Ch.6 Practice Answer Key
1) If a reaction is spontaneous above 500K, and ΔH° = 150 kJ/mol, what is the value of
ΔS°? Assume ΔH° and ΔS° don’t significantly change with temperature.
ΔG = ΔH – TΔS  At 500K ΔG = 0 because this is where it becomes spontaneous
0 = ΔH – TΔS  ΔH = TΔS
Then plug in values: 150 kJ/mol = 500K ΔS
ΔS = 0.3 kJ/molK or 300 J/molK
2) If a piston is pushed on a gas to change the volume from 75.0mL to 50.0mL, what are
the signs for ΔV and w? What happens to the energy of the system? The surroundings?
-ΔV and +w (work is being done on the system; in other words, the system is taking
in/absorbing work)
The energy of the system (the gas) increases because it absorbs work and work
influences the internal energy of the system: ΔE = q + w. No heat is changing here, so
an increase in work must increase the energy. Since the energy in the universe is
constant (law of conservation of energy), the surroundings are losing energy since it
is doing work on the system.
3) (a) When potassium reacts with water the temperature increases. What are the signs for
ΔH, ΔS, and ΔG?
-ΔH, +ΔS, and –ΔG (ΔH is negative because heat is released, ΔS is positive because of
the reaction occurring: K (s) + H2O (l)  KOH (aq) + H2 (g) solids and liquids have less
disorder than liquids and gases so entropy increases, and ΔG is negative because the
reaction occurs)
(b) Based on your answer for part a, does temperature have any effect on whether or
not this reaction will occur? Why or why not?
No because all signs are favorable which means that this reaction will always occur
regardless of the temperature. You can see this using the formula: ΔG = ΔH – TΔS. Both
terms will be negative (-ΔH and – TΔS), thus ΔG will always be negative.
4) Many salts dissolve more quickly/become more soluble when added to hot water vs.
cold water. Explain why this is from an enthalpy standpoint.
If they become more soluble in hot water this indicates that the salts absorb energy
(+ΔH ) in order to dissolve. Therefore, if there’s more energy available for them to
take in, as is the case with hot water, they will dissolve more readily.
5) Consider the following hypothetical reaction: AB (g)
A (g) + B (g). Under standard
conditions, the enthalpy change for the reaction going from left to right (forward
reaction) is H° = +48.0kJ.
a. Is the value of S° for this reaction positive or negative? Explain your answer.
ΔS° is positive in the forward direction because one mole of gas is forming two
moles of gas therefore there’s greater entropy.
b. The above reaction is spontaneous under standard conditions. Predict what will
happen to G for this reaction as the temperature is increased. Justify your
prediction.
Since ΔH is positive (unfavorable) and ΔS is positive (favorable), the sign of ΔG
will depend on T: ΔG = +ΔH – TΔS. Based on the signs, in order for ΔG to be
negative the TΔS term must win out. In order for this to happen T must be large
enough to ‘overtake’ the positive ΔH, which means the forward reaction will be
spontaneous at higher T. Therefore, ΔG will change from positive to negative to
more negative (more favorable) as T increases.
c. Show how the temperature at which the reaction changes from spontaneous to
nonspontaneous can be predicted. What additional information is necessary?
If ΔG = 0 and this is plugged into the formula ΔG = ΔH – TΔS, you can solve for
ΔG by plugging in the given ΔH to begin solving for T. The other piece of
information you would need is the value for ΔS. Once you have this value, you
can solve for T.
6) Consider the following reaction:
2S (g) + 2OF2 (g)  SO2 (g) + SF4 (g) ΔH ° = -320 kJ
a) What is the change in enthalpy per mole of sulfur added?
ΔH = -320 kJ/2mol S  ΔH = -160 kJ/mol
b) What is the change in enthalpy per gram of SO2 formed?
ΔH = (-320 kJ/mol SO2) x (1mol SO2/64.06g SO2)
7) Find H for this reaction : 4 NH3(g) + 7 O2 (g)  4 NO2 (g) + 6 H2O (l).
Use this data for H°f :
NH3 (g) = -46 kJ/mol
H2O (l) = -286 kJ/mol
NO2 (g) = 34 kJ/mol
ΔH = ΣH°f products - ΣH°f reactants
ΔH = [4(34 kJ/mol) + 6(-286 kJ/mol)] – 4(-46 kJ/mol)  WATCH SIG FIGS
ΔH = 140 kJ/mol -1720 kJ/mol + 180 kJ/mol
ΔH = -1400 kJ/mol