CHM2045exam3solutions

CHM2045exam3solutions - Solutions to CHM2045 Exam#3 Spring...

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Solutions to CHM2045 Exam #3, Spring 2008, Form Code A 1. N 2 (g) + 3H 2 (g) 2NH 3 (g) Using molar ratios: Consumption of 0.500 atm of N 2 would result in production of 1.00 atm NH 3 . Consumption of 0.500 atm of H 2 would result in production of 0.33 atm NH 3 . Therefore, H 2 is the limiting reactant and will be completely consumed. To consume 0.500 atm of H 2 , 0.170 atm of N 2 is used up, leaving (0.500 – 0.17) = 0.33 atm N 2 left over. So the pressure in the vessel is (0.33 atm NH 3 + 0.33 atm N 2 ) = 0.66 atm total . 2. As per Graham's Law: (Rate of S x O y ) / (Rate of O 2 ) = 0.707 = square root of (M of O 2 ) / (M of S x O y ). So, (0.707) 2 = (32 g/mol) / (? g/mol) (? g/mol) = 64 g/mol, and the only listed S x O y with that molar mass is SO 2 . 3. As per the Kinetic-Molecular Theory of Gases, the average kinetic energy is the same for all gases at the same temperature, and it is directly proportional to the temperature. So, if the temperature increases from 90.0K to 270.K, the average kinetic energy will triple (regardless of the gas). So the answer is 1:3 . 4. n of Ne = PV/RT = [(1.50 atm)(0.400 L)] / [(0.0821)(513K)] = 0.0142 mol Ne n of O 2 = PV/RT = [(550/760 atm)(0.150 L)] / [(0.0821)(403K)]=0.00328 mol O 2 T o t a l m o l e s n tot = 0.0175 mol Total P = nRT/V = [(0.0175)(0.0821)(298K)] / (0.500 L) = 0.856 atm 5. Attractive interparticular/intermolecular forces between gas particles/molecules in a real/non-ideal gas tend to decrease the pressure in the gas relative to that of an ideal gas. 6. Clausius-Clapeyron equation: ln (P 2 /P 1 ) = ( Δ H vap / R)[(1/T 1 ) – (1/T 2 )] T 1 = (56.2 + 273) = 329 K P 1 = 760 torr (at boiling point) T 2 = (25.0 + 273) = 298 K P 2 = 225 torr Solving for Δ H vap results in 32.0 kJ/mol 7. If Na metal is body-centered cubic, then there are two atoms of Na per unit cell.

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CHM2045exam3solutions - Solutions to CHM2045 Exam#3 Spring...

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