Name. CHEM 115 Exam #4 PRACTICE. Complete the following questions/calculations. (Show all of your work.)

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1 Name CHEM 115 Eam # PRACTICE Complete the following questions/calculations. (Show all of your work.) (12.5 points) 1. Use Hess law and the following data (a) CH (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) (b) CH (g) + CO2(g) 2 CO (g) + 2 H2 (g) (c) CH (g) + H2O (g) CO (g) + H2 (g) ΔH = -890 ΔH = +27 ΔH = +250 to determine ΔH for the reaction rn (a) CH (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) rn (b) CH (g) + CO2(g) 2 CO (g) + 2 H2 (g) 2*rn(c) 2CH (g) + 2H2O (g) 2CO (g) + 6 H2 (g) ΔH = -890 ΔH = +27 ΔH = (2)*(+250 ) CH (g) + 2 O2 (g) CO (g) + 8 H2 (g) ΔH = -1 (12.5 points) 2. Reaction of solid white phosphorus (P ) with chlorine gas (Cl 2 ) yields phosphorus trichloride as the sole product at 100 % yield. (a) Write a balanced reaction. P (s) + 6 Cl 2 (g) à PCl (l) (b) When 1.00 g of solid white phosphorus (P ) is reacted with ecess chlorine gas, the reaction gives off 9.27 of heat. What is the heat of reaction per mole of PCl formed? 9.27 q = 1.00gP 0.97gP 1molePCl = molepcl

2 (12.5 points). A bomb calorimetry eperiment is performed with isobutane [(CH) CH molar mass = g / mole] as the combustable substance. The data obtained are: mass of isobutane burned: g final calorimeter temperature: 2.7ºC heat capacity of calorimeter:.97 /ºC initial calorimeter temperature: 22.88ºC a. What is q v, in /g, for the combustion of isobutane? 0 = q cal + q rn so q cal = -q rn q rn.97 = o 1 C ( ) o C = g isobutane g isobutane b. What is q v, in /mole, for the combustion of isobutane? g q rn = 1 = 285 = 2.58 g isobutane mole mole 10 mole c. How many grams of isobutane would need to be burned to produce 1,000 of heat? 1,000 mass = = 20.6 g 9.11 g isobutane (10 points). Calculate the amount of energy required to (a) heat 20.0 grams of water from 50.0ºC to 100.0ºC (the normal boiling point) AND also (b) vaporize ONLY HALF of the water at 100ºC? Make sure the sign of the energy is correct [for water ΔH vap = 0.6 / mole, specific heat of liquid water =.18 J / gºc ] Both processes are endothermic so the heats should be positive. " (a) q heating = ( 20.0g).18 J % $ # g ' C& ( ) C =,18 J or.18 " (b) q vap = 0.6 % " $ ' ( 10.0g) 1mole % $ ' = 22.5 # mole& # g& 2

3 (6 points) 5. Draw Lewis structures for water (H 2 O) and n-heane (CH CH 2 CH 2 CH 2 CH 2 CH ). Discuss why water is a (highly) polar molecule while heane is non-polar. The polarity of a molecule is net combination of the polarities of the bonds in the molecule. Water has very polar bonds between H (electronegativity = 2.2) and O (electronegativity =.). The bond angles are arranged at ~10.5 and thus the dipoles from the two bonds do not cancel. The result is a large separation in charge (negative on the oygen). The bonds between C and H are slightly polar. The electronegativity of H is about 2.2 and that of C is about 2.5. The C-C bond is nonpolar. Each C-H bond will have the same polarity, but the molecule has nearly non-polar nature because the dipoles all nearly cancel. (6 points) 6. Discuss why ethanol (CH CH 2 OH) has a lower boiling point than water (H 2 O) even though ethanol is a larger molecule (with larger dispersion forces) AND is ALSO capable of H-bonding. The ethanol is not as capable of H-bonding (only one H-bonding H per molecule. Also, ethanol is larger as pointed out above. Thus the net amount of H-bonding per unit volume in water is greater than in ethanol. It is interesting to point out that all of the structural isomers of the five-carbon alcohol (C 5 H 12 O) have boiling points above that of water. In other words, the gain in dispersion forces as we go to larger molecules eventually wins out even over the very strong H-bonding found in water. (12.5 points, 2.5 points each) 7. a. the critical temperature is (in ºC) ºC b. the normal boiling point pressure (in atm) 1.00 atm c. the temperature at the triple point is (in ºC) -1ºC d. which letter in the diagram corresponds to the region of gas A e. which has the higher density the solid or liquid phase solid, C 16.5 [not to scale] C B 2 Pressure (atm) 0.22 A Temperature ( C) [not to scale]

4 ( points each, 16 points total) 8. a. Circle the choice with the lower surface tension: C 8 H 18 vs. CH (CH 2 ) 6 OH EXPLAIN your choice.c 8 H 18 has only dispersion forces as compared to the comparable size and shaped alcohol which will have similar dispersion forces but additional H-bonding ability. Fun facts: for C 8 H 18 viscosity = cp at 25 C, for CH (CH 2 ) 6 OH 5.81 cp at 25 C b. Circle the choice with the lower boiling point: CO vs. N 2 EXPLAIN your choice.n 2 is a small non-polar diatomic molecule with only dispersion forces. CO is also a small diatomic molecule, but it is also polar. Both have comparable dispersion forces, but the permanent dipole of CO will raise the b.p. Fun facts: for CO b.p. = 82 K, for N 2 b.p. = 77 K c. Select compound with the highest melting point: NaOH vs. CH OH vs. MgO EXPLAIN your choice. methanol, CH OH is definitely not the right answer b/c it is a small molecule and the other two are ionic compounds. The MgO has +2/-2 charges, the NaOH has +1/-1. The sizes matter less since they are comparable. Fun facts: for CH OH m.p. = 175 K, for NaOH m.p. = 591 K, for MgO m.p. = 125 K d. Circle the substance which has the highest solubility in water: chloroform (CHCl ) vs. benzoic acid (C 6 H 5 COOH) vs. propylene glycol (CH CHOHCH 2 OH) EXPLAIN your choice. The one that is most capable of H-bonding with water while not repelling water with a non-polar portion should have the highest solubility. Propylene glycol has two H-bonding alcohol groups and a small nonpolar part. Benzoic acid can H-bond, but the benzene ring is hydrophobic. Polar chloroform can be epected to dissolve somewhat in water. Fun facts: propylene glycol is fully miscible with water, chloroform is soluble at 8 g / L of water, while benzoic acid is only soluble to a limit of 2.9 g / L of water. (1 points) 9. a. A sample of metal with a mass of g absorbs 99.0 J of heat, and the temperature of the metal increases 1.0ºC. What is the specific heat of the metal in J /gºc a b c. 1.7 d e. none of the above b. Calculate the amount of energy required to (1) melt 18.0 grams of water at 0.0 ºC, (2) heat ALL of that water to ºC (the normal boiling point), and () vaporize HALF of that water at 100 C. Make sure the sign of the energy is correct [for water: ΔHfus = 6.01 /mole, ΔHvap = 0.0 /mole, C p =.18 J /gºc] (1) 18.0g $ # & 1mole $ # & 6.01 $ # & = " %" 18.0g %" mole % (2) 18.0g $ # &.18J $ 1 $ # &# & " %" g C %" 1, 000J % () 18.0g $ # & 1mole $ # &.0 $ # & = " 2 %" 18.0g %" mole % ( ) C = +7.5

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