Solutions to Homework Assignment #1
CHM 152
Spring 2002
6.12
6.37
(a)
∆H = 2(−1452.8kJ) = −2905.6 kJ
(b)
∆H = +1452.8 kJ
(c)
∆H = −1452.8 kJ + 4(44 kJ) = −1276.8 kJ
(a)
∆H° = 2 ∆H fD (H2O) − 2 ∆H fD (H2) − ∆H fD (O2)
∆H° = (2 mol)(−285.8 kJ/mol) − (2 mol)(0) − (1 mol)(0) = −571.6 kJ
(b)
∆H° = 4 ∆H fD (CO2) + 2 ∆H fD (H2O) − 2 ∆H fD (C2H2) − 5 ∆H fD (O2)
∆H° = (4 mol)(−393.5 kJ/mol) + (2 mol)(−285.8 kJ/mol) − (2 mol)(226.6 kJ/mol) − (5 mol)(0)
= −2599 kJ
6.43
The amount of heat given off is:
 1 mol NH 3   −92.6 kJ 
126
. × 104 g NH 3 

 = −3.43 × 104 kJ
 17.03 g NH 3   2 mol NH 3 
18.2
(a) spontaneous, (b) nonspontaneous, (c) spontaneous, (d) nonspontaneous
18.4
Entropy is a measure of the disorder or randomness of a system. The units are J/mol◊K.
18.5
(a) entropy increases, (b) entropy decreases, (c) entropy increases, (d) entropy decreases, (e) entropy
decreases, (f) entropy increases, (g) entropy increases
18.9
(a)
This is easy. The liquid form of any substance always has greater entropy (molecular level disorder).
(b)
This is hard. At first glance there may seem to be no apparent difference between the two substances
that might affect the entropy (molecular formulas identical). However, the first has the −O−H
structural feature which allows it to participate in hydrogen bonding with other molecules. This
allows a more ordered arrangement of molecules in the liquid state. The standard entropy of
CH3OCH3 is larger.
(c)
This is also difficult. Both are monatomic species. However, the Xe atom has a greater molar mass
than Ar. Xenon has the higher standard entropy.
(d)
Same argument as part (c). Carbon dioxide gas has the higher standard entropy (see Appendix 3).
(e)
O3 has a greater molar mass than O2 and thus has the higher standard entropy.
(f)
Using the same argument as part (c), one mole of N2O4 has a larger standard entropy than one mole
of NO2. Compare values in Appendix 3.
Use the data in Appendix 3 to compare the standard entropy of one mole of N2O4 with that of two
moles of NO2. In this situation the number of atoms is the same for both. Which is higher and why?
18.12
The standard entropy change for a reaction can be calculated using the following equation.
D
∆Srxn
= ΣnS°(products) − ΣmS°(reactants)
(a)
D
∆Srxn
= S°(Cu) + S°(H2O) − [S°(H2) + S°(CuO)]
= (1 mol) (33.3 J/K⋅mol) + (1 mol)(188.7 J/K⋅mol) − [(1 mol)(131.0 J/K⋅mol) + (1 mol)(43.5 J/K⋅mol)]
= 47.5 J/K
(b)
D
∆Srxn
= S°(Al2O3) + 3S°(Zn) − [2S°(Al) + 3S°(ZnO)]
= (1 mol)(50.99 J/K⋅mol) + (3 mol)(41.6 J/K⋅mol) − [(2 mol)(28.3 J/K⋅mol) + (3 mol)(43.9 J/K⋅mol)]
= −12.5 J/K
(c)
D
∆Srxn
= S°(CO2) + 2S°(H2O) − [S°(CH4) + 2S°(O2)]
= (1 mol)(213.6 J/K⋅mol) + (2 mol)(69.9 J/K⋅mol) − [(1 mol)(186.2 J/K⋅mol) + (2 mol)(205.0 J/K⋅mol)]
= −242.8 J/K
Why was the entropy value for water different in parts (a) and (c)?
18.13
18.17
All parts of this problem rest on two principles. First, the entropy of a solid is always less than the entropy
of a liquid, and the entropy of a liquid is always much smaller than the entropy of a gas. Second, in
comparing systems in the same phase, the one with the most complex particles has the higher entropy.
(a)
Positive entropy change (increase). One of the products is in the gas phase.
(b)
Negative entropy change (decrease). Liquids have lower entropies than gases.
(c)
Positive. Same as (a).
(d)
Positive. There are two gas−phase species on the product side and only one on the reactant side.
Using Equation (18.8) of the text to solve for the change in standard free energy,
(a)
∆G° = 2 ∆GfD (NO) − ∆GfD (N2) − ∆GfD (O2) = (2 mol)(86.7 kJ/mol) − 0 − 0 = 173.4 kJ
(b)
∆G° = ∆GfD [H2O(g)] − ∆GfD [H2O(l)] = (1 mol)(−228.6 kJ/mol) − (1 mol)(−237.2 kJ/mol) = 8.6 kJ
(c)
∆G° = 4 ∆GfD (CO2) + 2 ∆GfD (H2O) − 2 ∆GfD (C2H2) − 5 ∆GfD (O2)
= (4 mol)(−394.4 kJ/mol) + (2 mol)(−237.2 kJ/mol) − (2 mol)(209.2 kJ/mol) − (5 mol)(0) = −2470 kJ
18.19
Reaction A: First apply Equation (18.7) of the text to compute the free energy change at 25°C (298 K)
∆G = ∆H − T∆S = 10,500 J − (298 K × 30 J/K) = 1560 J
The +1560 J shows the reaction is not spontaneous at 298 K. The ∆G will change sign (i.e., the reaction
will become spontaneous) above the temperature at which ∆G = 0.
T =
∆H 10500 J
=
= 350 K
30 J / K
∆S
Reaction B: Calculate ∆G.
∆G = ∆H − T∆S = 1800 J − (−298 K × 113 J/K) = 35,500 J
The free energy change is positive, which shows that the reaction is not spontaneous at 298 K. Since both
terms are positive, there is no temperature at which their sum is negative. The reaction is not spontaneous
at any temperature.
18.20
Reaction A: Calculate ∆G from ∆H and ∆S.
∆G = ∆H − T∆S = −126,000 J − (298 K × 84 J/K) = −151,000 J
The free energy change is negative so the reaction is spontaneous at 298 K. Since ∆H is negative and ∆S is
positive, the reaction is spontaneous at all temperatures.
Reaction B: Calculate ∆G.
∆G = ∆H − T∆S = −11,700 J − [(298 K) × (−105 J/K)] = +19,600 J
The free energy change is positive at 298 K which means the reaction is not spontaneous at that
temperature. The positive sign of ∆G results from the large negative value of ∆S. At lower temperatures,
the −T∆S term will be smaller thus allowing the free energy change to be negative.
∆G will equal zero when ∆H = T∆S.
Rearranging,
T =
−11700 J
∆H
=
= 111 K
−105 J / K
∆S
At temperatures below 111 K, ∆G will be negative and the reaction will be spontaneous.
18.32
The standard free energy change is given by:
D
∆Grxn
= ∆GfD (graphite) − ∆GfD (diamond)
You can look up the standard free energy of formation values in Appendix 3 of the text.
D
∆Grxn
= (1 mol)(0) − (1 mol)(2.87 kJ/mol) = −2.87 kJ
Thus, the reaction is spontaneous under standard state conditions at 25°C. However, the rate of the
diamond to graphite conversion is very slow so that it will take millions of years before the process is
complete.
18.43
For a reaction to be spontaneous, ∆G must be negative. If ∆S is negative, as it is in this case, then the
reaction must be exothermic (why?). When water freezes, it gives off heat (exothermic). Consequently,
the entropy of the surroundings increases and ∆Suniverse > 0.
18.44
If the process is spontaneous as well as endothermic, the signs of ∆G and ∆H must be negative and
positive, respectively. Since ∆G = ∆H − T∆S, the sign of ∆S must be positive (∆S > 0) for ∆G to be
negative.
18.52
Assuming that both ∆H° and ∆S° are temperature independent, we can calculate both ∆H° and ∆S°.
∆H° = [ ∆H fD (CO) + ∆H fD (H2)] − [ ∆H fD (H2O) + ∆H fD (C)]
∆H° = [(1 mol)(−110.5 kJ/mol) + (1 mol)(0 kJ/mol)] − [(1 mol)(−241.8 kJ/mol) + (1 mol)( 0 kJ/mol)]
∆H° = 131.3 kJ
∆S° = [S°(CO) + S°(H2)] − [S°(H2O) + S°(C)]
∆S° = [(1 mol)(197.9 J/K⋅mol) + (1 mol)(131.0 J/K⋅mol)] − [(1 mol)(188.7 J/K⋅mol) + (1 mol)(5.69 J/K⋅mol)]
∆S° = 134.5 J/K
It is obvious from the given conditions that the reaction must take place at a fairly high temperature (in
order to have red−hot coke). Setting ∆G° = 0
0 = ∆H° − T∆S°
T =
∆H D
∆S D
 1000 J 
(131.3 kJ) 

 1 kJ  = 976 K = 703D C
=
134.5 J / K
The temperature must be greater than 703°C for the reaction to be spontaneous.
18.57
(a)
∆S > 0
(b)
∆S < 0
(c)
∆S > 0
(d)
∆S > 0