Question

1. Calculate the free energy change of the following processes at 298 K from the given information. At what temperature range would each of these be spontaneous?: a) ∆H˚ = 293 kJ; ∆S˚ = -695 J/K

b) ∆H˚ = -1137 kJ; ∆S˚ = 0.496 kJ/K

c) ∆H˚ = -86.6 kJ; ∆S˚ = -382 J/K

2-. Given the following chemical reactions, determine the heat of hydrogenation for C3H4(g) + 2 H2(g)  C3H8(g), and write the balanced thermochemical equation corresponding to the overall reaction.

1) 2 H2(g) + O2(g) .....> 2 H2O(l) ∆H = -571.6 kJ/mol

2) C3H4(g) + 4 O2(g) .....> 3 CO2(g) + 2 H2O(l) ∆H = -1937 kJ/mol

3) C3H8(g) + 5 O2(g) ......> 3 CO2(g) + 4 H2O(l) ∆H = -2220 kJ/mol

3-  A reaction is first order and obeys the Arrhenius equation. If the rate constant at 298 K is 7.74x10-3 s -1 and the activation energy is 98 kJ/mol, what is the rate constant at 275 K? Assume A is constant over this temperature range.

Answer 1

Q1.

In order to compare equilibirum vs. spontaneous/nonspontaneous reactions, we better use a criteria.

Recall that if dSuniverse > 0, this is spontaneous, if dSuniverse = 0, this is inequilbirium and if dSuniverse < 0 this is never possible.

Then, recall that

dSuniverse = dSsurroundings + dSsystem

dSsystem = Sproducts - Sreactants

dSsurroundings = Qsurroundings/T = -dHsystem/T

therefore

dSystem =  -dHsystem/T + dSsystem

If we multiply by -T

dGrxn = dHrxn - T*dSrxn

Now, analysis of dG value... which is the "free energy" available for a process to follow

if dG <0 , this will be spontaneous

if dG = 0 , this is in equilibrium

if dG > 0, this will not be spontaneous

Now...

dG = dH - T*dS

Possible values are, dH = +/- and dS = +/-; T is always positive ( absolute value)

Analysis of cases:

Case 1.

if dH is positive (-) and dS is positive (+) --> this favours always a negative value of dG; spontaneous

Case 2.

if dH is positive (+) and dS is positive (-) --> this favours always a positive value of dG; not spontaneous

Case 3.

if dH is positive (+) and dS is positive (+) --> dG = dH - T*dS --> analysis must be done

if T is very low... then dH > T*dS; then this will be Positive value in dG; i.e. not spontaneous

if T is very high... then  dH < T*dS; then this will be Negative value in dG; i.e. spontaneous

Case 4.

if dH is positive (-) and dS is positive (-) --> dG = dH - T*dS --> analysis must be done

if T is very low... then dH > T*dS; then this will be Negative value in dG; i.e. spontaneous

if T is very high... then  dH < T*dS; then this will be Positive value in dG; i.e. non spontaneous

a)

293*10^3 - (T) * (-695) < 0

293*10^3 / (-695) < T

-421 ; this will never be spontaneous (Case 2)

b)

case 1; this favours dG always T > 0 K

c)

-86.6*10^3 - T*(-382) < 0

-86.6*10^3 < -382*T

T > (-86.6*10^3)/(382) > 226.70 K