Chem 75 February, 2017 Practice Exam 2

Size: px

Start display at page:

Download "Chem 75 February, 2017 Practice Exam 2"

Frederick Perry
6 years ago
Views:

1 As before, here is last year s Exam 2 in two forms: just the questions, and then the questions followed by their solutions. 1. ( points) At high temperature, aluminum nitride, AlN(s), decomposes stoichiometrically to Al(g) and N 2 (g). One experiment starting from pure AlN and nothing else measured a total pressure of bar at 1919 K. (a) Write a balanced chemical reaction representing this process. (b) Write an equilibrium constant expression consistent with your reaction in (a), and find its numerical value at 1919 K. (c) The standard molar enthalpy of formation of AlN(s) is kj mol 1 at 1919 K, that for Al(g) is kj mol 1, while that for N 2 (g) is zero. Find the standard molar entropy change, ΔS R o, for this reaction, and justify the algebraic sign of your answer in terms of the way you wrote the reaction in part (a). 2. (18 points) Large South American beetles, known as bombardier beetles, spray their attackers with a hot, irritating solution of quinone (symbolized Q here) in water. The beetle stores an aqueous solution which is 25% by weight hydrogen peroxide (H 2 O 2 ) and 10% by weight hydroquinone (symbolized H 2 Q). When attacked or threatened, the beetle releases an enzyme into this solution to promote the oxidation of H 2 Q: H 2 Q(aq) + H 2 O 2 (aq) Q(aq) + 2 H 2 O(l) (reaction 1) and the decomposition of excess peroxide: H 2 O 2 (aq) H 2 O(aq) + 1 O 2(g) (reaction 2) 2 The O 2 (g) helps squirt the fluid onto the beetle s attacker. Assuming that the heat capacity of the spray mixture is the same as for pure water and assuming that the catalytic enzyme cause these reactions to go to completion very fast, calculate the spray s temperature. Assume the beetle s temperature is 35 C initially. Relevant data are collected below. Note that the heat capacity is given per gram of water, not per mole, and assume the beetle s reaction solution has this heat capacity, too. Molar masses: H 2 Q = 110 g mol 1 H 2 O 2 = 34 g mol 1 H 2 O = 18 g mol 1 o ΔH R(reaction 1) = 203. kj mol 1 o ΔH R(reaction 2) = 94.6 kj mol 1 C P o (H2 O) = 4.2 J K 1 g 1 3. ( points) On the front page of the exam, you ll find the Clapeyron expression for the constraint that relates P and T for any two-phase equilibrium of a pure substance. It is exact. Following this, you ll find the Clausius-Clapeyron equation that we derived from it, which holds for a gas-liquid equilibrium. Our derivation required three distinct conceptual steps. Tell me below what these three steps were, paying attention to the way each step is described and starting with the Clapeyron expression. You may use English as well as or in place of math in your answers! (The two expressions referenced here are at the top of the next page.)

2 Clapeyron: dp = ΔS φ Clausius-Clapeyon: ln P 1 = ΔH o vap 1 dt φ ΔV φ P 2 R T 1 1 T 2 (a) First, we invoked an exact relation, i.e., an identity. What was it? (b) Next, we invoked two very good approximations. What were they? (c) Finally, we made a reasonable assumption that led to the final Clausius- Clapeyron equation. What was it? 4. (3 points each) For each pair of quantities below, the one on the left is either greater than (>), less than (<), or equal to (=) the one on the right. Enter the appropriate symbol (>, <, =) in the blanks provided in the middle. (a) ΔS for the phase transition Ne(g) Ne(l) is zero. (b) Methane is more soluble in water than is neon. Thus, the Henry s Law constant k H for Ne in water is the Henry s Law constant for CH 4 in water. (c) ΔG for the creation of an ideal binary mixture is zero. (d) The osmotic pressure of a M solution of NaCl is the osmotic pressure of a M solution of sucrose. 5. (6 points) The free energies of formation, ΔG f o, of Fe2 O 3 (s) and Au 2 O 3 (s) are known. One is 126 kj mol 1 and the other is 742 kj mol 1. Which is which, and how can you tell? 6. (6 + 6 points) Two quick questions: (a) Here s a P-V plot showing a van der Waals gas isotherm for some temperature below the critical temperature T c. Draw, as accurately as you can, the isotherm for T = T c. The critical point is at the red dot, at (P c, V c ). P c P V c V

3 (b) In 1917, our ol friend G. N. Lewis teamed up with his colleague H. Storch to measure the vapor pressure of Br 2 over room-temperature solutions of Br 2 in CCl 4. Here are some of their data (and the notation 10 3 x Br 2 means the numbers in the table should be multiplied by 10 3 to yield the mole fraction): 10 3 x Br P Br 2 /Torr The vapor pressure of pure Br 2 is P o Br2 = 228 Torr. Does Br 2 in these solutions follow Raoult s or Henry s law (or maybe neither!) across this concentration range?

4 Solutions 1. ( points) At high temperature, aluminum nitride, AlN(s), decomposes stoichiometrically to Al(g) and N 2 (g). One experiment starting from pure AlN and nothing else measured a total pressure of bar at 1919 K. (a) Write a balanced chemical reaction representing this process. Any balanced net reaction works, but what we write here will affect everything to follow! Let s write it this way: 2 AlN(s) 2 Al(g) + N 2(g) (b) Write an equilibrium constant expression consistent with your reaction in (a), and find its numerical value at 1919 K. Based on our answer in (a), the K expression has to be 2 PN2 K = P Al P o 3 where I ve included the standard pressure P o just to remind us that we want the gas partial pressures in bar units. The total pressure is the sum of the Al gas and N 2 gas partial pressures, and their ratio is determined by reaction stoichiometry: every N 2 molecule has two Al atoms associated with it: P Al = 2P N 2. Thus, we have P = bar = P Al + P N2 = 2P N2 + P N2 = 3P N2 P N2 = bar and P Al = bar K = P 2 AlPN2 = = (c) The standard molar enthalpy of formation of AlN(s) is kj mol 1 at 1919 K, that for Al(g) is kj mol 1, while that for N 2 (g) is zero. Find the standard molar entropy change, ΔS R o, for this reaction, and justify the algebraic sign of your answer in terms of the way you wrote the reaction in part (a). Now we calculate the thermodynamic quantities that lead to the reaction entropy change: ΔG o R = RTln K = 388 kj mol 1 ΔH o R = 2ΔHf o (Al(g)) + ΔHf o (N2 (g)) 2ΔH o f (AlN(s)) = kj mol kj mol 1 = 1252 kj mol 1 o ΔS o ΔH R = R ΔGR o = 450 J mol T 1 K 1 and we see that ΔS for the reaction as we wrote it is positive, as expected, reflecting the production of three moles of gaseous products per two moles of solid reactants. 2. (18 points) Large South American beetles, known as bombardier beetles, spray their attackers with a hot, irritating solution of quinone (symbolized Q here) in water. The beetle stores an aqueous solution which is 25% by weight hydrogen peroxide (H 2 O 2 )

5 and 10% by weight hydroquinone (symbolized H 2 Q). When attacked or threatened, the beetle releases an enzyme into this solution to promote the oxidation of H 2 Q: H 2 Q(aq) + H 2 O 2 (aq) Q(aq) + 2 H 2 O(l) (reaction 1) and the decomposition of excess peroxide: H 2 O 2 (aq) H 2 O(aq) + 1 O 2(g) (reaction 2) 2 The O 2 (g) helps squirt the fluid onto the beetle s attacker. Assuming that the heat capacity of the spray mixture is the same as for pure water and assuming that the catalytic enzyme cause these reactions to go to completion very fast, calculate the spray s temperature. Assume the beetle s temperature is 35 C initially. Relevant data are collected below. Note that the heat capacity is given per gram of water, not per mole, and assume the beetle s reaction solution has this heat capacity, too. Molar masses: H 2 Q = 110 g mol 1 H 2 O 2 = 34 g mol 1 H 2 O = 18 g mol 1 ΔH o R(reaction 1) = 203. kj mol 1 ΔH o R(reaction 2) = 94.6 kj mol 1 C P o (H2 O) = 4.2 J K 1 g 1 Here s our plan of attack. Since we re interested in the final temperature, an intensive quantity, we can consider the fate of any amount of solution. Let s pick 1 g. In this gram, we have 0.25 g or 7.35 mmol of H 2 O 2 and 0.10 g or mmol of H 2 Q. For reaction 1, H 2 Q is the limiting reagent, and consuming all of it leads to an enthalpy change of ( 203. kj mol 1 ) (0.909 mmol) = J. The left-over H 2 O 2 is consumed in reaction 2, leading to an enthalpy change of ( 94.6 kj mol) (7.35 mmol mmol) = J. Because the reaction is fast, we can assume it happens adiabatically, which means here that the enthalpy released in the reaction has no place to go but into our little gram of solution. The total released enthalpy, J in magnitude, is a heat q that raises the solution temperature. We write q = C P ΔT with C P = 4.2 J K 1 (since we can assume we still have roughly 1 g of solution with a specific heat capacity of 4.2 J K 1 g 1 ) and with ΔT = the temperature rise. We find ΔT = 189 K and go WHAT? That can t be right! That would make the final temperature 224 C! Water is no longer a liquid at that temperature! So what happens? Simple: the beetle produces a boiling hot solution! The final temperature must be 100 C (OK - there would be a touch of boiling point elevation, but let s ignore that here). But do we boil away all the water? Raising 1 g of solution from 35 C to 100 C takes only (65 K) (4.2 J K 1 ) = 273 J of our total reaction enthalpy release, leaving ( ) J = J left to boil some water. You might recall that the molar enthalpy of vaporization of water is about 40 kj mol 1 (but you weren t expected to remember this). To boil all of our 1 g sample (which is 1/18 mol = 55.6 mmol of solution) would take about (40 kj mol 1 ) (55.6 mmol) = 2.2 kj. So we don t have enough released enthalpy to boil all our sample, but some of it boils, releasing hot, pressurized steam to help expel the hot, nasty solution from the beetle. (Just the fact that the beetle squirts some liquid solution tell us that it didn t all boil away!) What about the O 2 product? We have to heat it, too, but it s easy to show that we produce so little of it to significantly change our conclusion. This is one impressive little beetle; it can stand an internal temperature of 100 C!

6 3. ( points) On the front page of the exam, among the Potentially useful things to know material, you ll find the Clapeyron expression for the constraint that relates P and T for any two-phase equilibrium of a pure substance. It is exact. Following this, you ll find the Clausius-Clapeyron equation that we derived from it, which holds for a gas-liquid equilibrium. Our derivation required three distinct conceptual steps. Tell me below what these three steps were, paying attention to the way each step is described and starting with the Clapeyron expression. You may use English as well as or in place of math in your answers! (a) First, we invoked an exact relation, i.e., an identity. What was it? In the Clapeyron equation, we replaced the phase transition entropy change, ΔS φ, with the enthalpy change divided by the transition temperature. This is exact: ΔS φ = ΔH φ T φ where we want "φ" to be "vap," for the l g transition. (b) Next, we invoked two very good approximations. What were they? Our Clausius equation now looks like this, and we make the two approximations as indicated (where we also take each quantity to be per mole, as needed): dp dt φ = ΔS φ ΔV φ = dp ΔH vap dt φ T vap V(g) dp ΔH vapp dt φ T 2 vap R ΔH vap T vap ΔV vap exact ΔV vap = V(g) V(l) V(g), because V(g) >> V(l) using the ideal gas approximation, V(g) = RT vap P (c) Finally, we made a reasonable assumption that led to the final Clausius- Clapeyron equation. What was it? To integrate the Clausius equation, we assumed that the enthalpy of vaporization was independent of temperature over the range from T 1 to T 2 : dp = ΔH vapp dt φ T vap RT becomes dp = ΔH vap dt P T 2 vap R and we integrate: P 2 dp = ln P 2 = P P 1 P 1 T 2 T 1 ΔH vap dt T 2 R ΔH vap R T 2 T 1 dt T 2 = ΔH vap R 1 T 2 1 T 1

7 4. (3 points each) For each pair of quantities below, the one on the left is either greater than (>), less than (<), or equal to (=) the one on the right. Enter the appropriate symbol (>, <, =) in the blanks provided in the middle. (a) ΔS for the phase transition Ne(g) Ne(l) is < zero. Liquid entropies are less than gas entropies for any one substance. (b) Methane is more soluble in water than is neon. Thus, the Henry s Law constant k H for Ne in water is > the Henry s Law constant for CH 4 in water. For equal partial pressures of the two gases, the solubility (given by the gas solute molality) is inversely proportional to its Henry s Law constant, k H. So k H for Ne must be larger than that for methane to make methane more soluble. (c) ΔG for the creation of an ideal binary mixture is < zero. Ideal binary mixture formation is spontaneous (at fixed P and T). Thus ΔG < 0. (d) The osmotic pressure of a M solution of NaCl is > the osmotic pressure of a M solution of sucrose. Sucrose doesn t dissociate, but NaCl does. So the number of solute particles in the NaCl solution is twice that in the sucrose solution, giving it the greater osmotic pressure. 5. (6 points) The free energies of formation, ΔG f o, of Fe2 O 3 (s) and Au 2 O 3 (s) are known. One is 126 kj mol 1 and the other is 742 kj mol 1. Which is which, and how can you tell? Easy! Iron rusts, but gold doesn t! That means 2Fe(s) + (3/2)O 2 (g) Fe 2 O 3 (s) has a negative ΔG R, which here equals the ΔG f o of Fe2 O 3. So its ΔG f o = 742 kj mol 1. The Au 2 O 3 (s) ΔG f o must be positive. Its oxidation reaction has a positive ΔGR. 6. (6 + 6 points) Two quick questions: (a) Here s a P-V plot showing a van der Waals gas isotherm for some temperature below the critical temperature T c. Draw, as accurately as you can, the isotherm for T = T c. The critical point is at the red dot, at (P c, V c ).

8 P Here s the accurately drawn critical temperature isotherm. Note how it goes right through the critical point with zero slope and zero second derivative. Pretty much anything that came close to these properties was accepted! (b) In 1917, our ol friend G. N. Lewis teamed up with his colleague H. Storch to measure the vapor pressure of Br 2 over room-temperature solutions of Br 2 in CCl 4. Here are some of their data (and the notation 10 3 x Br 2 means the numbers in the table should be multiplied by 10 3 to yield the mole fraction): 10 3 x Br P Br 2 /Torr The vapor pressure of pure Br 2 is P o Br2 = 228 Torr. Does Br 2 in these solutions follow Raoult s or Henry s law (or maybe neither!) across this concentration range? If we assume Raoult s law, we can predict the following partial pressures from P i = x i P * i with Pi * = 228 Torr: 10 3 x Br P Br 2 /Torr These predicted partial pressures are not at all close to the observed values! Let s try Henry s law. We can calculate an observed Henry s law constant, k H, for each experiment, and if these are close, then we can say Henry s law works! Here s what we get: V

9 10 3 x Br k H /Torr Those values are acceptably close to each other; so, we can conclude that Henry s law seems to be followed, at least over this concentration range. That s what Lewis concluded, too (but from many more measurements than the four I ve shown here)!

The Second Law of Thermodynamics (Chapter 4)

The Second Law of Thermodynamics (Chapter 4) First Law: Energy of universe is constant: ΔE system = - ΔE surroundings Second Law: New variable, S, entropy. Changes in S, ΔS, tell us which processes made