Dr. JD Laposa Due December 2, 1998 - Chemistry 1E03-01: Tutorials

CHEMISTRY 1E03-01 ASSIGNMENT #3 Dr. J. D. Laposa Due December 2, 1998

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1. a) The U. S. government recently required that automobile fuels consist of a renewable component. The fermentation of glucose, C6H12O6(s), from corn is the key process to fulfill this requirement:

C6H12O6(s) --> 2 C2H5OH(l) + 2 CO2(g)

Calculate Delta Ho, Delta So, and Delta Go for this reaction at 25oC. Note: for glucose, use Delta Hof = - 1274.5 kJ/ mol, So = 212.1 J/(mol K), Delta Gof = - 910.56 kJ/mol.

Delta Ho = 2 Delta Hof(C2H5OH, l) + 2 Delta Hof(CO2, g) - Delta Hof(C6H12O6, s)

= [2(-277.7) + 2(-393.5) - (-1274.5)] kJ = -67.9 kJ

Delta So = 2 So(C2H5OH, l) + 2 So(CO2, g) - So(C6H12O6, s)

= [2(160.7) + 2(213.7) - (212.1)] J/K = +536.7 J/K = 0.5367 kJ/K

Delta Go = Delta Ho - T x Delta So = [-67.9 - (298 x 0.5367)] kJ = -227.8 kJ

Alternate method:

Delta Go = 2 Delta Gof(C2H5OH, l) + 2 Delta Gof(CO2, g) - Delta Gof(C6H12O6, s)

= [2(-174.9) + 2(-394.4) - (-910.56)] kJ = -228 kJ

b) Is the spontaneity of this reaction temperature dependent (assume that Delta Ho and Delta So are temperature independent) ? Explain.

No, the spontaneity is not temperature dependent. Delta Go is negative at all temperatures. (Recall that, under standard conditions of 1 atm, Delta Go is negative for a spontaneous reaction.) Here Delta Go = Delta Ho - T x Delta So= (-) - [T x (+)] = - always.

2. Industrially, steam and methane at 30 atm are heated over a nickel catalyst at 800 oC to produce hydrogen and carbon monoxide. If the hydrogen is to be then used in the synthesis of ammonia, the carbon monoxide must be removed, since it poisons the catalysts. The CO is removed at about 300 oC in the presence of a copper-zinc catalyst:

Reaction A: CO(g) + H2O(g) --> CO2(g) + H2(g)

(then the CO2 is finally removed by

H2O(l) + CO2(g) + K2CO3(aq) --> 2 KHCO3(aq)) (a) Use thermodynamic data to calculate Kp at 25 oC and 300 oC for Reaction A. Assume Delta Ho and Delta So are temperature independent.

CO(g) + H2O(g) --> CO2(g) + H2(g)

Since Delta Go = -RT ln Kp, we need Delta Go at 25oC and at 300oC.

Delta Ho = Delta Hof(CO2, g) + Delta Hof(H2, g) - Delta Hof(CO, g) - Delta Hof(H2O, g)

= [(-393.5) + 0 - (-110.5) - (-241.8)] kJ = -41.2 kJ

Delta So = So(CO2, g) + So(H2, g) - So(CO, g) - So(H2O, g)

= [213.7 + 130.6 - (197.6) - (188.7)] J/K = -42.0 J/K = - 0.0420 kJ/K

So, at 25oC = 298 K, Delta Go = Delta Ho - T x Delta So = [-41.2 - (298 x -0.0420)] kJ = -28.7 kJ

Thus, ln Kp = - Delta Go/(RT)

= -[-28.7 kJ/(mole of reaction)]/[8.314 J/(mole K) x 10- 3 kJ/J x 298 K] = 11.6 at 25oC

Kp = 1.09 x 105 at 298 K (25oC).

At 300 oC ( = 573 K), ln Kp = - Delta Go/(RT) = -[Delta Ho - T x Delta So]/RT

= -[41.2 - 573 x 0.0420 kJ]/RT

Note that both Delta Ho and T x Delta So in the above expression are in kJ units.

Thus, ln Kp = - [-17.1 kJ/(mole of reaction)]/[8.314 J/(mole K) x 10-3 kJ/J x 573 K] = 3.59 at 300oC

Kp = 36.2 at 573 K (300oC).

(b) Account for the fact that Reaction A is actually carried out at 300 oC and not at 25 oC.

Even though Kp is much larger (more products) at 25oC, the reaction rate is too slow. Thus, this reaction, lke many industrial reactions, is carried out at elevated temperatures (300oC), in spite of the fact that Kp is much smaller.

(c) About what temperature will Reaction A begin to favor the reactants CO and H2O?

Delta Go = Delta Ho - T x Delta So. When the system is at equilibrium, Delta G = 0.

Then (Delta H)/(Delta S) = T. At higher temperatures than this value of T, Delta G will be positive, and the backward reaction, favoring reactants, will be spontaneous.

T = (-412.8 kJ)/(0.0420 kJ/K) = 981 K.

Thus, at T > 981 K, treact A will begin to favor reactants.

3. The equilibrium constant Kc for the decomposition of phosgene, COCl2, is 4.63 x 10-3 mol/L at 527 oC:

COCl2(g) = CO(g) + Cl2(g)

Calculate the equilibrium partial pressures of all the components, starting with pure phosgene at 0.760 atm.

Since the problem gives pressure data, we will set up a pressure array, form a Kp expression, and then relate that expression to the given Kc.

Pressure	COCl2	CO	Cl2
Initial	0.760	0	0
Change	-x	x	x
Equilibrium	0.760 - x	x	x

Kp = P(CO) x P(Cl2)/P(COCl2) = x2/(0.760 - x)

Note that Delta n(gases) = 1 here. So, Kp = Kc (RT)1 = 4.63 x 10-3 mol/L x 0.0821 {(L atm)/(mol K)} x 800 K.

Kp = 0.304 atm = x2/(0.760 - x)

x2 + 0.304 x -0.231 = 0

x = 0.352 (we reject the physically-meaningless negative value for x from the solution of the quadratic equation.)

P(CO) = P(Cl2) = 0.352 atm P(COCl2) = 0.760 - 0.352 = 0.408 atm

4. a) Write a balanced equation for the reaction of iron(II) ions with liquid bromine.

2 Fe2+(aq) + Br2(l) --> 2 Fe3+(aq) + 2 Br-(aq)

b) Use Table 21.1 to calculate Eocell for the reaction of part a.

Br2(l) + 2 e- --> 2 Br-(aq) Eored = 1.09 V 2 Fe2+(aq) --> 2 Fe3+(aq) + 2 e- Eoox = -0.77 V

----------------------------------------------------------------- 2 Fe2+(aq) + Br2(l) --> 2 Fe3+(aq) + 2 Br-(aq) Eocell = 0.32 V

c) Calculate the standard Gibbs free energy change for the reaction of part a.

Delta Go = - n F Eo = [-(2)(96500)(0.32)] J x 10-3 kJ/J Delta Go = - 62 kJ

Note: if a different "n" is used in part b, part c will have a different answer. Whatever value of n is used in part b, the electrons in the two half-reactions of part b must cancel.

d) When the reaction of part a is multiplied by two, how does this affect the cell voltage and the Gibbs free energy change?

The cell voltage is unaffected by doubling the reaction of part a, but Delta Go is multiplied by 2.

e) Calculate the cell voltage when the [Br-] is reduced to 1.00 x 10 - 1 M.

E = Eo - 0.0257/2 ln {[Fe3+]2 [Br-]2/[Fe2+]2}

= [0.32 -0.257/2 ln 0.010] V = 0.38 V

5. Use reduction potential data to account for the following observations:

a) Copper pennies dissolve in nitric acid (NO2(g) is evolved), but not in hydrochloric acid. The standard reduction potential for nitrate ion being reduced to NO2 in acid solution is +0.80 V; this value is not tabulated in Kotz.

i) When copper dissolves in nitric acid, the oxidizing agent is NO3-:

You observed this in Experiment 2!

2 NO3-(aq) + 4 H+(aq) + 2 e- --> 2 NO2(g) + 2 H2O(l) Eored = 0.80 V Cu(s) --> Cu2+(aq) + 2 e- Eoox = -0.34 V ----------------------------------------------------------------- 2 NO3-(aq) + 4 H+(aq) + Cu(s) --> Cu2+(aq) + 2 NO2(g) + 2 H2O(l) Eocell = 0.46 V

A positive Eocell signifies spontaneity under standard conditions.

ii) When we write a proposed reaction for copper dissolving in hydrochloric acid, we are considering H+ to be the oxidizing agent:

2 H+(aq) + 2 e- --> H2(g) Eored = 0.00 V Cu(s) --> Cu2+(aq) + 2 e- Eoox = -0.34 V ----------------------------------------------------------------- 2 H+(aq) + Cu(s) --> Cu2+(aq) + H2(g) Eocell = -0.34 V

A negative Eocell signifies non-spontaneity under standard conditions; H+(aq) does not oxidize copper.

b) Cu+ disproportionates in aqueous solution according to

2 Cu+(aq) = Cu2+(aq) + Cu(s)

Cu+(aq) + e- --> Cu(s) Eored = 0.52 V Cu+(aq) --> Cu2+(aq) + e- Eoox = -0.15 V ----------------------------------------------------------------- 2 Cu+(aq) --> Cu2+(aq) + Cu(s) Eocell = +0.37 V

A positive Eocell signifies spontaneity under standard conditions.

6. A blast furnace uses Fe2O3 and CO(g) to produce 8.4 x 103 metric tonnes of iron per day (1 metric tonne = 1000 kg).

a) What mass of carbon dioxide would be produced per day by the blast furnace?

Fe2O3(s) + 3 CO(g) --> 2 Fe(s) + 3 CO2(g)

8.4 x 103 tonnes x 103 kg/tonne x 103 g/kg = 8.4 x 109 g

mol Fe = [8.4 x 109 g]/[55.85 g/mol] = 1. 5 x 108 mol

mol CO2 = 1.5 x mol Fe = 2.25 x 108 mol

g CO2 = 2.25 x 108 mol x 44.01 g/mol = 9.9 x 109 g

b) Compare this amount of carbon dioxide with that produced by one million automobiles, each burning 20 L of gasoline per day. Assume that gasoline has the formula C8H18, that it has a density of 0.74 g/mL, and that it burns completely. (For reference, you may be interested to know that gasoline consumption in the U. S. is about 1.6 x 109 L/day.)

20 L gas/auto x 103 mL/L x 0.74 g/ml x 106 auto = 1.5 x 109 g gasoline

mol gasoline = [1.5 x 109 g]/[114 g/mol] = 1. 3 x 107 mol

C8H18(l) + 25/2 O2(g) --> 8 CO2(g) + 9 H2O(l)

mol CO2 = 8 x mol gasoline = 1.0 x 108 mol

g CO2 = 1.0 x 108 mol x 44 g/mol = 4.4 x 109 g

The blast furnace produces more CO2 than 2 million autos do!

7. Typically, metal sulfides are first converted to oxides by roasting in air (SO2 is also a product), and then reduced with carbon to produce the metal and carbon dioxide. (Call this Process A, which consists of two steps.) Why are metal sulfides not reduced directly by carbon to yield liquid CS2 and the metal (Process B)? Give a thermodynamic analysis of both processses for a typical sulfide, ZnS. Use information in the textbook and the following data for CS2(l): Delta Hof = 87.9 kJ/ mol, So = 151.0 J/(mol K), Delta Gof = 63.6 kJ/mol.

NOTE: I have used thermodynamic data from a different text, so the individual answers will be slightly different. Nevertheless, the conclusions will be the same.

PROCESS A

Step 1

2 ZnS(s) + 3 O2(g) --> 2 SO2(g) + 2 ZnO(s)

Delta Go = 2 Delta Gof(SO2, g) + 2 Delta Gof(ZnO, s) - 2 Delta Gof(ZnS, s) - 3 Delta Gof(O2, g)

= [2(-300.3) + 2(-320.5) - 2(-198) - 3(0)]kJ = -845.4 kJ

Note that the same result is obtained if use Delta Go = Delta Ho - T x Delta So.

Step 2

2 ZnO(s) + C(graphite) --> 2 CO2(g) + 2 Zn(s)

Delta Go = 2 Delta Gof(Zn, s) + Delta Gof(CO2, g) - 2 Delta Gof(ZnO, s) - Delta Gof(C, graphite)

= [2(0) + (-394.4) - 2(-320.5) - (0)]kJ = +246.6 kJ

For the total process A (the sum of the two steps),

2 ZnS(s) + 3 O2(g) + C(graphite) --> 2 SO2(g) + 2 Zn(s) + CO2(g),

Delta Go = -598.8 kJ

So the overall process A is sopntaneous.

PROCESS B

2 ZnS(s) + C(graphite) --> 2 CS2(l) + 2 Zn(s)

Delta Go = Delta Gof(CS2, l) + 2 Delta Gof(Zn, s) - 2 Delta Gof(ZnS, s) - Delta Gof(C, graphite)

= [2(63.6) + 2(0) - 2(-198) - (0)]kJ = +459.6 kJ

So process B is not spontaneous.

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