Chemistry 2000 (Fall 2008) Problem Set #5: Entropy, Free Energy and Equilibrium Solutions Textbook Questions • See solutions guide for answers Additional Practice Problems 1. Hydrogen gas has ∆fH˚ = 0 J·mol-1 and ∆fG˚ = 0 J·mol-1, but its S˚ = 130.7 J·mol-1·K-1. The enthalpy and free energy values are zero; why isn’t the entropy value also zero? Enthalpy of formation (and free energy of formation) is the enthalpy change (and free energy change) involved in producing a compound from its elements in their standard states. Since hydrogen gas is an element in its standard state, its enthalpy of formation (and free energy of formation) is zero. Entropy is derived from the number of possible microstates (positional and energetic) for a molecule, and is measured in comparison to a perfect crystal of the substance at 0 K. There are many more ways to distribute energy in a sample of hydrogen gas under standard conditions (298 K) than there are in a sample of crystalline hydrogen at 0 K. As such, S˚ for hydrogen has a positive value – not zero. 2. (a) Suppose that you have two opaque jars which contain a total of 8 identical marbles. What is the entropy of this system? Express your answer as a multiple of kB. S = kB ln(9) There are nine possible microstates: • Jar A = 8 marbles, Jar B = 0 marbles • Jar A = 7 marbles, Jar B = 1 marble • Jar A = 6 marbles, Jar B = 2 marbles • Jar A = 5 marbles, Jar B = 3 marbles • Jar A = 4 marbles, Jar B = 4 marbles • Jar A = 3 marbles, Jar B = 5 marbles • Jar A = 2 marbles, Jar B = 6 marbles • Jar A = 1 marble, Jar B = 7 marbles • Jar A = 0 marbles, Jar B = 8 marbles (b) Suppose that you have two opaque jars which contain 4 red marbles and 4 blue marbles. What is the entropy of this system? S = kB ln(52) = kB ln(25) Now, it is necessary to consider the colours of the marbles in each jar. There are 52 = 25 possible microstates for this system. A sample microstate is: • Jar A = 1 red marble/1 blue marble, Jar B = 3 red marbles/3 blue marbles (c) Does having two different kinds of marbles make a difference to the entropy of the system? Yes. While the number of marbles was the same, having to discriminate between red and blue marbles increased the entropy of the system by increasing the number of “available microstates”. 3. There are two isomers of C3H6. Both are gases under standard conditions. (a) Draw the Lewis structure of each isomer. H H H C H C H C C H H C C H H H H H propene cyclopropane (b) Which isomer would you expect to have a larger entropy under standard conditions? (assuming the same volume of sample) Why? Propene should (and does) have a larger entropy at room temperature. It is possible to rotate about the C-C single bond, giving a large number of possible rotational microstates for the CH3 group of each propene molecule. In cyclopropane, the three carbon atoms are locked into a triangle, so there is only one equivalent microstate. Both molecules also have microstates associated with position of the gas molecules in the sample (both location and orientation) and energies of the gas molecules in the sample. Because we are talking about gases, we do not need to consider the ordering effect of intermolecular forces. (c) What would be the sign of the entropy change for the condensation of either of these gases? Condensation of a gas to either a liquid or solid always has a negative entropy change. 4. The reaction 4HCl(g) + O2(g) → 2Cl2(g) + 2H2O(g) comes to equilibrium at 400oC in a 2L flask. Analysis of the equilibrium mixture shows that it contains 60g of Cl2, 12g of H2O, 20g of HCl and 8g of O2. Calculate the equilibrium constant. K = 0.51 5. The equilibrium constant for the reaction H2(g) + I2(s) → 2HI(g) is 0.352 at 25oC. Suppose than an excess of solid iodine is placed in a rigid flask with 0.400 bar of hydrogen gas and 0.300 bar of hydrogen iodide. In what direction will the reaction proceed? The reaction will proceed forward (as written). 6. Calculate the equilibrium constant for the reaction CO(g) + Cl2(g) → COCl2(g) at 25oC. K = 6.4 × 1011 7. Using the solubility product of barium sulfate and the standard free energies of formation of solid barium sulfate and of the aqueous sulfate ion, calculate the standard free energy of formation of the aqueous Ba2+ ion and compare the value obtained to that given in the data tables in your textbook. See value in Appendix D 8. What is the vapour pressure of a solution made by dissolving 50g of ammonium sulphate into 300g of water at 40oC? The vapour pressure of pure water at this temperature is 7373 Pa. PH2O-solution = 6902 Pa 9. In lab, we often rinse wet glassware with acetone to remove the water then use a stream of air to evaporate off the acetone. The structure of acetone and vapour pressure curves for acetone and water are shown below. (1 atm = 1.01325 bar = 760 mmHg) Vapour Pressure Curves for Acetone and Water 800 Equilibrium Vapour Pressure (mmHg) 700 600 500 Acetone 400 Water 300 200 P20'C(acetone) = ~178 mmHg 100 0 -80 -60 -40 -20 0 20 40 60 80 100 120 P20'C(H2O) = 17 mmHg o Tobp(acetone) = 56.5oC Tobp(H2O) = 100oC Temperature ( C) (a) What type(s) of intermolecular forces are responsible for water’s solubility in acetone? dipole-dipole forces (including hydrogen bonding between H of water and O of acetone as well as hydrogen bonding between H of one water molecule and O of another water molecule) (b) What are the normal boiling points of acetone and water? see graph above; normal boiling point is the vapour pressure at 1 atm (760 mmHg); interpolation on the graph gives the normal boiling point of acetone as ~56 ˚C and the normal boiling point of water as ~100 ˚C. (c) What are the vapour pressures of acetone and water at room temperature (20 ˚C)? see graph above; interpolation gives a vapour pressure of ~17 mmHg = 0.023 bar for water and a vapour pressure of ~178 mmHg = 0.237 bar for acetone. (d) Briefly, justify the relative boiling points and vapour pressures of water and acetone. Water molecules can hydrogen bond with each other. Acetone has no hydrogen atoms bonded to N, O or F so acetone molecules cannot hydrogen bond with each other. As such, the intermolecular forces between water molecules are stronger than the intermolecular forces between acetone molecules. If the intermolecular forces are stronger, fewer molecules will have enough energy to escape the liquid phase. Thus, the vapour pressure of water is lower. (e) Why does blowing a stream of air over acetone-wet glassware accelerate evaporation? The liquid acetone and acetone vapour exist in equilibrium. The stream of air reduces the vapour pressure of acetone so that more acetone must evaporate to reach the equilibrium vapour pressure. In other words, the air pushes the acetone vapour away, so more acetone has to evaporate to restore the equilibrium. (f) Why is blowing a stream of air over water-wet glassware much less effective? Water has a much lower equilibrium vapour pressure than acetone. As such, there are fewer water vapour molecules to push away and fewer water molecules need to evaporate to restore the equilibrium. This method would eventually work but it might take hours! (g) If you need dry glassware for the following week’s lab, is this method of drying necessary? Why or why not? No, it isn’t necessary. As long as the wet glassware is not stored in a sealed environment, the water (or acetone) will slowly evaporate until the glassware is dry.