Using the following values for entropy determine if a reaction would be spontaneous.

I) ΔSsys = 30 J/K , ΔSsurr = 50 J/K

II) ΔSsys = 60 J/K, ΔSsurr = -85 J/K

A reaction will be spontaneous if the total entropy change is positive. It will not be spontaneous if the total entropy change is negative.

ΔStotal = ΔSsys + ΔSsurr

where

ΔStotal is the total entropy change

ΔSsys is the entropy change of the system

ΔSsurr is the entropy change of the surroundings

I.) ΔStotal = ΔSsys + ΔSsurr

ΔStotal = 30 J/K + 50 J/K

ΔStotal = 80 J/K

The total entropy change is positive, therefore the reaction will be spontaneous.

II.) ΔStotal = ΔSsys + ΔSsurr

ΔStotal = 60 J/K + -85 J/K

ΔStotal = -25 J/K

The total entropy change is negative, therefore the reaction will not be spontaneous.

Under what conditions are the following reactions spontaneous, and why?

Calculate the change in Gibbs free energy.

HCl+NaOCH3 → NaCl+HOCH3

pKa HCl=-7pKa HOCH3=16 T=298K

Keq=[NaCl][HOCH3][H+]/[HCl][NaOCH3][H+]

Keq=KaHCl/KaHOCH3

Keq=10^7/10^-16= 10^23

∆G=-RTlnK

∆G=-8.314x298xln10^23

∆G=-13.1kJ ← favored towards products

In the synthesis of gaseous methanol from carbon monoxide gas and hydrogen gas, the following equilibrium concentrations were determined at 444 K: [CO(g)]=0.0977M, [H2O(g)]=0.0799M, and [CH3OH(g)]=0.00799M. Calculate the equilibrium constant and Gibbs energy for this reaction.

CO(g)+2H2(g) → CH3OH(g)

Keq=[CH3OH]/([CO][H2]^2)

Keq=[0.00799]/([0.0977][0.0799]^2)

Keq=12.8

ΔGº=-RTln(Keq)

ΔGº=-(8.314J/Kmol)(444K)ln(12.8)(kJ/1000J)

ΔGº=-9.4kJ/mol

For the following reaction, what would ΔGº be at 298K ?

Fe3O4(s) → 3 Fe (s) + 2O2(g)

ΔHfº (kj/mol)-1118.4

ΔSº (J/molK) 146.427.8205.1

ΔHº 0 - (-1118.4) = +1118.4 kJ

ΔSº (3(27.8) + 2(205.1)) - 146.4 = 347.2 J/K

ΔGº = ΔHº-TΔSº = 1118.4kJ-(298.15K)(.3472kJ/K)= 1015kJ

If delta H = 158 kJ and delta S = 411 J/k. At what temperature will this reaction be spontaneous?

deltaG= deltaH-TdeltaS

0> 158000J – T(411 J/K)

T(411 J/K)/ 411 J/K > 158000 J/ 411 J/K

T > 384 K

Calculate ∆G of this reaction

∆H=-537.22kJ∆S=13.74J/KT=25ºC

H2(g)+F2(g) → 2HF

∆G=∆H-T∆S

∆G=(-537.22kJ)-(298K)(13.74J/K)

∆G=(-537.22kJ)-(298K)(.01374kJ/K)

∆G=-541.31kJ ← SPONTANEOUS

What must be the temperature if the following reaction has ΔGº=-52.9 kJ, ΔHº=-34.7kJ, and ΔSº=12.4J/K?

Fe2O3(s)+3CO(g) → 2Fe(s)+3CO2(g)

ΔGº=ΔHº-TΔSº

(-52.9 kJ)(1000J/kJ)=(-34.7 kJ)(1000J/kJ)-T(12.4J/K)

T=1470K

Label which reaction is spontaneous or nonspontaneous, and compute the overall reaction, given that it is spontaneous.

Cu2O (s) → 2 Cu(s) + ½ O2 (g) ΔGº = 125 kJ

C(s) + ½ O2(g) → CO(g) ΔGº = -175

Reaction 1: nonspontaneous; reaction 2: spontaneous

Net reaction:

Cu2O (s) + C (s) → 2 Cu (s) + CO(g)

ΔGº= 125+ (-175) = -50 kJ

Net reaction is spontaneous.