07/26/16
CHEM 1A
ANSWER KEY
Name _________________________________
Quiz 7 (30 points)
1. (6 points) Consider the flask diagramed below.
What will be the final partial pressures and mole fractions of each gas in the mixture after the
stopcock between the two flasks is opened? Show work.
PH2 = 475 torr / 1.5 = 317 torr = 0.417 atm
(1 atm = 760 torr)
PN2 = 0.200 atm / 3 = 0.0667 atm = 50.7 torr
XH2 = PH2 / Ptotal = 317 torr / 367.7 torr = 0.861
XN2 = 1 − 0.861 = 0.139
2. (8 points) In a laboratory experiment, 0.153 g of aluminum were allowed to react with 30.0 mL
of 0.260 M sulfuric acid. The gas produced in the reaction was collected over water at 23.0°C
and 756 mm Hg. What was the gas and what was the volume it occupied? Show work.
2 Al(s)
+
3 H2SO4(aq)
→
Al2(SO4)3(aq)
+
3 H2(g)
0.153 g
(26.98 g/mol)
30.0 mL × 0.260 M
5.67 mmol
7.80 mmol [7.80 mmol × (2/3) = 5.20 mmol Al required]
L.R = H2SO4
Molar Volume of Gas:
PH2 = 756 – 21.1 = 735 torr
T = 23.0 + 273 = 296 K
VM = V/n = RT/P = 62.36 torr∙L/(mol∙K) × 296 K / 735 torr = 25.1 L/mol = 25.1 mL/mmol
7.80 mmol H2SO4 ×
3 mmol H2
25.1 mL H2
×
= 196 mL H2
3 mmol H2SO4
1 mmol H2
PLEASE TURN OVER!!!
3. (8 points) Fructose, C6H12O6, an isomer of glucose, is a carbohydrate that plays an important
role in metabolism. The heat of combustion, ∆H°c, of fructose determined in a bomb
calorimeter experiment is −2810.4 kJ/mol.
(a) Calculate the caloric value of fructose in Cal/g. Show work.
2810.4 kJ
×
1 mol
1 mol
1 Cal
×
180.16 g
4.184 kJ
= 3.728 Cal/g
(b) Calculate the enthalpy of formation of fructose, ∆H°f, from the heat of combustion given.
Use the table of standard enthalpies of formation for any additional information. Show work.
∆Hf°
(kJ/mol)
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l); ∆H = −2810.4 kJ
X
0
−393.5
−285.8
6 mol×(−393.5 kJ/mol) + 6 mol×(−285.8kJ/mol) – 1 mol × X kJ/mol = −2810.4 kJ
X = 2810.4 – 4075.8
X = 1265.4
(c) Write the thermochemical equation for the reaction of formation (not of combustion) of
fructose.
6C(s, graphite) + 6H2(g); + 3O2(g) → C6H12O6(s); ∆H = −1265.4 kJ
4. (8 points) Ammonium nitrate is a substance that is used in cold packs, the other substance
being water.
(a) Using the table of standard enthalpies of formation, calculate the heat of solution, , ∆H°soln,
of ammonium nitrate. Show work.
∆Hf°
(kJ/mol)
NH4NO3(s)
−365.6
→
NH4+(aq)
−132.5
+
NO3−(aq); ∆H = ? kJ
−205.0
∆Hsoln° = −132.5 − 205.0 – (−365.6) = +28.1 kJ/mol
(b) If 5.00 g of ammonium nitrate is quickly stirred into 100 mL of water at 23.0°C, what will
be the final temperature of the solution? Show work.
5.00 g NH4NO3 / (80.05 g/mol) = 0.0625 mol
qrxn = 0.0625 mol × (+25.1 kJ/mol) × (1000 J / kJ) = +1756 J
qsolution = −qrxn = −1756 J = 4.184 J/(g∙°C) × 105 g × ∆T
∆T = −4.0°C
Tf = Ti + ∆T = 23.0°C – 4.0°C = 19.0°C