Chem 121 Winter 2016: Section 03, Sample Problems
Thermodynamics
1
When a 9.55g sample of solid sodium hydroxide dissolves in 100.0g of water in a coffee-cup
calorimeter the temperature rises from 21.6 to 37.8oC.
(a) Calculate the quantity of heat released in the reaction.
(b) Using your result from part (a), calculate H in kJ/mol of NaOH for the solution process.
Assume that the specific heat of the solution is the same as that of pure water.
(a)
The specific heat capacity of water is 4.18 J K-1 g-1. The 9.55g of NaOH dissolves in 100 mL of
water (whose density is 1 g mL-1) to give 109.55 g of solution, with the same heat capacity as
water. The heat required to raise the temperature of this from 21.6 to 37.8 oC is therefore
109.55 x 4.18 x (37.8 – 21.6) = 7418 J. (Note that we don’t need to change oC to K in this
instance, as we are just subtracting one temperature from another.)
(b)
The molecular weight of NaOH is 23 + 16 + 1 = 40 g mol-1, so 9.55 is 9.55/40 = 0.239 moles.
Thus the enthalpy change when one mole of sodium hydroxide is dissolved is 7,418/0.239 =
31,040 J.
2
Which of the following processes are spontaneous and which are non-spontaneous?
(a) the melting of ice at -5oC and 1 bar pressure; not spontaneous (since this is below the
normal freezing point of water)
(b) dissolution of sugar in a cup of hot coffee; spontaneous
(c) the reaction of nitrogen atoms to form N2 molecules at 25oC and 1 bar; spontaneous (if
that were not true, the atmosphere would be full of nitrogen atoms)
(d) alignment of iron filings in a magnetic field; spontaneous
(e) formation of CH4 and O2 molecules from CO2 and H2O at room temperature and 1 bar. Not
spontaneous (what is described is the reverse of the combustion of methane)
3
The element caesium, Cs, freezes at 28.4oC and its molar enthalpy of fusion at that
temperature is 2.09 kJ mol-1.
(a) When molten caesium solidifies at its normal melting point is S positive or negative?
(a)
Negative. We can explain this in two ways: qualitatively, liquids are always more
disorganised than solids (with the sole, but interesting exception of one isotope of liquid
helium), so if molten caesium solidifies its entropy must fall. Quantitatively we could use the
definition of S: S = qrev/T. The freezing of a solid is always exothermic (i.e., heat must be
removed to make it happen). That means qrev is negative, which in turn means that S is
negative.
(b) Calculate the value of S when 15.0g of Cs solidifies at 28.4oC.
(b)
Use the formula just given. Since the enthalpy of fusion that’s given is the molar enthalpy,
we start by finding the number of moles of Cs. The AW of Cs is 132.9 g mol-1, so 15.0g is
15.0/132.9 = 0.1129 moles. Thus the heat released upon freezing is 0.1129 x (-2090) = -236 J.
S = qrev/T = -236/(273+28.4) = -0.78 J K-1.
(c) What is the value of S when 30.0g Cs melts at the same temperature and 1 bar
pressure?
(c)
Easy. This is just the reverse process, with twice as much Cs, so we change the sign of the
answer from Part (b), because the process is now going backwards, and double the number:
+ 1.56 J K-1.
4
Predict the sign of the entropy change for the following reactions; justify your answer in each
case.
(a)
2SO2(g) + O2(g)  2SO3 (g) Minus. We are going from 3 moles of gas to 2, and
also going from a mixture to a pure material; both would make entropy diminish
(b)
Ba(OH)2(s)  BaO(s) + H2O(g)
gas phase.
5
Plus. Increase in the number of moles in the
(c)
CO(g) + 2H2(g) 
CH3OH(l) Minus. Decrease in the number of moles of gas.
(d)
gas.
FeCl2(s) + H2(g) 
Fe(s) + 2HCl(g)
Plus. Increase in the number of moles of
Use the data in the appendix of your text book (or look up data online) to determine the
standard enthalpy change, standard entropy change and standard Gibb’s energy change for
each of the following reactions:
(a)
Ni(s) + Cl2(g) 
(b)
CaCO3 (s, calcite)  CaO(s) + CO2(g)
(c)
P4O10(s) + 6H2O(l)  4H3PO4(aq)
(d)
2CH3OH(l) + 3O2  2CO2(g) + 4H2O(l)
NiCl2(s)
Comment on the values of each of the thermodynamic parameters for each reaction.
In each case the change in State Function is (products) – (reactants), taking into account the number
of moles of each species. Thus, for example, in equation (c) we take 4 x the enthalpy of
phosphoric acid and subtract from it the enthalpy of one mole of P4H10 and 6 x the enthalpy
of water. The same process is used for each State Function.
Comments might include whether the reaction is exo- or endo-thermic, and possible reasons for this,
why the change in entropy has the sign it has, and what the value of G tells us about the
direction of spontaneity of the reaction.
6
From the values given for Ho and So calculate Go for each of the reactions given below. If
the reaction is not spontaneous at 298K and 1 bar, determine at what temperature (if any)
the reaction would become spontaneous.
(a)
2PbS(s) + 3O2(g)  2PbO(s) + 2SO2(g)
Ho=-844 kJ,
So=0165 JK-1
(b)
2POCl3(g)  2PCl3(g) + O2(g)
Ho=572 kJ,
So=179 J K-1
G = H – T S, so use the values given to determine G at 298K.
A reaction flips from spontaneous to non-spontaneous, or from non-spontaneous to spontaneous,
as G reaches a value of zero, so the second part is done by setting G=0 and solving for the
temperature.
7
Consider the following reaction between oxides of nitrogen: NO2(g) + N2O(g)  3NO(g)
(a) Use data from the appendix, or online, to predict how Go for the reaction changes with
temperature.
Look up values for H and S and use G = H – T S to find G.
(b) Calculate Go at 800K, assuming that Ho and So do not change with temperature. Is the
reaction spontaneous at 800K? [Note: You might be surprised to see that there’s a little
superscript zero in this question when the reaction is not being carried out at 298K, which is,
as you know, part of the definition of standard conditions that we’ve met. In fact we often
expand the definition of standard conditions to include a temperature or pressure different
from those to which the normal definition applies. Don’t worry about this; as far as we are
concerned, standard conditions will normally mean 298K and 1 bar.)
(c) Write down the expression for the equilibrium constant for the reaction.
K = [NO]3 / { [NO2] [N2O] }