CHM 152 Thermodynamics page 1 of 7
CHM 152 - Thermodynamics
Spontaneity
1. True or false? All exothermic reactions are spontaneous? _false _
2. Which variable can tell you if a process is spontaneous without exception? a. ΔHrxn b. ΔSrxn c. ΔHsurr d. ΔSsurr e. ΔStot
Enthalpy and Entropy
3. Circle the correct words: Nature tends towards (higher or lower) energy and more (order or disorder)? 4. Which of the following reactions will have the most positive ΔHo ? (Hint: Draw Lewis structures.)
a. N2(g) → 2 N(g)
b. F2(g) → 2 F(g)
c. O2(g) → 2 O(g)
d. These reactions would all have the same ΔHo .
Notice that all reactions are 1 mol of a diatomic gas going to 2 mol of a monoatomic gas, i.e., the bonds between the two atoms are broken. This is a tricky question and really relates to CHM151 knowledge, i.e., a single bond is weaker than a double bond is weaker than a triple bond. If a triple bond is the strongest bond, it takes more energy to break it. N2 has a triple bond, O2 has a double bond, and F2 has a single bond.
5. Which state of matter has the highest entropy? ___gas _____
6. Predict whether the entropy change will be positive or negative for the following: a. H2O (g) g H2O (l) ΔS___-__
b. C6H12O6(s) g 2C2H5OH(l) + 2CO2(g) ΔS___+__
c. 2NH3(g) + CO2(g) g H2O(l) + NH2CONH2(aq) ΔS___-__
d. NaCl(s) g NaCl(aq) ΔS___+_
e. Cu(s) (100oC) g Cu(s) (25oC) ΔS___-_ f. 2NH3(g) D N2(g) + 3H2(g) ΔS___+_
7. If a process is endothermic and the process creates more order than existed before, the process is: a. always spontaneous b. never spontaneous c. spontaneous at high T d. spontaneous at low T 8. Which of the following substances has the greatest entropy per mole?
a. O2(g)
b. N2(g)
c. CO(g) d. CO2(g)
e. C4H10(g)
All choices are gases and, assuming they are all at the same temperature, the largest molecule will have the highest entropy because of the greater possible movement of atoms within each molecule.
CHM 152 Thermodynamics page 2 of 7 9. Which of the following reactions will have an increase in entropy? Choose all that apply.
a. SO3(g) → 2SO2(g) + O2(g)
b. H2O(l) → H2O(s)
c. Br2(l) → Br2(g)
d. H2O2(l) → H2O(l) + ½ O2(g)
a. SO3(g) → 2SO2(g) + O2(g) ΔS > 0 because 1 mol gas produces 3 mol gas
b. H2O(l) → H2O(s) ΔS < 0 because liquid turns into solid
c. Br2(l) → Br2(g) ΔS > 0 because liquid turns to gas
d. H2O2(l) → H2O(l) + ½ O2(g) ΔS > 0 because 1 mol liquid produces 1 mol liquid + ½ mol gas
10. Without consulting entropy tables, predict the sign of ΔS for the following process and choose the correct reasoning for your prediction: The mass of nitrogen remains constant.
N2(g, 10 atm) → N2(g, 1 atm)
a. positive; there is an increase in the number of gas molecules
b. positive; the gas expands into a larger volume
c. negative; the gas is compressed into a smaller volume d. negative; the gas expands into a larger volume
e. negative; there is a decrease in the number of gas molecules.
The only way to change the pressure of a gas from a high to a lower pressure, without reducing temperature or removing gas atoms/molecules, is to increase the volume. In doing so, the gas molecules have a larger space to move and hence there is greater randomness or disorder.
11. Without consulting entropy tables, predict the sign of ΔS for the following process: Pb(s) + Cl2(g) → PbCl2(s).
a. ΔS > 0
b. ΔS < 0
c. ΔS = 0
d. More information is needed to make a reasonable prediction. A solid and a gas turn into a solid.
Standard Molar Entropies
12. Which of the following compounds has the lowest entropy at 25 oC? a. CH3OH(l)
b. CO(g)
c. MgCO3(s)
d. H2O(l)
e. H2O(g)
CHM 152 Thermodynamics page 3 of 7 13. Using the data below, calculate ΔSorxn for the following reaction:
4 Cr(s) + 3 O2(g) → 2 Cr2O3(s)
a. 548.1 J/K b. 147.7 J/K c. -147.7 J/K
d. -548.1 J/K
e. None of the above.
ΔS°rxn = [ 2 x 81.2 ] – [ ( 4 x 23.77 ) + ( 3 x 205.138 ) ] = -548.1 J/K
14. Calculate the standard entropy change, ΔS°, for the following reaction at 25 °C: 4Al(s) + 3O2(g) → 2Al2O3(s)
S°[Al(s)] = 28.32 S°[O2(g)] = 205 S°[Al2O3(s)] = 51.0
ΔS = (2) - (4) - (3) = -626
*note that entropy decreases as expected based on decrease in #molecules of gas
Free Energy
15. In 1774 Joseph Priestly prepared oxygen by heating mercury(II) oxide according to the reaction HgO(l) →
Hg(l) + ½O2, for which ΔHo = 90.84 kJ/mol and ΔSo = 108 J/K.mol. Which of the following statements is true
for this reaction?
a. The reaction is spontaneous only at low temperatures. b. The reaction is spontaneous at all temperatures. c. ΔGo becomes less favorable as temperature increases.
d. The reaction is spontaneous only at high temperatures.
e. The reaction is at equilibrium at 25 oC and 1 atm pressure.
Plug the enthalpy and entropy values into ΔGo = ΔHo− TΔSo and you’ll find that ΔGo < 0 only when the temperature is
above approximately 841 K ( 568 °C ); anything lower than this temperature makes ΔGo > 0 ( non-spontaneous).
16. For the decomposition reaction of hydrogen peroxide:
H2O2(g) → H2O(l) + 1/2 O2(g) , ΔHo = -106 kJ/mol; ΔSo = 58 J/K
Is H2O2(g) stable?
a. Yes, under all conditions.
b. Yes, if the temperature is low enough. c. Yes, if the rate of decomposition is low.
d. Yes, if the O-O bond energy is greater than the O-H bond energy.
e. No.
Note that ΔHo is negative and ΔSo is positive. ΔGo will always be negative (even when T = 0 K). Therefore, the decomposition of hydrogen peroxide will always be spontaneous and hence never stable.
SubstanceΔ S, J/K·mol
Cr(s) 23.77
O2(g) 205.138 Cr2O3(s) 81.2
CHM 152 Thermodynamics page 4 of 7 17.a. Calculate the standard free energy change, ΔG°, for the following at 25 °C:
MgO(s) + C(graphite) → Mg(s) + CO(g)
ΔH° = 491.18 kJ ΔS° = 197.67 J/K ΔG° = ΔH° - TΔS°
ΔG° = 491.18 kJ - (298 K) = + 432.27 kJ
b. Is this reaction spontaneous at 25 °C? If not, at what temperature can we make this reaction spontaneous? No, it is not spontaneous at 25 °C (ΔG° is a large positive value)
Set ΔG = 0: 0 = ΔH - TΔS ⇒ T =
T = = 2484.8 K ⇐ This is T at equilibrium, spont at T > 2484.8 K
18.Calculate ΔHorxn, ΔSorxn andΔGorxn for 2 Mg(s) + O2(g) D 2 MgO(s)) (Hint – balance the rxn)
ΔHorxn = [ 2(-601.7)] – [0] = -1203.4 kJ
ΔSorxn = [ 2(26.9)] – [ 2(32.7) + 1(205.0) ] = -216.6 J/K ΔGorxn = [ 2(-569.4) ] – [0] = -1138.8 kJ
19.From # 18 the results for ΔHorxn means (exothermic or endothermic), the results for ΔSorxn means (disorder
or order) increased, andΔGorxn means (products or reactants) are favored under standard conditions.
20.Can you measure enthalpy (H)? no entropy (S)? yes Gibb’s free energy (G)? no
Standard Free Energy of Formation
21.Which of the following does not have a standard enthalpy of formation of zero? A. I2(l) B. Cl2(g) C. Au(s) D. Ne(g) E. Hg(l)
I2 is a solid under standard conditions
22.The standard enthalpy of formation equation for NH3(g) is:
a) N(g) + 3H(g) → NH3(g) b) N2(g) + 3H2(g) → 2NH3(g)
c) N2(l) + H2(g) → NH3(g) d) N2(g) + H2(g) → NH3(g)
23.Calculate ΔH for the following reaction at 25 °C. 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
CHM 152 Thermodynamics page 5 of 7 ΔH [NO(g)] = 90.3 ΔH [H2O(g)] = -241.8 ΔH [NH3(g)] = -45.9
= (4)(90.3 ) + (6)(-241.8 ) - [(4)(-45.9 ) + (5)(0)] = -906.0 kJ / mol
24. Use the following data to calculate ΔGo at 298 K for the combustion of propane: C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
a. 2109.6 kJ/mol b. 608.8 kJ/mol c. -608.8 kJ/mol
d. -2109.6 kJ/mol
e. None of the above.
Remember that ΔGo means the change in free energy under standard state conditions, i.e., 25°C or 298 K. Therefore, ΔGo
can be calculated as follows.
ΔGo = [ ( 3 x – 394.6 ) + ( 4 x – 237.2 ) ] – [– 23.0 ] = -2109.6 kJ/mol 25. Calculate ΔGo
rxn at 25oC for this reaction: 2 NaHCO3(s) g Na2CO3(s) + CO2(g) + H2O(g). Given: ΔHorxn = 128.9 kJ/mol and ΔSorxn = 321 J/molK
ΔGorxn = 128.9 kJ/mol – (298 K)(0.321 kJ/molK) = 33.2 kJ/mol
Free Energy and Equilibrium
26. a) Calculate ΔG° for the following reaction. b) What is the value of the equilibrium constant at 298 K? c) Is this reaction spontaneous at 298 K?
3C2H2(g) → C6H6(g) ΔG = 209.2 ; ΔG = 129.7
a) ΔG° = (1) - (3) =ΔG° = -497.9 kJ / mol
b) K = e = e = 1.89 x 1087
c) ΔG° is (-) and K is very large, so the reaction is spontaneous at 298 K
SubstanceΔ Go
f, kJ/mol
C3H8(g) -23.0 O2(g) 0 CO2(g) -394.6 H2O(l) -237.2
CHM 152 Thermodynamics page 6 of 7 27.a) Calculate ΔG° for the following reaction at 298 K. Kb = 1.8x10-5
NH3(aq) + H2O(l) D NH (aq) + OH-(aq)
ΔG° = -RT ln K = - (8.314 )(298 K)(ln 1.8 x 10-5) = 2.7x104 or 2.7x101
b) Calculate ΔG at 298 K, when [NH ] = 0.10 M, [OH-] = 0.050 M and [NH3] = 0.10 M.
ΔG = ΔG° + RTln Q Q =
ΔG = 2.7x104 + (8.314 )(298 K)
ΔG = 2.7x104 - 7.4x103 ΔG = 2.0x104 or 2.0x101
28.If ΔG is a negative number fill in the following: i. Is the reaction spontaneous or not? _ spontaneous
ii. Q relates how to K? Q __<___K
iii. Is the reaction going forwards or backwards? __ forwards ___ iv. Eventually the value of ΔG will reach __zero__
29. Circle the correct sign: When ice melts ΔS is ( + or - ) and ΔH is ( + or - ). Under what conditions will this process be spontaneous, if ever? Be specific and refer to the appropriate temperature. Spontaneous ABOVE the melting point of zero Celcius. In other words ice melts naturally above 0oC. 30. For carbon disulfide, CS2, the enthalpy and entropy for vaporization is 27.7 kJ/mol and 86.4 J/molK,
respectively. What is the boiling point (°C) for CS2? Will CS2 boil above or below this temperature?
0 = ΔH – TΔS = 27.7 kJ/mol – T(.0864 kJ/mol K)
T = (27.7 kJ/mol) / (.0864 kJ/molK) = 321 K = boiling point, it boils ABOVE this
31.At 25ºC the equilibrium constant for this reaction CO(g) + 2H2(g) à CH3OH(g) has the value Kp = 2.1 x
10 4 . Calculate ΔGºrxn for this reaction at this temperature.
CHM 152 Thermodynamics page 7 of 7 32.For the unbalanced reaction 2 SO2(g) + O2(g) → 2 SO3(g) calculate ΔG at 25.0ºC when the reactants
and product are at the following partial pressures: 10.0 atm SO2 , 10.0 atm O2 , and 1.00 atm SO3.
ΔG = ΔGo + RT ln Q Q = 12 / (102 * 10) = 1/1000 ΔGo = [ 2mol(-371.1 kJ/mol)] – [ 2mol(-300.2 kJ/mol) + 0 ] = -141.8 kJ ΔG = -141.8 kJ + (8.314 J/mol·K) (298 K) ( kJ/1000J) (ln 1/1000) = -159 kJ 33.Calculate K for MgCO3(s) à MgO(s) + CO2(g)
ΔGorxn = [ -569.4 + -394.4 ] – [ -1012 ] = 48.2 kJ ΔGorxn = -RT ln K
48.2 kJ (1000J / kJ) = (-8.314 J/molK) (298 K) ln K ln K = -19.45 so K = 3.6 x 10-9
34.Calculate the temperature at which this reaction changes from being spontaneous to non spontaneous: Mg(s) + O2(g) à MgO(s). Is the reaction spontaneous above or below this temperature?
Set ΔG to zero to find the temp at which the rxn changes between spont and not spont 0 = ΔH - TΔS
0 = (-1203 kJ) – T (-0.2166 kJ/K) T = -1203 kJ / -0.2166 kJ/K
T = 5554 Kelvin or 5281oC rxn is spont BELOW this temp
35.Calculate ΔG for Mg(s) + O2(g) à MgO(s) at 30.5oC if the pressure of oxygen gas is 1.33 atm.
ΔG = ΔGorxn + RT ln Q Q = 1/1.33 = 0.75188
ΔG = (-1139 kJ) + (8.314 J/molK) (kJ / 1000J) (303.5 K) ln 0.75188 ΔG = -1140 kJ = -1.140 x 103 kJ
