# Homework Help: Reaction Problems

1. Feb 11, 2008

### predentalgirl1

Just basically need my answers looked over and checked. Thanks.

(a) If we started with 1 mL of 2-propanol and reacted it all to form acetone then approximately how much heat should evolve (use DH°= -117 kJ/mol, and density of 2-propanol= 0.789 g/mL)?
Molar mass of 2-propanol=60.10 g/mol
Density=mass/volume
Mass=0.789 x 1=0.789g
No. of moles of 2-propanol=0.789/60.10=0.0131moles

Heat evolved= 0.0131 x -117= - 1.535 kJ

(b) If there were 5 mL of water (density = 1g/mL) mixed with the 1 mL of 2-propanol and all of the propanol formed acetone what is the mole ratio of acetone (Xl, a) and of water (Xl, w) in the liquid product mixture?
Mass of water= 5g
Molar mass of water= 18.016 g/mole
No.of moles of water=5/18.016=0.278 moles

Molar mass of 2-propanol=60.10 g/mol
Mass=0.789 x 1=0.789g
No. of moles of 2-propanol=0.789/60.10=0.0131moles
mole ratio of acetone (Xl, a) and of water (Xl, w)= 0.278/0.0131=21.19

c) Given the pressure of 765 mm Hg and at an experimental boiling point of the mixture of Tboil = 70° C, we find the vapor pressure of pure water (P°w) from a table to be 233.7 mm Hg. What is the partial pressure of water (Pw) above the mixture at Tboil?
No.of moles of water=5/18.016=0.278 moles
Mole fraction of water= 0.278/21.19= 0.0131
partial pressure of water= vapor pressure of pure water x mole fraction of water

= 233.7x0.0131 =3.06mm Hg

(d) From part c) what is the partial pressure of acetone (Pa) and what is the mole ratio of acetone in the vapor phase above the mixture at Tboil?

Vapour pressure of acetone =1191.55mmHg
No. of moles of acetone=0.789/60.10=0.0131moles
Mole fraction of = 0.0131/21.19=6.181x 10-4
Partial fraction of acetone = 1191.55x6.181x 10-4=0.736mmHg

(e) Correspondingly what is the expected mole ratio of the distillate upon collecting the first few drops and does this show more or less concentration of the acetone?