Due: Tuesday (at start of lecture),November 28, 2017 (60 pts) Chemistry 178-Mill

Question

Due: Tuesday (at start of lecture),November 28, 2017 (60 pts) Chemistry 178-Miller: HW Set #6 (8 bonus points if turned in Nov. 17 at Recitation) Chemical Thermodynamies. Please show work, label all answers clearly, and write legibly On the TOP of your paper, please write (a) your full name: (b) the full name of your TA: and (c) your section as "SECTION #". Failure to do this will result in a deduction of2 pts. (1) (20 pts total) 18.00g of ice at -10.00 C are placed in a room at 2500 C. The atmosphere of the room is an enormous reservoir that can reversibly absorb or release energy via heat at a constant temperature of 25.00°c (a) (4 pts) In a single sentence, describe the spontaneous process that takes place. What is the initial state and what is the final state of the system? (b) (4 pts) The entire process can be divided into 3 separate steps. The first step is (1 ) H:O(s) at-10.00- H:O(s) at 0.00 Write the second and third steps of the total process. (c) (4 pts) Evaluate AHayem (kJ) for the entire process, from initial state to final state of the system. For a pure substance, AH mo AT (see Section S.5 of the textbook). Useful information include Glice) 2.092 J/g-K; cwater) 4.184 Jg-K for the temperature ranges considered; and AHlH0)- 6.008 kJ/mol at its normal melting freezing point. (d) (4 pts) Evaluate /.mnings (k) and ASumonip (JK) for the catire process. (c) (4 pts) From your answer to part (d), what can you conclude about AS for the entire process? (2) (24 pts total) Ultrapure germanium, for use in semiconductors, is prepared by the following four steps: (1) Ge of"ordinary purity is obtained by the high-temperature reduction of GeO: with carbon: (2) this Ge is converted to GeCl, by treatment with Cl and then purified by distillation; (3) GeCl is then hydrolyzed in water to GeO and (4) GeO is reduced with H to ultrapure germanium, which is then zone-refined (a) (8 pts) Write balanced chemical equations for each of these four chemical steps using the smallest integer coefficients. (b) (8 pts) Evaluate AS (J/K) and AG (kJ) at 298.15 K for each reaction in part (a). (c) (8 pts) with respect to enthalpy and entropy changes, which reactions in part (a), Le-1H4), benefit from using high-temperatures and which benefit from using low-temperatures? Explain your choices Useful information not found in Appendix C of the textbook Substance r(kJ/mole)(/mol-K) Ge(s) GeClal) GeOfs) 31.1 245.6 39.7 -531.8 -580.0 (3) (16 pts total) Sugars can be digested acrobically or anaerobically (eg. by yeast). During aerobic digestion, sugars react with oxygen to form carbon dioxide and water. During anacrobic digestion, sugars are decomposed by enzymes, acting as catalysts, in the presence of water to create carbon dioxide and alcohols (a) (4 pts) Write the balanced chemical equation for aerobic digestion of one mole of sucrose, CH Ous) (b) (4 pts) Evaluate Go (k) and Aee (V) for the aerobic reaction in (a) for one mole of sucrose at 25°C, (INFO not in Appendix C: Me(sucrose, s)--2221.2 klinol; S(sucrose, s)-3924 Jin1-K) (c) (4 pts) Write the balanced chemical equation for anacrobic digestion of one mole of sucrose. Assume that ethanol, C HOH), is the only alcohol product for anacrobic digestion. (d) (4 pts) Evaluate Go (kl) and AE" (V) for the anaerobic reaction in (c) for one mole of sucrose at 25°C.

Explanation / Answer

According to the statement reaction A will be

GeO2 + C (s) ==== Ge (s) + CO2 (g)

b) Ge (s) + 2 Cl2 ===== GeCl4

c) GeCl4 + 2H2O ==== GeO2 + 4HCl

d) GeO2 + 2 H2 ====== Ge + 2 H2O

to calculate enthalpy use the equation

H rxn = H products - H reactants

To calcualte entropy use the equation

S rxn = S products - S reactants

Then to calculate gibbs use equation

G = H - T S

Entropy of Carbon is 5.8

Entropy of HCl g is 186.9

Entropy of CO2 213.6

Entropy of water 69.95

this information was taken from the NIST webpage

To calculate enthalpy for first reaction it will be

GeO2 + C (s) ==== Ge (s) + CO2

H = 2 * -393.5 - 580*1 = 186.5 KJ / mol

Entropy will be

S = 213.6 + 31.1 - (39.7 + 5.8) = 199.2 J / K mol

Gibbs will be

=186.5*1000 - 298.15*199.2 = 127 138 J / K mol

You have to multiply by 1000 the enthalpy to get Joules

Repeat the procedure for the other 3 reactions

Ge (s) + 2 Cl2 ===== GeCl4

Enthalpy = -531.8 - (2*0 + 0)

Enthalpy = -531.8 KJ / mole

Entropy = 245.6 - (31.1 - 2*223)

Entropy = -231.5 J / k mole

Gibbs = Enthalpy - Temperature * Entropy

Gibbs = -531 800 - (298*-231.5)

gibbs = -462 813 J / mole

c) GeCl4 + 2H2O ==== GeO2 + 4HCl

Enthalpy = (4 * -92.31 - 580) - (-531.8 + 2*-285.8)

Enthalpy = 154.16 KJ / mole

Entropy = (4 * 186.9 + 39.7) - (245.6 + 2*69.95)

Entropy = 401.8 J / k mole

Gibbs = 154.16 * 1000 - 298*(401.8) =

gibbs = 34 423 J / mole

GeO2 + 2 H2 ====== Ge + 2 H2O

Entalphy = (0 + 2 * -285.8) - (2*0 - 580) =

Enthalpy = 8.4 KJ / mole

Entropy = (31.1 + 2* 69.95) - (39.7 + 2*130.7)

Entropy = -130.1 J / k mole

Gibbs = 8.4*1000 - (298 * -130.1) =

gibbs = 47 169 J / mole

reactions a and b will benefit from a high temperature

reaction c will benefit from a low temperature

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