F^2/0.2 = 7.83 P_F = 1.25 As pressure is directly proportional to concentration, we can make an ICE table: F_2 rightarrow 2F 1.45....0 -1.25...+ 2(1.25) 0.2.....2.5 The dissociation percent is: %d = 1.25/1.45 * 100% = 86.21%

The Kp expression is:

$K_p=\frac{P_F^2}{P_{F_2}}=7.83$

Using the data you given:

$\frac{P_F^2}{0.2}=7.83$

As pressure is directly proportional to concentration, we can make an ICE table:

$\\ F_2\rightarrow 2F\\ 1.45...........0\\ -1.25......+2(1.25)\\ 0.2.......2.5$

The dissocation percent is:

$\%d=\frac{1.25}{1.45}*100\%=86.21\%$