12.E: Thermodynamics - Homework
Spontaneity
-
Indicate whether the following processes are spontaneous or nonspontaneous.
- Liquid water freezing at a temperature below its freezing point
- Liquid water freezing at a temperature above its freezing point
- The combustion of gasoline
- A ball thrown into the air
- A raindrop falling to the ground
- Iron rusting in a moist atmosphere
- Answer
-
- spontaneous
- nonspontaneous
- spontaneous
- nonspontaneous
- spontaneous
- spontaneous
-
Many plastic materials are organic polymers that contain carbon and hydrogen. The oxidation of these plastics in air to form carbon dioxide and water is a spontaneous process; however, plastic materials tend to persist in the environment. Explain.
- Answer
-
Although the oxidation of plastics is spontaneous, the rate of oxidation is very slow. Plastics are therefore kinetically stable and do not decompose appreciably even over relatively long periods of time.
Entropy
-
Determine the entropy change for the combustion of gaseous propane, C
3
H
8
, under the standard conditions to give gaseous carbon dioxide and water.
- Answer
- 100.6 J/K
-
By calculating Δ
S
univ
at each temperature, determine if the melting of 1 mole of NaCl(
s
) is spontaneous at 500 °C and at 700 °C.
- Answer
- As Δ S univ < 0 at each of these temperatures, melting is not spontaneous at either of them. The given values for entropy and enthalpy are for NaCl at 298 K.
-
What assumptions are made about the thermodynamic information (entropy and enthalpy values) used to solve this problem?
- Answer
- It is assumed that these do not change significantly at the higher temperatures used in the problem.
Free Energy
-
A reaction has
= 100 kJ/mol and
Is the reaction spontaneous at room temperature? If not, under what temperature conditions will it become spontaneous?
- Answer
-
The reaction is nonspontaneous at room temperature.
Above 400 K, Δ G will become negative, and the reaction will become spontaneous.
-
Consider a system similar to the one in
Figure 16.8
, except that it contains six particles instead of four. What is the probability of having all the particles in only one of the two boxes in the case? Compare this with the similar probability for the system of four particles that we have derived to be equal to
What does this comparison tell us about even larger systems?
- Answer
- The probability for all the particles to be on one side is 1/32. This probability is noticeably lower than the 1/8 result for the four-particle system. The conclusion we can make is that the probability for all the particles to stay in only one part of the system will decrease rapidly as the number of particles increases, and, for instance, the probability for all molecules of gas to gather in only one side of a room at room temperature and pressure is negligible since the number of gas molecules in the room is very large.
-
Consider the system shown in
Figure 16.9
. What is the change in entropy for the process where the energy is initially associated with particles A and B, and the energy is distributed between two particles in different boxes (one in A-B, the other in C-D)?
- Answer
-
There is only one initial state. For the final state, the energy can be contained in pairs A-C, A-D, B-C, or B-D. Thus, there are four final possible states. \( \Delta \) S = 1.91 x 10 -23 J/K
-
At room temperature, the entropy of the halogens increases from I
2
to Br
2
to Cl
2
. Explain.
- Answer
- The masses of these molecules would suggest the opposite trend in their entropies. The observed trend is a result of the more significant variation of entropy with a physical state. At room temperature, I 2 is a solid, Br 2 is a liquid, and Cl 2 is a gas.
-
Indicate which substance in the given pairs has the higher entropy value. Explain your choices.
- C 2 H 5 OH( l ) or C 3 H 7 OH( l )
- C 2 H 5 OH( l ) or C 2 H 5 OH( g )
- 2H( g ) or H( g )
- Answer
-
- C 3 H 7 OH( l ) as it is a larger molecule (more complex and more massive), and so more microstates describing its motions are available at any given temperature.
- C 2 H 5 OH( g ) as it is in the gaseous state.
- 2H( g ), since entropy is an extensive property, and so two H atoms (or two moles of H atoms) possess twice as much entropy as one atom (or one mole of atoms).
-
Predict the sign of the entropy change for the following processes. Give a reason for your prediction.
- Na + ( aq ) + Cl - ( aq ) \( \rightarrow \) NaCl ( s )
- Answer
-
- Negative. The relatively ordered solid precipitating decreases the number of mobile ions in solution.
- Positive. There is a net increase of seven moles of gas from reactants to products.
-
Consider the decomposition of red mercury(II) oxide under standard state conditions.
- Is the decomposition spontaneous under standard state conditions?
- Above what temperature does the reaction become spontaneous?
- Answer
-
- The reaction is nonspontaneous;
- Above 566 °C the process is spontaneous
-
Calculate Δ
G
° for each of the following reactions from the equilibrium constant at the temperature given.
- Answer
-
- The reaction is nonspontaneous;
- Above 566 °C the process is spontaneous.
-
Calculate the equilibrium constant at 25 °C for each of the following reactions from the value of Δ
G
° given.
- Answer
-
- K = 41;
- K = 0.053
-
The evaporation of one mole of water at 298 K has a standard free energy change of 8.58 kJ.
- Is the evaporation of water under standard thermodynamic conditions spontaneous?
- Determine the equilibrium constant, K P , for this physical process.
- By calculating ∆ G , determine if the evaporation of water at 298 K is spontaneous when the partial pressure of water, is 0.011 atm.
- If the evaporation of water were always nonspontaneous at room temperature, wet laundry would never dry when placed outside. In order for laundry to dry, what must be the value of in the air?
- Answer
-
-
(a) Under standard thermodynamic conditions, the evaporation is nonspontaneous; (b)
- K p = 0.031
- The evaporation of water is spontaneous
- P H2O must always be less than K p or less than 0.031 atm. 0.031 atm represents air saturated with water vapor at 25 °C, or 100% humidity.
-
-
One of the important reactions in the biochemical pathway glycolysis is the reaction of glucose-6-phosphate (G6P) to form fructose-6-phosphate (F6P):
- Is the reaction spontaneous or nonspontaneous under standard thermodynamic conditions?
- Standard thermodynamic conditions imply the concentrations of G6P and F6P to be 1 M , however, in a typical cell, they are not even close to these values. Calculate Δ G when the concentrations of G6P and F6P are 120 μ M and 28 μ M respectively, and discuss the spontaneity of the forward reaction under these conditions. Assume the temperature is 37 °C.
- Answer
-
- Nonspontaneous as ΔG < 0
- ΔG = Δ G o + RT lnQ. ΔG=-2.1 kJ. The forward reaction to produce F6P is spontaneous under these conditions.
-
When ammonium chloride is added to water and stirred, it dissolves spontaneously and the resulting solution feels cold. Without doing any calculations, deduce the signs of Δ
G
, Δ
H
, and Δ
S
for this process, and justify your choices.
- Answer
-
Δ G is negative as the process is spontaneous. Δ H is positive as with the solution becoming cold, the dissolving must be endothermic. Δ S must be positive as this drives the process, and it is expected for the dissolution of any soluble ionic compound.
-
What happens to
(becomes more negative or more positive) for the following chemical reactions when the partial pressure of oxygen is increased?
- Answer
-
- Increasing the oxygen partial pressure will yield a decrease in Q and ΔG thus becomes more negative.
- Increasing the oxygen partial pressure will yield a decrease in Q and ΔG thus becomes more negative.
- Increasing the oxygen partial pressure will yield an increase in Q and ΔG thus becomes more positive