Entropy+and+Gibbs+Free+Energy+Assignment

__Yellow Fish (Jenna, Jennice, Reza, Mackenzie, Jessica):__
N2(g) + 2O2(g) -> 2NO2
 * a) Based on the given information, above what temperature is the reaction favourable?**

ΔS = S(nitrogen dioxide) - S(nitrogen gas, oxygen gas) = [240.1 J/mol*K x 2 mol] - [(191.5 J/mol*K x 1 mol) + (205.1 J/mol*K x 2 mol)] = 480.2 J/K - 601.7 J/K ΔS = -121.5 J/K

ΔH = H(nitrogen dioxide) - H(nitrogen gas, oxygen gas) = [33.2 kJ/mol x 2 mol] - [(0 kJ/mol x 2 mol) + (0 kJ/mol x 1 mol)] = 66.4 kJ - 0 kJ ΔH = 66.4 kJ x 1000J/1kJ ΔH = 66400 J

ΔG = ΔH -TΔS Set ΔG = 0 J and solve for T T = [ΔH - ΔG]/ΔS T = [(66400 J) - (0 J)]/(-121.5 J/K) T = -546.50 K T = -546.50 K - 273 T = -819.5 °C

Therefore, the reaction is favourable when the temperature is greater than -819.5 °C.

We are at risk of forming it in air here in Toronto in Canada Day, because the average temperature is 18°C which is equal to 291 K (18°C + 273). Since the reaction is favourable when the temperature is greater than -819.5 °C , the reaction is favourable however the rate at which it occurs is fairly slow.
 * b) Are we at risk of forming it in air here in Torontoduring the Victoria Day Weekend? (Average temperature of 18°C) Explain.**

ΔG = ΔH - TΔS = 66400 J - (18°C + 273)*(-121.5 J/K) = 66400 J - (291 K)(-107.9 J/K) ΔG = 101756.5 J x 1 kJ/1000J = 101.756 kJ
 * c) Calculate the Gibbs Free Energy of this reaction on the day mentioned above.**


 * d) Calculate the equilibrium constant for the reaction on the day mentioned above.**

__Green Fish Question (Cylita, Catherine, Stephanie, Christina):__

 * A) At what temperature is the reaction favorable:** above 442 K or 168 ˚C

N2 (g) + 2H2 (g) --> N2H4 (g)

Hydrazine: ΔH˚f = + 95.4 KJ/mol ΔS˚f = + 237.11 J/K · mol

Hydrogen: ΔS˚f = + 130.7 J/K · mol

Nitrogen: ΔS˚f = + 191.5 J/K · mol

ΔSrxn = ∑ n ΔS˚f (products) - ∑ n ΔS˚f (reactants) ΔSrxn = ∑ [(1 x 237.11 J/K · mol)] - ∑ [(1 x 191.5 J/K · mol) + (2 x 130.7 J/K · mol)] ΔSrxn = -215.79 J/K · mol ΔSrxn = -0.21579 KJ/K · mol

ΔG = ΔH – TΔS 0 = (95.4 KJ/mol) – (-0.21579 KJ/K · mol)T T = -95.4 KJ/mol / 0.21579 KJ/K · mol T = 442.0964 K --> 168.946 ˚C


 * B)Are we at risk of forming it here during the Summer Solstice**: No we are not. The reaction for the formation of hydrazine is favorable above 168˚C. The temperature during the Summer Solstice is only 24˚C, which is much lower than the temperature needed to form this molecule.


 * C) Calcuate the Gibbs Free Energy for this day**: 160 KJ

ΔG = ΔH – TΔS ΔG = (+ 95.4 KJ/mol) – (302.15 K)( -0.21579 KJ/K · mol) ΔG = 160.6009 KJ


 * D) Calculate the Equilibrium Constant**:

__Red Fish Question (Hasnain, Grey, Brandon, Ho Jun)__
S8 (s) + 8O2(g) --> 8SO2(g)


 * 1) Based on the given information, above what temperature is the reaction favourable?**

ΔS = S(Sulfur Dioxide) - S(Sulfur + Oxygen Gas) ΔS = -1400.504 J/mol*K

ΔH = H(Sulfur Dioxide) - H(Sulfur + Oxygen Gas) ΔH = -296.8 KJ/mol - 0 ΔH = -296800 J/mol

ΔG = ΔH -ΔST Set ΔG =0 KJ 0 KJ = -296800 J/mol - T(- 1400.504 J/mol*K) T =211 K T = - 62 * C

We are at risk of forming it in air in Fort MacMurray, because the average temperature is -4 °C, which is greater than the favorable temperature of -62 °C.
 * 2) Are we at risk of forming it in air in Fort MacMurray during the Winter Solstice? ( Average temperature of -4°C) ? Explain.**

ΔG = ΔH -TΔS ΔG = -296800 J/mol - (-4°C + 273) (-1400.504 J/mol*K) ΔG = 79935.576 J ΔG = 79.9 KJ
 * 3) Calculate the Gibbs Free Energy of this reaction on the day mentioned above.**


 * 4) Calculate the equilibrium constant for the reaction on the day mentioned above.**

???

Orange and Yellow Fish 2. a)

__Blue Fish:__

 * 1) Based on the given information, above what temperature is the reaction favourable?**

ΔS = S(ammonia) - S(nitrogen gas, hydrogen gas) ΔS =192.78 J/mol*K - 0 ΔS = 0.19278 KJ/mol*K

ΔH = H(ammonia) - H(nitrogen gas, hydrogen gas) ΔH = 45.9 KJ/mol - 0

ΔG = ΔH -ΔST Set ΔG =0 KJ 0 KJ = 45.9 KJ/mol - (0.19278 KJ/mol*K)T T = 238.1 K T = -35.05 ° C ( This value is a little strange)

We are at risk of forming it in air here in Toronto in Canada Day, because the average temperature is 25 °C, which is far more than the favorable reaction temperature.
 * 2) Are we at risk of forming it in air here in Toronto in Canada Day? ( Average temperature of 25°C) Explain.**

ΔG = ΔH -ΔST ΔG = 45.9 KJ/mol - (0.19278 KJ/mol*K)(25°C + 273) ΔG = - 11.5 KJ
 * 3) Calculate the Gibbs Free Energy of this reaction on the day mentioned above.**

4) Calculate the equilibrium constant for the reaction on the day mentioned above.

Orange and Yellow Fish 2. a)

C(graphite) à C(diamond) a. C(diamond) + O2 (g) à CO2 (g) ΔHcomb = -395.4 kJ/mol Flip C(graphite) + O2 (g) à CO2 (g) ΔHcomb = -393.5 kJ/mol CO2 (g) à C(diamond) + O2 (g) ΔHcomb = 395.4 kJ/mol C(graphite) + O2 (g) à CO2 (g) ΔHcomb = -393.5 kJ/mol = C(graphite) à C(diamond) ΔHrxn = 1.9 kJ/mol 1.9 kJ/mol x 1000J/kJ = 1900 J/mol ΔS(graphite) = 5.7 J/mol.K ΔS(diamond) = 2.4 J/mol.K ΔSrxn = ΔSproducts - ΔSreactants = 2.4 J/mol.K - 5.7 J/mol.K = -3.3 J/mol.K ΔGrxn = ΔHrxn - TΔSrxn 0 = (1900 J/mol) – T(-3.3 J/mol.K) - 1900 J/mol = (3.3 J/mol.K)T T= (-1900 J/mol) / (3.3 J/mol.K) = 575.76 K The reaction is favourable above 575.76 K.
 * Orange and Red Fish (Leigh, Kami, Saptarshi and Ahsan)**

b. Diamonds will not spontaneously convert into coal on the September 3rd (at 22oC). This is because coal is a more stable allotrope of carbon than coal. This goes against the second law of thermodynamics which states that the entropy of the universe is greater than zero, or that the universe is becoming more random. This reaction would be endothermic, meaning that energy would need to be added, and therefore the reaction is non-spontaneous.

c. ΔGrxn = ΔHrxn - TΔSrxn = (1900 J/mol) – 295 K(-3.3 J/mol.K) = 1900 J/mol + 973.5 J/mol = 2873.5 J/mol