* The actual temperature dependence for the enthalpy and entropy of reaction is given by the following two equations:

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1 CHM 3400 Problem Set 5 Due date: Tuesday, October 7 th Do all of the following problems. Show your work. "The first essential in chemistry is that you should perform practical work and conduct experiments, for he who does not perform practical work nor makes experiments will never attain the least degree of mastery." - Abu Musa Jabir ibn Hayyan, AD) 1) Consider the following two chemical reactions i. Fe 2O 3(s) + 3 CO(g) 2 Fe(s) + 3 CO 2(g) (reduction of iron) ii. 2 H 2O 2(l) 2 H 2O(l) + O 2(g) (decomposition of hydrogen peroxide) a) Using the information in the appendix of Atkins, find G rxn, H rxn, and S rxn for each of the above reactions, carried out for standard conditions and at T = 25. C. b) Use your answer in part a to confirm that G rxn = H rxn - T S rxn for each reaction. c) Are either of these reactions spontaneous for standard conditions? Justify your answer. 2) As discussed in class, we expect both H rxn, and S rxn to be approximately independent of temperature.* Because of this, we can use the relationship G rxn = H rxn - T S rxn and the condition that G rxn = 0. for a system at equilibrium to find the temperature at which the reactants and products in a chemical reaction are at equilibrium for standard conditions. An interesting application of the above result is estimating from room temperature thermochemical data the temperature at which a phase transition occurs. a) For each of the following processes use the data in the appendix of Atkins to find H rxn, and S rxn at T = 25. C. i Br 2(l) Br 2(g) (vaporization of liquid bromine) ii Al(s) Al(l) (melting of solid aluminum) b) Based on your answer in a, estimate the temperature at which each of the above phase transitions occurs reversibly when p = atm. c) Compare your results in b to experimental results for the temperatures at which the normal phase transitions occur for the above substances, T vap(br 2) = C, and T fus(al) = C. * The actual temperature dependence for the enthalpy and entropy of reaction is given by the following two equations: H rxn(t 2) = H rxn(t 1) T2 + T1 C p dt (2.1) S rxn(t 2) = S rxn(t 1) T2 + T1 ( C p/t) dt (2.2) 3) Neon is a gas at room temperature, but liquid neon forms at low temperatures. The following experimental data are for the vapor pressure of liquid neon T ( C) p (torr) T ( C) p (torr) a) Plot the above data according to the Clausius-Clapeyron equation (that is, plot ln p vs 1/T). b) Find the line that best fits your data plot in a. c) Based on your results above find H vap(ne) and T vap(ne), the enthalpy of vaporization and normal boiling point for neon. (NOTE: You may use EXCEL to plot the data above. This is essentially the same sort of plot you will do in the enthalpy of vaporization of water experiment in CHM 3400L lab.)

2 4) A phase diagram is given below for some ficticious pure chemical substance, and can be used to answer the questions given below. Note that this substance has two different solid phases, and that the pressure axis is logarithmic. a) Which phase is more dense: s 1 or s 2? s 2 or l? l or g? Justify your answer. b) How many triple points are there in the phase diagram? Where is each triple point located (T and p), and what three phases are at equilibrium at each triple point? c) How many normal phase transition points are there in the phase diagram? Where is each of these points located (T and p), and what phase transition is occurring at each point? d) The temperature of 1.00 mol of the above substance is changed reversibly from 270. K to 370. K, at a constant pressure p = 10.0 atm. Describe what happens as the process is carried out. 5) Consider the system pictured below. On the left side of the container there is mol of O 2 (M = g/mol), confined to a volume V = L, while on the right side there is mol of N 2 (M = g/mol), confined to a volume V = L. The barrier separating the gases is removed, and the gases are allowed to mix isothermally until equilibrium is achieved. Find G mix, H mix, and S mix for the process. (HINT: The gases can be assumed to behave ideally for the conditions of the problem. You can use the procedure followed in class for mixing of ideal gases, but note that the initial pressures of gas on the two sides of the barrier will be different, and so you cannot simply use the equations in Chapter 6 of Atkins that apply to ideal mixing of gases at the same initial temperature and pressure.)

3 Solutions. 1) a) i) G rxn = [ 2 G f(fe(s)) + 3 G f(co 2(g)) ] [ G f(fe 2O 3(s)) + 3 G f(co(g)) ] = [ 2 (0.0 kj/mol) + 3 ( kj/mol) ] [ ( kj/mol) + 3 ( kj/mol) ] = kj/mol H rxn = [ 2 H f(fe(s)) + 3 H f(co 2(g)) ] [ H f(fe 2O 3(s)) + 3 H f(co(g)) ] = [ 2 (0.0 kj/mol) + 3 ( kj/mol) ] [ ( kj/mol) + 3 ( kj/mol) ] = kj/mol S rxn = [ 2 S (Fe(s)) + 3 S (CO 2(g)) ] [ S (Fe 2O 3(s)) + 3 S (CO(g)) ] = [ 2 (27.28 J/mol. K) + 3 ( J/mol. K) ] [ (87.40 J/mol. K) + 3 ( J/mol. K) ] = J/mol. K ii) G rxn = [ 2 G f(h 2O( )) + G f(o 2(g)) ] [ 2 G f(h 2O 2( )) ] = [ 2 ( kj/mol) + (0.0 kj/mol) ] [ 2 ( kj/mol) ] = kj/mol H rxn = [ 2 H f(h 2O( )) + H f(o 2(g)) ] [ 2 H f(h 2O 2( )) ] = [ 2 ( kj/mol) + (0.0 kj/mol) ] [ 2 ( kj/mol) ] = kj/mol S rxn = [ 2 S (H 2O( )) + S (O 2(g)) ] [ 2 S (H 2O 2( )) ] = [ 2 (69.91 J/mol. K) + ( J/mol. K) ] [ 2 (109.6 J/mol. K) ] = J/mol. K b) i) G rxn = H rxn T S rxn = ( kj/mol) (298. K) ( kj/mol. K) = kj/mol The result agrees with the value found directly from the free energy data to within roundoff error. ii) G rxn = H rxn T S rxn = ( kj/mol) (298. K) ( kj/mol. K) = kj/mol The result agrees with the value found directly from the free energy data to within roundoff error. c) Since for both reactions G rxn < 0, both reactions are spontaneous for standard conditions. 2) a) i) H rxn = [ H f(br 2(g))] [ H f(br 2(l))] = kj/mol S rxn = [S (Br 2(g))] [S (Br 2(l))] = ( ) J/mol K = J/mol K ii) H rxn = [ H f(al(l))] [ H f(al(s))] = kj/mol S rxn = [S (Al(l))] [S (Al(s))] = ( ) J/mol K = J/mol K b) T b*(br 2) = ( J/mol)/(93.23 J/mol K) = K = 58.4 C T f*(al) = ( J/mol)/(11.22 J/mol K) = K = C

4 c) For Br 2, T b* = C vs the value 58.4 C found above, a difference of 0.4 C For Al, T f* = C vs the value C found above, a difference of 7.8 C Both calculated values are close to the true values. The Br2 value is closer than the value for Al because the temperature at which the phase transition occurs is closer to 25.0 C. and so the neglect of accounting for the temperature dependence of H rxn, and S rxn is less important. 3) a) T ( C) p (torr) T(K) (1/T) (K -1 ) ln(p) b) The data are plotted below: Plot of ln(p) vs (1/T) 8.5 ln(p) (1/T) (K-1) Based on the plot, I get slope = K, intercept = c) H vap = - mr = - ( K)(8.314 J/mol K) = J/mol To find T b we can solve our equation for the line for 1/T ln(p) = (224. K)(1/T) Note that at the normal boiling point p =760. torr (ln(p) = 6.633) (1/T) = ln(p) = = K -1, and so T b = K ( K) ( K)

5 4) a) If dp/dt > 0 then the phase to the right of the boundary between phases is lower in density, while if dp/dt < 0 then the phase to the left of the boundary between phases is lower in density. Using this s 1 or s 2 dp/dt > 0, and so (s 1) > (s 2) s s or l l or g dp/dt > 0, and so (s 2) > (l) dp/dt > 0, and so (l) > (g) b) There are two triple points shown in the phase diagram. Their locations are p = 0.1 atm, T = 265 K p = 0.3 atm, T = 290 K s 1, s 2, g s 2, l, g There is clearly a third triple point that will occur, with s 1, s 2, and l existing simultaneously at equilbrium, but it falls outside of the range of values shown in the phase diagram. c) A normal phase transition occurs when two phases exist simultaneously at equilibrium when p = 1.00 atm. For this phase diagram three such points occur p = 1.00 atm, T = 275 K p = 1.00 atm, T = 292 K p = 1.00 atm, T = 332 K s 1 s 2 transition (solid to solid) s 2 l transition (solid to liquid, or melting) l g transition (liquid to gas, or vaporization) d) The substance is in the s 1 solid phase at p = 10.0 and T = 270 K. The substance warms until T = 284 K, when the s 1 s 2 phase transition occurs. Between 284 K and 300 K the substance remains in the s 2 solid phase. At T = 300 K the s 2 l phase transition occurs. Between 300 K and 360 K the substance remains in the liquid phase. At T = 360 K the l g phase transition occurs. Between 360 K and 370 K, the final temperature, the substance remains a gas. 5) We can do this problem the same way we proceeded in class. However, in this problem the initial pressures are not equal, and so we will need to take that into account. First of all, since the gases are ideal and the process is isothermal, H mix = 0. G mix = G f - G i where G = n O2 O2 + n N2 N2 and j = j + RT ln(p j) We have enough information to calculate pressure, since we know n, V, and T, and can assume the gases behave ideally. From the ideal gas law p = nrt/v

6 Using this, we can find the initial and final pressures for O 2 and N 2. p O2,i = (1.000 mol) ( L.atm/mol. K) (300.0 K) = atm (20.00 L) p O2,f = (1.000 mol) ( L.atm/mol. K) (300.0 K) = atm (30.00 L) p N2,i = (1.000 mol) ( L.atm/mol. K) (300.0 K) = atm (10.00 L) p N2,f = (1.000 mol) ( L.atm/mol. K) (300.0 K) = atm (30.00 L) So G i = (1.000 mol) ( O2 + RT ln(1.2309)) + (1.000 mol) ( N2 + RT ln(2.4618)) G f = (1.000 mol) ( O2 + RT ln(0.8206)) + (1.000 mol) ( N2 + RT ln(0.8206)) and so G mix = G f - G i = (1.000 mol) ( O2 + RT ln(0.8206)) + (1.000 mol) ( N2 + RT ln(0.8206)) = J - (1.000 mol) ( O2 + RT ln(1.2309)) + (1.000 mol) ( N2 + RT ln(2.4618)) Finally, since T is constant G mix = H mix - T S mix Since H mix = 0., it follows that S mix = - G mix/t = - ( J)/(300.0 K) = J/K