ChE102 Chemistry for Engineers Final Exam Review package Waterloo SOS
Fall 2010
Phases Q1 A gaseous compound known to contain only carbon, hydrogen, and nitrogen is mixed with exactly the volume of oxygen required for its complete combustion to CO 2, H2O, and N2. Burning 20 litres litres of the gaseous reaction mixture (the unknown compound plus oxygen) produces 14 litres of CO 2, 10 litres of H2O vapour, and 2 litres of N 2, all at the same temperature and pressure. (a) Of the 20 litres of reaction mixture, how many litres are oxygen? (b) What is the molecular formula of the unknown compound?
Basis: 20 litres of the the reaction mixture containing containing (20- a) litres of CxHyNz plus a volumes of O2. For an ideal gas at constant temperature and pressure, the volume of the ideal gas is directly proportional to the number of moles of the ideal gas. Therefore, the volume ratios given in the problem statement can also be interpreted as molar ratios. r atios. The chemical equation for the reaction r eaction is therefore:
(20 − a ) C x H y N z + a O 2 → 14 CO 2 + 10 H 2 O + 2 N 2 Balancing this chemical equation for C, H, N, and O gives four equations:
Carbon (C ) : (20 − a ) ⋅ x = 14 Hydrogen( H ) : (20 − a ) ⋅ y = 20 Nitrogen ( N ) : (20 − a ) ⋅ z = 4 Oxygen(O) : 2 ⋅ a = 38 The solution to these equations is:
a = 19
∴ x = 14 , y = 20 , z = 4 Of the 20 litres of reaction mixture, 19 litres are O 2, and the molecular formula of the unknown compound is C14H20N4. Phases Q2 Dew point is defined as the temperature at which the vapour pressure of water is equal to the partial pressure of water vapour in an air sample. Exactly 0.105 g of water was evaporated into 10.0 10.0 L of dry air at 60.0ºC and 100 kPa. Using the vapour pressure data below, estimate estimate the dew point of the moist air sample if the total pressure is maintained at 100 kPa. Data: [K] T [K] o
P
[kPa]
280 0.9912
285 1.388
o
290 1.919
295 2.620
Basis: 0.105 g of H2O in 10.0 L of dry air at 60.0 C and 100 kPa. Data: Molecular mass of H2O, µ H 2O = 18.01528 g mol
298.15 3.197
The number of moles of H2O evaporated is, 0.105 g H2O n = H 2O
The number of moles of dry air is, P ⋅V 100 kPa n dry air = = R ⋅ T
mol H2O 18.01528 g H 2O
10.0 L 333.15 K
= 0.0058284 mol H2O
mol K 8.31451 kPa L
= 0.36101 mol dry air
The partial pressure of H2O in the humid air is therefore, 0.0058284 mol × 100 kPa = 1.5888 kPa PH 2O = y H 2O ⋅ Ptotal = 0.0058284 mol + 0.36101 mol The dew point temperature of the humid air is the temperature at which this partial pressure of H 2O becomes equal to the vapour pressure of H 2O. Inspection of the given vapour pressure data shows that the dew point temperature should fall between 285 K and 290 K. Since the Clausius-Clapeyron equation states that, ∆ H vap 1 P o 1 − ln 1o = − P R T 1 T 2 2 we find that between 285 K and 290K, ∆ H vap 1 1.388 kPa 1 = − ln − R 285 K 290 K 1.919 kPa ∆ H vap ∴ = 5354.734 K R
Therefore,
1 1.5888 kPa 1 = −(5354.734 K ) ⋅ − T 1 . 388 kPa 285 K dew point ∴ T dew point = 287.06 K
ln
The dew point temperature of the humid air is 287.06 K or 13.91 oC. Phases Q3 At 25.0ºC, the vapour pressure of trichloroethene (C 2HCl3) is 73.00 mm-Hg. At the same temperature, the vapour pressure of chloroform (trichloromethane, CHCl 3) is 199.1 mm-Hg. a) What is the pressure exerted by a solution containing 32.05% by mass of C 2HCl3 (the balance being CHCl3) at 25.0ºC? b) What is the vapour phase mole fraction of CHCl 3 that is in equilibrium with the liquid solution specified in (a)?
Basis: 100 g of solution containing 32.05 g of C 2HCl3 and 67.95 g of CHCl 3 at 25.0ºC. Data: Molecular masses in g/mol: µ C2 HCl3 = 131.388 µ CHCl 3 = 119.377
