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Thermodynamics - Worksheet 12 - Fall 2007 | CH 301, Exams of Chemistry

The University of Texas at AustinChemistry
Prof. David A. Laude

WORKSHEET 12 Material Type: Exam; Professor: Laude; Class: PRINCIPLES OF CHEMISTRY I; Subject: Chemistry; University: University of Texas - Austin; Term: Fall 2007;

Typology: Exams

Pre 2010

Uploaded on 08/30/2009

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Fall 2007 CH 301 Worksheet 12--Thermodynamics
1. A small (74 g) serving of French fries is burned in a bomb calorimeter containing 3 L of water. The
temperature of the water increases by 82°C. The calorimeter has a heat capacity of 200 J/°C, the density
of water is 1 g/mL, and the heat capacity of water is 4.18 J/g°C. How much heat is evolved per gram of
french fries?
14.1 kJ/g
2. The same calorimeter as in number 1 is used to measure the enthalpy of dissolving 2.5 mol of potassium
chloride (KCl) in water. If the enthalpy of the process is ∆H = +15 kJ/mol and the initial temperature of
the water is 298 K, what is the final temperature of the water?
295.1 K
3. Given
C2H2 + 5/2 O2 →2 CO2 + H2O ∆H = -1300 kJ/mol
C2H2 + H2 →C2H4 ∆H = -175 kJ/mol
C2H4 + H2O →CH3CH2OH ∆H = -44 kJ/mol
H2 + ½ O2 →H2O ∆H = -286 kJ/mol
Find ∆H for the combustion of ethanol,
CH3CH2OH + 3 O2 → 2 CO2 + 3 H2O
-1367 kJ/mol
4. Given the following information, calculate the enthalpy change for the reaction
4 HNO3(l) + 5 N2H4(l) → 7 N2(g) + 12 H2O(l)
∆Hf° (HNO3(l)) = -174.10 kJ/mol
∆Hf° (N2H4(l)) = +50.63 kJ/mol
∆Hf° (H2O(l)) = -285.83 kJ/mol
-2987 kJ/mol
5. Calculate the free energy change of the same reaction at 298 K, given
Sm° (HNO3(l))= 155.60 J/K•mol
Sm° (N2H4(l))= 121.21 J/K•mol
Sm° (N2(g))= 191.61 J/K•mol
Sm° (H2O(l))= 69.91 J/K•mol
-3269 kJ/mol
5. Calculate the enthalpy change for the combustion of ethanol using bond energies.
C-H = 412 kJ/mol
C-C = 348 kJ/mol
C-O = 360 kJ/mol
C=O = 743 kJ/mol
O=O = 497 kJ/mol
O-H = 463 kJ/mol
-1028 kJ/mol
6. Using bond energies, calculate the enthalpy change for the reaction
CH4(g) + 4 Cl2(g) → CCl4(g) + 4 HCl(g)
C-H = 412 kJ/mol
C-Cl = 338 kJ/mol
Cl-Cl = 242 kJ/mol
Cl-H = 431 kJ/mol
-460 kJ/mol
7. Use enthalpies of formation to find the enthalpy change for the same reaction.
∆Hf° (CH4(g)) = -74.81 kJ/mol
∆Hf° (CCl4(g)) = -163.78 kJ/mol
∆Hf° (HCl(g)) = -92.31 kJ/mol
-458.21 kJ/mol
8. Find the approximate work done for the following reaction at 400 K.
H2 (g) + ½ O2 (g) → H2O (g) 1.67 kJ
pf3

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Fall 200 7 CH 301 Worksheet 12--Thermodynamics

  1. A small (74 g) serving of French fries is burned in a bomb calorimeter containing 3 L of water. The temperature of the water increases by 82°C. The calorimeter has a heat capacity of 200 J/°C, the density of water is 1 g/mL, and the heat capacity of water is 4.18 J/g°C. How much heat is evolved per gram of french fries? 14.1 kJ/g
  2. The same calorimeter as in number 1 is used to measure the enthalpy of dissolving 2.5 mol of potassium chloride (KCl) in water. If the enthalpy of the process is ∆H = +15 kJ/mol and the initial temperature of the water is 298 K, what is the final temperature of the water? 295.1 K
  3. Given C 2 H 2 + 5/2 O 2 → 2 CO 2 + H 2 O ∆H = -1300 kJ/mol C 2 H 2 + H 2 → C 2 H 4 ∆H = -175 kJ/mol C 2 H 4 + H 2 O → CH 3 CH 2 OH ∆H = -44 kJ/mol H 2 + ½ O 2 → H 2 O ∆H = -286 kJ/mol Find ∆H for the combustion of ethanol, CH 3 CH 2 OH + 3 O 2 → 2 CO 2 + 3 H 2 O -1367 kJ/mol
  4. Given the following information, calculate the enthalpy change for the reaction 4 HNO 3 (l) + 5 N 2 H 4 (l) → 7 N 2 (g) + 12 H 2 O(l) ∆Hf ° (HNO 3 (l)) = -174.10 kJ/mol ∆Hf ° (N 2 H 4 (l)) = +50.63 kJ/mol ∆Hf ° (H 2 O(l)) = -285.83 kJ/mol -2987 kJ/mol
  5. Calculate the free energy change of the same reaction at 298 K, given S (^) m° (HNO 3 (l))= 155.60 J/K•mol S (^) m° (N 2 H 4 (l))= 121.21 J/K•mol

S (^) m° (N 2 (g))= 191.61 J/K•mol S (^) m° (H 2 O(l))= 69.91 J/K•mol -3269 kJ/mol

  1. Calculate the enthalpy change for the combustion of ethanol using bond energies. C-H = 412 kJ/mol C-C = 348 kJ/mol

C-O = 360 kJ/mol C=O = 743 kJ/mol

O=O = 497 kJ/mol O-H = 463 kJ/mol -1028 kJ/mol

  1. Using bond energies, calculate the enthalpy change for the reaction CH 4 (g) + 4 Cl 2 (g) → CCl 4 (g) + 4 HCl(g) C-H = 412 kJ/mol C-Cl = 338 kJ/mol

Cl-Cl = 242 kJ/mol Cl-H = 431 kJ/mol -460 kJ/mol

  1. Use enthalpies of formation to find the enthalpy change for the same reaction. ∆Hf ° (CH 4 (g)) = -74.81 kJ/mol ∆Hf ° (CCl 4 (g)) = -163.78 kJ/mol ∆Hf ° (HCl(g)) = -92.31 kJ/mol -458.21 kJ/mol
  2. Find the approximate work done for the following reaction at 400 K. H 2 (g) + ½ O 2 (g) → H 2 O (g) 1.67 kJ
  1. For the reaction 3 H 2 (g) + N 2 (g) → 2 NH 3 (g) how many moles of hydrogen gas must be reacted for the work to be 8 kJ at 300 K? 2.4 mol
  2. A reaction occurs in a beaker. You touch the beaker and it feels cold. What is the sign of ∆H for the reaction? What can you say about the sign of ∆S? ∆H > 0, ∆S < 0
  3. A reaction happens in a balloon, and in the end, the volume of the balloon has doubled. What is the sign of work for the reaction? negative
  4. Give the sign of the entropy change of the system for the following processes: a. Dr. Laude pours hot water in a tub of liquid N2 and makes a thundercloud. positive b. Water freezes. negative c. Two cars are in a head-on collision. positive d. Sugar is dissolved in a drink. positive
  5. For the following reaction, what happens to the entropy of the system? 2 CH 4 (g) + 3 Cl 2 (g) → 2 CHCl 3 (l) + 3 H 2 (g) It decreases.
  6. The reaction above happens. What can you say about its enthalpy change? It is exothermic.
  7. From your knowledge of the temperature dependence of (a) water boiling and (b) water freezing, predict the sign of ∆H and ∆S. (a) ∆H > 0, ∆S > 0; (b) ∆H < 0, ∆S < 0
  8. The reaction H 2 (g) + ½ O 2 (g) → H 2 O (g) is exothermic. Is its spontaneity temperature-dependent? In what way? Yes. It is only spontaneous at low temperatures, where favorable decrease in enthalpy overcomes the unfavorable decrease in entropy.
  9. For each of the following combinations of enthalpy and entropy change, tell whether it occurs always , never , at high temperature , or at low temperature. a. ∆H < 0, ∆S < 0 at low temperature b. ∆H < 0, ∆S > 0 always c. ∆H > 0, ∆S < 0 never d. ∆H > 0, ∆S > 0 at high temperature
  10. Explain why, although water has three vibrational degrees of freedom, carbon dioxide must have four. (Hint: both molecules have 9 total degrees of freedom)