Welcome to Chem 1B
Thermo-Chemistry
Chapters 9 & 10
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Thermodynamics - ch 9
1. For a particular process q = – 10 kJ and w = 25 kJ. Which of the
following statements is true?
a. Heat flows from the surroundings to the system.
b. The system does work on the surroundings.
c. ∆E = – 15 kJ
d. All of these are true.
e. None of these is true.
Thermodynamics - ch 9
Sign Notation
q > 0 ⇒ endothermic ⇒ heat is flowing from the surroundings into the system
q < 0 ⇒ exothermic ⇒ heat is flowing out of the system to the surroundings
w > 0 ⇒ work is being done on the system by the surroundings when a gas is
getting compressed by an external pressure
w < 0 ⇒ the system is doing work on the surroundings when a gas is
expanding against an external pressure
∆E > 0 ⇒ internal energy of a gas increases when the temperature of the gas
increases
∆E < 0 ⇒ internal energy of a gas decreases when the temperature of the gas
decreases
∆E = 0 ⇒ internal energy of a gas remains constant if isothermal
Thermodynamics - ch 9
2. A gas absorbs 4.8 J of heat and then performs 13.0 J of work. What is
the change in internal energy of the gas?
Thermodynamics - ch 9
3. Which of the following statements is correct?
a. The internal energy of a system increases when more work
is done by the system than heat is flowing into the system.
b. The internal energy of a system decreases when work is done on
the system and heat is flowing into the system.
c. The system does work on the surroundings when an ideal gas
expands against a constant external pressure.
d. All the statements are true.
e. All the statements are false.
Thermodynamics - ch 9
4. Which of the following are always endothermic?
a. Melting
b. Combustion
c. Condensation
d. All of the above
Thermodynamics - ch 9
Breaking bonds requires energy ⇒ q > 0
Making bonds releases energy ⇒ q < 0
Thermodynamics - ch 9
5. The following reaction occurs at constant temperature and pressure.
4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g)
a.
b.
c.
d.
e.
The system does work on the surroundings and w < 0
The surroundings do work on the system and w < 0
The system does work on the surroundings and w > 0
The surroundings do work on the system and w > 0
No work is done, w = 0
Thermodynamics - ch 9
3 Equations for Work
1) w = – Pext∆V
2) w = – ∆nRT
3) w = – nR∆T
equations 2 & 3
are derived from PV = nRT
choose the equation based on the
variables that are given in the problem
Thermodynamics - ch 9
6. There are two containers, one has 1 mole of CO2 (g) and the other
has 1 mole of Ne (g). Both gases are heated from 25 oC to 30 oC at
constant pressure. For each of the following indicate true or false.
a. No work is done in heating either gas
b. More work is done in heating CO2 (g) than for Ne (g)
c. The change in internal energy of CO2 (g) is greater than it
is for Ne (g)
Thermodynamics - ch 9
7. For nitrogen gas at 25 oC, Cv = 20.8 J K-1 mol-1 and Cp = 29.1 J K-1
mol-1. When a sample of nitrogen gas is heated at constant
pressure, what fraction of the energy is converted to heat?
a. 1.00
b. 0.417
c. 0.285
d. 0.715
e. 0.588
Thermodynamics - ch 9
At constant pressure the fraction of
energy converted to heat
⇒ Cv /Cp
At constant volume the fraction
of energy converted to heat
⇒ Cv /Cv or 1
Thermodynamics - ch 9
8. Which of the following are state functions?
a. work, heat
b. work, heat, enthalpy, energy
c. enthalpy, energy
d. work, heat, enthalpy
e. heat, enthalpy, energy
Thermodynamics - ch 9
State Functions
are pathway independent ⇒ since ∆ is defined as (final state –
initial state) then ∆ anything can not depend on the states in
between the initial and final states a.k.a. the “path” ⇒ if
dealing with a state function you can use any path to solve
for it ⇒ the easiest paths for ∆E and ∆H are…
∆E = qv
∆H = qp
Thermodynamics - ch 9
State Functions
Think of all the
possible ways you
could go from peak A
to peak B – no matter which
path you choose the value of
Δh will be the same
Peak B
Peak A
Δh
Thermodynamics - ch 9
9. Which one of the following statements is false?
a. The change in internal energy, ∆E, for a process is equal to
the amount of heat absorbed at constant volume, qv.
b. The change in enthalpy, ∆H, for a process is equal to the
amount of heat absorbed at constant pressure, qp.
c. A bomb calorimeter measures ∆H directly.
d. If qp for a process is negative, the process is exothermic.
e. The freezing of water is an example of an exothermic
reaction.
Thermodynamics - ch 9
10. When a gas is cooled at constant volume from an initial
temperature of 50 oC to 25 oC it loses 60 J of heat. When the same
gas is cooled from 50 oC to 25 oC at constant pressure, it loses 100
J of heat. What is the value of the enthalpy change, ∆H, when the
gas is heated from 25 oC to 50 oC at constant volume?
a. + 60 J
b. – 60 J
c. – 100 J
d. + 100 J
e. – 40 J
Thermodynamics - ch 9
11. Suppose you add 45 J of heat to a system, let it do 10. J of
expansion work, and then return the system to its initial state by
cooling and compression. Which statement is true for this process?
a. ∆H < ∆E
b. The work done in compressing the system must exactly equal
the work done by the system in the expansion step.
c. ∆H = 70. J
d. The change in the internal energy for this process is zero.
Thermodynamics - ch 9
12. Consider a gas in a 1.0-L bulb at STP that is connected via a valve
to another bulb that is initially evacuated. After the valve is
opened…
a. q > 0
q<0
q=0
b. w > 0 w < 0 w = 0
c. ∆H > 0 ∆H < 0 ∆H = 0
d. ∆E > 0 ∆E < 0 ∆E = 0
Thermodynamics - ch 9
13. For the following reactions at constant pressure, predict if
∆H > ∆E, ∆H < ∆E, or ∆H = ∆E. Why is there a difference? Is
work being done by the system or the surroundings?
a. H2 (g) + Cl2 (g) → 2 HCl (g)
b. N2 (g) + 3 F2 (g) → 2 NF3 (g)
Thermodynamics - ch 9
14. Consider a sample containing 2 moles of xenon that undergoes the
following changes:
Pa = 10.0 atm
Va = 10.0 L
Step 1
Pb = 10.0 atm
Vb = 5.0 L
Step 2
Pc = 20.0 atm
Vc = 25.0 L
For each step, assume that the external pressure is constant
and equals the final pressure of the gas for that step. Calculate q, w, ΔE
and ΔH (in kJ) for each step and calculate the total values. Which values
would change if you went directly from state a to state c?
Thermodynamics - ch 9
15. Calculate ∆E (in kJ) when 1 mol of liquid is vaporized at 1 atm and
its boiling point (73°C). ∆Hvap = 28.6 kJ/mol at 73°C.
Thermodynamics - ch 9
16. Consider the combustion of ethanol:
C2H5OH (l) + 3O2 (g) → 2CO2 (g) + 3H2O (l)
∆H = –1.37×103 kJ
When a 15.6-g sample of ethanol (molar mass = 46.1 g/mol) is burned,
how much energy is released as heat?
A) 87.8 kJ
B) 2.14 × 104 kJ
C) 4.64 × 102 kJ
D) 4.05 × 103 kJ
E) 4.75 kJ
Thermodynamics - ch 9
17. Consider the following reaction:
CH4 + 4 Cl2 (g) → CCl4 (g) + 4 HCl (g), ∆H = – 434 kJ
How many grams of methane must be react to raise the temperature
of 1.2 L of water 50°C (CH2O = 4.184 J/g°C)
Thermodynamics - ch 9
18. A 25.0 g piece of unknown metal was transferred from an oven at
115 °C into a coffee cup calorimeter containing 150. mL of water at
24 °C and allowed to come to equilibrium where the temperature
was measured to be 28 °C. Calculate the specific heat capacity of
the metal. (Ccal = 25 J/K and Cwater = 4.18 J/g°C)
Thermodynamics - ch 9
19. A 5.00 g of aluminum pellets (CAl = 0.89 J°C-1g-1) and 10.0 g of
iron pellets (CFe = 0.45 J°C-1g-1) are heated to 100.0°C. The
mixture of hot aluminum and iron is then dropped into 97.3 mL of
water at 22.0°C. Calculate the equilibrium temperature assuming
no heat is lost to the surroundings. CH2O = 4.18 Jg-1°C-1
Thermodynamics - ch 9
20. In a coffee cup calorimeter 50.0 mL of 0.10 M NaOH and 20.0 mL
of 0.30 M HCl are mixed. The temperatures before and after the
reaction were measured to be 23.0°C and 31.0°C. Calculate the
molar enthalpy change for the neutralization reaction. Assume no
heat is absorbed by the calorimeter. The density of each solution
is 1.0 g/mL and the specific heat capacity for each solution is 4.18
J/g°C.
Thermodynamics - ch 9
21. Calculate the final temperature when 38 mL of 0.15 M Pb(NO3)2 is
mixed with 42 mL of 0.23 M KI both initially at 23 °C. Assume no
heat is lost to the surroundings. The specific heat capacity for the
solution is 4.18 J/g°C and the density is 1.0 g/mL.
Pb2+ + 2 I- → PbI2
∆H = – 297 kJ/mol
Thermodynamics - ch 9
22. Given the following data calculate the ∆H° for the reaction:
NO + O → NO2
Rxn 1: 2 O3 → 3 O2
Rxn 2: O2 → 2 O
Rxn 3: NO + O3 → NO2 + O2
∆H° = – 427 kJ
∆H° = +495 kJ
∆H° = – 195 kJ
Thermodynamics - ch 9
Hess’ 3 Laws
1. If a reaction is flipped ⇒
change the sign of ∆H
2. If a reaction is multiplied by a number ⇒
multiply ∆H by the same number
3. If two or more reactions are added ⇒
take the sum of the ∆Hs
Thermodynamics - ch 9 - Feldwinn
23. Calculate ΔH for the reaction
N2H4 (l) + O2 (g) → N2 (g) + 2 H2O (l)
given the following data:
2 NH3 (g) + 3 N2O (g) → 4 N2 (g) + 3 H2O (l)
N2O (g) + 3 H2 (g) → N2H4 (l) + H2O (l)
2 NH3 (g) + ½ O2 (g) → N2H4 (l) + H2O (l)
H2 (g) + ½ O2 (g) → H2O (l)
ΔH = – 1010. kJ
ΔH = – 317 kJ
ΔH = – 143 kJ
ΔH = – 286 kJ
Thermodynamics - ch 9
24. Choose the correct equation for the standard enthalpy of formation
of CO (g), where ∆H°f for CO = –110.5 kJ/mol.
a. 2 Cgraphite (s) + O2 (g) → 2 CO (g)
∆H° = –110.5 kJ
b. Cgraphite (s) + O (g) → CO (g)
∆H° = –110.5 kJ
c. Cgraphite (s) + ½ O2 (g) → CO (g)
∆H° = –110.5 kJ
d. Cgraphite (s) + CO2 (g) → 2 CO (g)
∆H° = –110.5 kJ
e. CO (g) → Cgraphite (s) + O (g)
∆H° = –110.5 kJ
Thermodynamics - ch 9
25. The combustion of methanol takes place according to the reaction
2 CH3OH (l) + 3 O2 (g) → 2 CO2 (g) + 4 H2O (l)
Calculate ∆H for the combustion of 1 mol of methanol under
standard conditions. Use the following standard enthalpies of
formation:
∆H°f for CH3OH (l) = – 238.5 kJ/mol
∆H°f for CO2 (g) = – 393.5 kJ/mol
∆H°f for H2O (l) = – 285.6 kJ/mol
Thermodynamics - ch 9
Hess’ Equation:
∆H°rxn = Σ ∆H°f (products) – Σ ∆H°f (reactants)
Thermodynamics - ch 9
26. The heat combustion of acetylene, C2H2 (g), at 25°C, is –1299
kJ/mol. At this temperature, ∆H°f values for CO2 (g) and H2O (l)
are –393 and –286 kJ/mol, respectively. Calculate ∆H°f for
acetylene.
a. 2376 kJ/mol
b. 625 kJ/mol
c. 227 kJ/mol
d. –625 kJ/mol
e. none of these
Thermodynamics - ch 10
27. Which has the greatest entropy?
a. 1 mol of He at STP or 1 mol of He at 25°C
b. 1 mol of Ne at STP or 1 mol of CH4 at STP
c. 1 mol of Cl2 at STP or 1 mol of F2 at STP
Thermodynamics - ch 10
Entropy (S) Considerations:
1. S(s) < S(l) << S(g)
2. There is more entropy at higher temperatures and/or larger volumes
(lower pressures)
3. The more bonds per molecule the greater the positional probability
ex: CH4 > H2
4. If there are the same number of atoms in the molecules/elements;
then the one with more electrons has the greater the positional
probability ex: Ar > He
5. For the same atom but different structures (allotropes) the positional
probability is greater in the more disordered structure ex: C
(graphite) > C (diamond)
Thermodynamics - ch 10
28. Predict if ∆Ssys and ∆Ssurr is positive or negative for the following
under standard conditions.
a. melting ice
b. photosynthesis ⇒
6 CO2 (g) + 6 H2O (l) → C6H12O6 (s) + 6 O2 (g)
c. precipitation of AgCl
Thermodynamics - ch 10
29. One mole of an ideal gas at 25°C is expanded isothermally and
reversibly from 125.0 L to 250.0 L. Which statement is correct?
a. ∆Sgas = 0
b. ∆Ssurr = 0
c. ∆Suniv = 0
Thermodynamics - ch 10
30. One mole of an ideal gas is compressed isothermally and reversibly
at 607.4 K from 5.60 atm to 8.90 atm. Calculate ∆S for the gas.
a. 2.34 J/K
b. – 2.34 J/K
c. – 3.85 J/K
d. 3.85 J/K
e. 0 J/K
Thermodynamics - ch 10
31. In a certain reversible expansion, a system at 300. K absorbs
exactly 6.00 × 102 J of heat. In the irreversible recompression to the
original state of the system, twice as much work is done on the
system as is performed on the surroundings in the expansion. What
is the entropy change of the system in the recompression step?
a) – 4.00 J/K
b) – 2.00 J/K
c) 0.00 J/K
d) 2.00 J/K
e) 4.00 J/K
Thermodynamics - ch 10
32. A system composed of a ideal gas expands spontaneously in one
step from an initial volume of 1.00 L to a final volume of 2.00 L at
a constant temperature of 200 K. During the process the gas does
200 J of work. What conclusion can be reached about the value of
the entropy change, ΔS, for this process?
a) ΔS = + 1.00 J/K
b) ΔS = – 1.00 J/K
c) ΔS is less than +1.00 J/K
d) ΔS is greater than +1.00 J/K
e) None of the above
Thermodynamics - ch 10
33. Calculate the change in entropy for a process in which 3.00 moles
of liquid water at 0°C is mixed with 1.00 mole of liquid water at
100°C in a perfectly insulated container. The molar heat capacity of
liquid water is 75.3 Jmol-lK-1
Thermodynamics - ch 10
34. Calculate ∆S when 54 g of water is heated from – 22°C to 156 °C at
a constant pressure of 1 atm. The heat capacities for solid, liquid
and gaseous water are 2.03 J/g°C, 4.18 J/g°C and 2.02 J/g°C
respectively. The enthalpies of fusion and vaporization are 6.01
kJ/mol and 40.7 kJ/mol respectively.
Thermodynamics - ch 10 - Felwinn
35. Consider the process
A (l) at 75°C → A (g) at 155°C
which is carried out at constant pressure. The total ΔS for this
process is 75.0 Jmol-1K-1. For A (l) and A (g) the Cp values are 75.0
Jmol-1K-1 and 29.0 Jmol-1K-1 respectively. Calculate ΔHvap at
125°C (its boiling point).
Thermodynamics - ch 10
36. Indicate true or false for each of the following statements.
a. Spontaneous reactions must have a positive ΔSº for the
reaction.
b. When the change in free energy is less than zero for a
chemical reaction, the reaction must be exothermic.
c. For a spontaneous reaction, if ΔSº < 0 then the reaction must
be exothermic.
Thermodynamics - ch 10
Spontaneous ⇒ wants to go forward on its own K > Q
∆Suniverse > 0 or ∆Gsystem < 0
Non-spontaneous ⇒ wants to go backward on its own K < Q
∆Suniverse < 0 or ∆Gsystem > 0
Equilibrium ⇒ doesn’t prefer one direction over the other K = Q
∆Suniverse = 0 or ∆Gsystem = 0
Thermodynamics - ch 10
∆H
∆S
Spontaneous Temperatures
∆H < 0
negative
∆S > 0
positive
All
Temperatures
∆H > 0
positive
∆S < 0
negative
No
Temperatures
∆H > 0
positive
∆S > 0
positive
“High” Temperatures
Tspont > ∆H/ ∆S or Tspont > Teq**
∆H < 0
negative
∆S < 0
negative
“Low” Temperatures
Tspont< ∆H/ ∆S or Tspont < Teq**
** Note ⇒ If ∆H and ∆S are the same sign Teq = ∆H /∆S
Thermodynamics - ch 10
37. A 100-mL sample of water is placed in a coffee cup calorimeter.
When 1.0 g of an ionic solid is added, the temperature decreases
from 21.5°C to 20.8°C as the solid dissolves. Which of the
following is true for the dissolving of the solid?
a. ∆H < 0
b. ∆Suniv > 0
c. ∆Ssys < 0
d. ∆Ssurr > 0
e. none of these
Thermodynamics - ch 10
38. At what temperatures (high or low) would the dissociation of
hydrogen be spontaneous?
H2 (g)  2 H (g)
Thermodynamics - ch 10
39. For the freezing of water at – 10 °C and 1 atm, predict whether ∆H,
∆S, and ∆G should be positive, negative or zero.
Thermodynamics - ch 10
40. Given the following data, calculate the normal boiling point for
formic acid (HCOOH).
∆Hfo (kJ/mol)
S° (J/K·mol)
HCOOH (l)
– 410.
130.
HCOOH (g)
– 363
251
a. 2.57 K
b. 1730°C
c. 388°C
d. 82°C
e. 115°C
Thermodynamics - ch 10
41. Consider the following reaction at 25 oC.
CO (g) + H2O (g) → H2 (g) + CO2 (g)
For this reaction ΔHo = –5.36 kJ and ΔSo = –109.8 J /K. At what
temperatures will the reaction be spontaneous?
a. T > 48.8 K
b. T < 48.8 K
c. T > 20.5 K
d. T < 20.5 K
e. Spontaneous at all temperatures.
Thermodynamics - ch 10
42. Consider the following reaction.
2 POCl3 (g) → 2 PCl3 (g) + O2 (g)
The free energies of formation at 25 °C are given below. ΔGf°
POCl3 (g) = –502 kJ/mol, ΔGf° PCl3 (g) = –270 kJ/mol
Indicate true or false.
a. The entropy change for the reaction is positive.
b. The reaction is not spontaneous at standard conditions and
25 °C but will eventually become spontaneous if the
temperature is increased.
c. The equilibrium constant for the reaction at 298 K is less than 1.
d. The enthalpy change for the reaction is positive.
Thermodynamics - ch 10
ΔG° verses Keq
ΔG° tells you where the equilibrium lies
If ΔG° < 0 the equilibrium favors the products or Keq > 1
If ΔG° > 0 the equilibrium favors the reactants or Keq < 1
ΔG° = –RTlnKeq
Thermodynamics - ch 10
43. Consider the following reaction at 800 K.
2 NF3 (g) → N2 (g) + 3 F2 (g)
At equilibrium, the partial pressures are PN2= 0.040 atm, PF2=
0.063 atm and PNF3= 0.66 atm. Which of the following is true
about the value of ∆G°?
a. is a positive number
b. is a negative number
c. is equal to zero
d. is independent of the temperature
e. can not be predicted from this data
Thermodynamics - ch 10
44. Use the following reaction to answer the following.
2 SO2 (g) + O2 (g) → 2 SO3 (g)
a. Calculate ∆G°
b. What is the equilibrium constant at 25 °C?
c. At 25 °C the initial pressures for SO2, O2 and SO3 are 0.001
atm, 0.002 atm and 40 atm respectively. Will SO3 be
consumed or will SO3 be formed?
Substance
∆Gf° (kJ/mol)
SO2 (g)
–300
SO3 (g)
–321
Thermodynamics - ch 10
45. For a certain reaction the Kp is 50 at 25 °C and 9 x 103 at 110 °C.
a. Is the reaction endothermic or exothermic?
b. Calculate Kp at 275 °C?
Thermodynamics - ch 10 - Feldwinn
46. For a particular reaction a graph was made by plotting ln(Keq) as a
function of 1000/K. The x and y intercepts were found to be 3.0
and 40. respectively. Determine the values of ΔH° and ΔS° for this
reaction.
Answer key – ch. 9
1. For a particular process q = – 10 kJ and w = 25 kJ. Which of
the following statements is true?
a. Heat flows from the surroundings to the system.
b. The system does work on the surroundings.
c. ∆E = – 15 kJ
d. All of these are true.
e. None of these is true. If q < 0 heat flows out of the system.
If w > 0 work is done on the system by the surroundings.
∆E = q + w = -10 kJ + 25 kJ = 15 kJ
Answer key – ch. 9
2. A gas absorbs 4.8 J of heat and then performs 13.0 J of work.
What is the change in internal energy of the gas?
∆E = q + w
∆E = (4.8 J) + (-13.0 J) = -8.2 J
Answer key – ch. 9
3. Which of the following statements is correct?
a. The internal energy of a system increases when more work
is done by the system than heat is flowing into the system.
b. The internal energy of a system decreases when work is
done on the system and heat is flowing into the system.
c. The system does work on the surroundings when an ideal
gas expands against a constant external pressure.
w = -Pext∆V (“expand” => ∆V > 0 => w < 0)
d. All the statements are true.
e. All the statements are false.
Answer key – ch. 9
4. Which of the following are always endothermic?
a. Melting => heat is absorbed when bonds are broken
b. Combustion => always exothermic
c. Condensation => heat is released when bonds are formed
d. All of the above
Answer key – ch. 9
5. The following reaction occurs at constant temperature
and pressure.
4 NH3 (g) + 5 O2 (g)  4 NO (g) + 6 H2O (g)
a. The system does work on the surroundings and w < 0
The reaction goes from 9 moles of gas to 10 moles of
gas => ∆V > 0 => w < 0
b. The surroundings do work on the system and w < 0
c. The system does work on the surroundings and w > 0
d. The surroundings do work on the system and w > 0
e. No work is done, w = 0
Answer key – ch. 9
6. There are two containers; one has 1 mole of CO2 (g) and the other
has 1 mole of Ne (g). Both gases are heated from 25 oC to 30 oC
at constant pressure. For each of the following indicate true or
false.
a. No work is done when heating either gas False => w = 0 only
at constant volume or if expanding in a vacuum
b. More work is done in heating CO2 (g) than for Ne (g)
False => work is dependent on the change in temperature
which is the same for both gases
c. The change in internal energy of CO2 (g) is greater than it is
for Ne (g) True => the heat capacity for CO2 is higher
than Ne due to the bonds in the molecule
Answer key – ch. 9
7. For nitrogen gas at 25 oC, Cv = 20.8 J K-1 mol-1 and Cp = 29.1
J K-1 mol-1. When a sample of nitrogen gas is heated at
constant pressure, what fraction of the energy is converted to
heat?
a. 1.00
b. 0.417
c. 0.285
-1mol-1/29.1JK-1mol-1 = 0.715
C
/C
=
20.8JK
v p
d. 0.715
e. 0.588
Answer key – ch. 9
8. Which of the following are state functions?
a. work, heat
b. work, heat, enthalpy, energy
c. enthalpy, energy
d. work, heat, enthalpy
e. heat, enthalpy, energy
Answer key – ch. 9
9. Which one of the following statements is false?
a. The change in internal energy, ∆E, for a process is equal to
the amount of heat absorbed at constant volume, qv.
b. The change in enthalpy, ∆H, for a process is equal to the
amount of heat absorbed at constant pressure, qp.
c. A bomb calorimeter measures ∆H directly. In a bomb
calorimeter the ∆H is measured indirectly by observing
how much heat is absorbed by the calorimeter
d. If qp for a process is negative, the process is exothermic.
e. The freezing of water is an example of an exothermic
reaction.
Answer key – ch. 9
10. When a gas is cooled at constant volume from an initial
temperature of 50 oC to 25 oC it loses 60 J of heat. When the
same gas is cooled from 50 oC to 25 oC at constant pressure, it
loses 100 J of heat. What is the value of the enthalpy change,
∆H, when the gas is heated from 25 oC to 50 oC at constant
volume?
a. + 60 J
∆H is a state function
b. – 60 J
which means we can pick any path we
c. – 100 J
want to calculate ∆H => the easiest path
d. + 100 J
is constant pressure =>
e. – 40 J
∆H = qp = +100 J
Answer key – ch. 9
11. Suppose you add 45 J of heat to a system, let it do 10. J of
expansion work, and then return the system to its initial state
by cooling and compression. Which statement is true for this
process?
a. ∆H < ∆E
b. The work done in compressing the system must exactly
equal the work done by the system in the expansion step.
c. ∆H = 70. J
d. The change in the internal energy for this process is zero.
Since it has returned to it’s initial state there has been no
overall change to the system
Answer key – ch. 9
12. Consider a gas in a 1.0-L bulb at STP that is connected via a
valve to another bulb that is initially evacuated. After the
valve is opened…
a. q > 0 q < 0 q = 0
b. w > 0 w < 0 w = 0
No energy is required to
c. ∆H > 0 ∆H < 0 ∆H = 0
expand into a vacuum
d. ∆E > 0 ∆E < 0 ∆E = 0
Answer key – ch. 9
13. For the following reactions at constant pressure, predict if
∆H > ∆E, ∆H< ∆E, or ∆H = ∆E. Why is there a
difference? Is work being done by the system or the
surroundings?
a. H2 (g)
+
Cl2 (g) 
2 HCl (g)
∆H = ∆E + ∆(PV) or ∆H = ∆E + ∆(nRT)
since the moles of gas remain constant ∆H = ∆E
b. N2 (g) + 3 F2 (g)  2 NF3 (g)
since the moles of gas decrease => ∆H < ∆E
Answer key – ch. 9
14. Consider a sample containing 2 moles of xenon that
undergoes the following changes:
Pa = 10.0 atm
Va = 10.0 L
Step 1
Pb = 10.0 atm
Vb = 5.0 L
Step 2
Pc = 20.0 atm
Vc = 25.0 L
For each step, assume that the external pressure is constant and
equals the final pressure of the gas for that step. Calculate q, w, ΔE
and ΔH (in kJ) for each step and calculate the total values. Which
values would change if you went directly from state a to state c?
Step 1 => constant pressure => ΔH = q = nCp ΔT => if you use
PV=nRT to find the temperatures you get Ta = 609K, Tb = 305K
and Tc = 3047 K (for step 2) => continued on next slide…
Answer key – ch. 9
14. …continued
ΔH = q = (2mol)(5/2)(8.3145x10-3kJmol-1K-1)(305K-609K) = -12.6 kJ
w = -PΔV => w = -(10atm)(5.0L-10L)(0.1013kJatm-1L-1) = 5.1 kJ
ΔE = q + w = (-12.6 kJ) + (5.1 kJ) = -7.5 kJ
Step 2 => neither constant pressure or constant volume =>
w = -(20atm)(25L-5L)(0.1013kJatm-1L-1) = -40.5 kJ
ΔH = nCp∆T =(2mol)(5/2)(8.3145x10-3kJmol-1K-1)(3047K-305K)= 114 kJ
ΔE = nCv ∆T =(2mol)(3/2)(8.3145x10-3kJmol-1K-1)(3047K-305K)= 68.4 kJ
q = ΔE – w = (68.4 kJ) – (-40.5 kJ) = 109 kJ
Answer key – ch. 9
15. Calculate ∆E (in kJ) when 1 mol of liquid is vaporized at 1
atm and its boiling point (73°C). ∆Hvap = 28.6 kJ/mol at
73°C.
∆E = ∆H - ∆(PV) => ∆E = ∆H - ∆(nRT)
since T and R are constants
∆E = ∆H - ∆nRT
∆E = (1 mol)(28.6 kJ/mol) –
(1mol-0mol )(8.3145x10-3kJmol-1K-1)(73 + 273K)
= 25.7 kJ
Answer key – ch. 9
16. Consider the reaction:
C2H5OH (l) + 3O2 (g) → 2CO2 (g) + 3H2O (l) ∆H = –1.37×103 kJ
When a 15.6-g sample of ethyl alcohol (molar mass = 46.1 g/mol)
is burned, how much energy is released as heat?
A) 87.8 kJ
B) 2.14 × 104 kJ
C) 4.64 × 102 kJ
D) 4.05 × 103 kJ
E) 4.75 kJ
15.6 g C2H5OH x
1 mol C2H5OH
46.1 g
x
1370 kJ
1 mol C2H5OH
= 464 kJ
Answer key – ch. 9
17. Consider the following reaction:
CH4 + 4 Cl2 (g) → CCl4 (g) + 4 HCl (g), ∆H = –434 kJ
How many grams of methane must be burned to raise the
temperature of 1.2 L of water 50°C (CH2O = 4.184 J/g°C)
qH2O = mC ∆T = (1.2 kg)(4.184 J/g°C)(50°C) = 251 kJ
251 kJ x 1 mol CH4 x 16 g CH4 = 9.3 g CH4
434 kJ
1 mol CH4
Answer key – ch. 9
18. A 25.0 g piece of unknown metal was transferred from an
oven at 115 °C into a coffee cup calorimeter containing 150.
mL of water at 24 °C and allowed to come to equilibrium
where the temperature was measured to be 28 °C. Calculate
the specific heat capacity of the metal. (Ccal = 25 J/K and
Cwater = 4.18 J/g°C)
heat lost by metal = heat absorbed by water and the calorimeter
-qmetal = qwater + qcal
-mC∆T = mC∆T + C∆T
-(25g)(C)(28°C-115°C) = (150g)(4.18 J/g°C)(28°C- 24°C) + (25
J/K)(28°C- 24°C)
Cmetal = 1.2 J/g°C
Answer key – ch. 9
19. A 5.00 g of aluminum pellets (CAl = 0.89 J°C-1g-1) and 10.0 g
of iron pellets (CFe = 0.45 J°C-1g-1) are heated to 100.0°C.
The mixture of hot aluminum and iron is then dropped into
97.3 mL of water at 22.0°C. Calculate the equilibrium
temperature assuming no heat is lost to the surroundings.
CH2O = 4.18 Jg-1°C-1
-(qAl +qFe) = qwater
-(mC∆T + mC∆T) = mC∆T
-((5g)(0.89 J°C-1g-1)(Tf-100°C) + (10g)(0.45 J°C-1g-1)(Tf-100°C))
= (97.3g)(4.18 Jg-1°C-1)(Tf-22°C)
Tf = 23.7°C
Answer key – ch. 9
20. In a coffee cup calorimeter 50.0 mL of 0.10 M NaOH and
20.0 mL of 0.30 M HCl are mixed. The temperatures before
and after the reaction were measured to be 23.0°C and
31.0°C. Calculate the molar enthalpy change for the
neutralization reaction. Assume no heat is absorbed by the
calorimeter. The density of each solution is 1.0 g/mL and the
specific heat capacity for each solution is 4.18 J/g°C.
-qrxn = +qsoln
-qrxn = mC∆T = (50g+20g)(4.18 J/g°C)(31°C-23°C) = 2341J
qrxn = -2.34 kJ
continued on next slide…
Answer key – ch. 9
24. …continued
HCl = > (20 mL)(0.3M) = 6 mmol
NaOH => (50 mL)(0.1M) = 5 mmol => Limiting reagent
since
HCl + NaOH  NaOH + H2O
qrxn = -2.34 kJ/0.005mol = -468 kJ/mol
Answer key – ch. 9
21. Calculate the final temperature when 38 mL of 0.15 M
Pb(NO3)2 is mixed with 42 mL of 0.23 M KI both initially at
23 °C. Assume no heat is lost to the surroundings. The
specific heat capacity for the solution is 4.18 J/g°C and the
density is 1.0 g/mL.
Pb2+ + 2 I-  PbI2
∆H = -297 kJ/mol
-qrxn = qsoln
-qrxn = mC∆T
to find qrxn we need to identify the limiting reagent
Continued on next slide…
Answer key – ch. 9
21. …continued
Pb2+ => (38 mL)(0.15 M) = 5.7 mmol
I- => (42 mL)(0.23 M) = 9.7 mmol => Limiting Reagent
qrxn = (0.0097 mol I-)(-297kJ/2 mol I-) = 1440J
now using
-qrxn = mC∆T
-(-1440J) = (38g + 42g)(4.18 J/g°C)(Tf - 23°C)
Tf = 27.3 °C
Answer key – ch. 9
22. Given the following data calculate the ∆H° for the reaction:
NO + O  NO2
2 O3  3 O2
O2  2 O
NO + O3  NO2 + O2
∆H° = -427 kJ
∆H° = +495 kJ
∆H° = -195 kJ
Rxn 1 => flip and divide by 2
Rxn 2 => flip and divide by 2
Rxn 3 => keep as is
∆H°total = (-427 kJ)(-0.5) + (495 kJ)(-0.5) + (-195 kJ) = -49 kJ
Answer key – ch. 9
23. Calculate ΔH for the reaction
N2H4 (l) + O2 (g)  N2 (g) + 2 H2O (g)
given the following data:
2 NH3 (g) + 3 N2O (g)  4 N2 (g) + 3 H2O (l) ΔH = -1010. kJ
N2O (g) + 3 H2 (g)  N2H4 (l) + H2O (l)
ΔH = -317 kJ
2 NH3 (g) + ½ O2 (g)  N2H4 (l) + H2O (l)
ΔH = -143 kJ
H2 (g) + ½ O2 (g)  H2O (l)
ΔH = -286 kJ
continued on next slide…
Answer key – ch. 9
23. …continued
Rxn 1 => multiply by 1/4
Rxn 2 => flip and multiply by 3/4
Rxn 3 => flip and multiply by 1/4
Rxn 4 => multiply by 9/4
∆Htotal = (1/4)(-1010. kJ) + (-3/4)(-317 kJ) + (-1/4)(-143 kJ) +
(9/4)(-286 kJ) = -622.5 kJ
Answer key – ch. 9
24. Choose the correct equation for the standard enthalpy of
formation of CO (g), where ∆H°f for CO = –110.5 kJ/mol.
a. 2 Cgraphite (s) + O2 (g) → 2 CO (g)
∆H° = –110.5 kJ
b. Cgraphite (s) + O (g) → CO (g)
∆H° = –110.5 kJ
c. Cgraphite (s) + ½ O2 (g) → CO (g)
∆H° = –110.5 kJ
d. Cgraphite (s) + CO2 (g) → 2 CO (g)
∆H° = –110.5 kJ
e. CO (g) → Cgraphite (s) + O (g)
∆H° = –110.5 kJ
Answer key – ch. 9
25. The combustion of methanol takes place according to the
reaction
2 CH3OH (l) + 3 O2 (g) → 2 CO2 (g) + 4 H2O (l)
Calculate ∆H° for the combustion of 1 mol of methanol under
standard conditions. Use the following standard enthalpies of
formation:
∆H°f for CH3OH (l) = -238.5 kJ/mol
∆H°f for CO2 (g) = -393.5 kJ/mol
∆H°f for H2O (l) = -285.6 kJ/mol
∆H°rxn = Σ∆H°(products) - Σ∆H°(reactants)
∆H°rxn = (2 mol CO2)(-393.5 kJ/mol) + (4 mol H2O)(-285.6
kJ/mol) – (2 mol CH3OH)(-238.5 kJ/mol) = -1452 kJ
if only 1 mol of CH3OH => (0.5)(-1452 kJ) = -726 kJ
Answer key – ch. 9
26. The heat combustion of acetylene, C2H2 (g), at 25°C, is –1299
kJ/mol. At this temperature, ∆H°f values for CO2 (g) and H2O
(l) are –393kJ/mol and –286 kJ/mol, respectively. Calculate
∆H°f for acetylene.
a. 2376 kJ/mol
C2H2 + 5/2O2  2CO2 + H2O
b. 625 kJ/mol
c. 227 kJ/mol
∆H°rxn = Σ∆H°(products) - Σ∆H°(reactants)
d. –625 kJ/mol
(-1299kJ) = (2mol CO2)(-393kJ/mol) +
e. none of these
(1mol H2O)(-286kJ/mol) – (1mol C2H2)(∆H°f )
∆H°f = 227 kJ/mol
Answer key – ch. 10
27. Which has the greatest entropy?
a. 1 mol of He at STP or 1 mol of He at 25°C
as T ↑ S ↑
b. 1 mol of Ne at STP or 1 mol of CH4 at STP
as #atoms/molecule ↑ S ↑
c. 1 mol of Cl2 at STP or 1 mol of F2 at STP
as #subatomic particles/molecule ↑ S ↑
Answer key – ch. 10
28. Predict if ∆Ssys and ∆Ssurr is positive or negative for the
following under standard conditions.
a. melting ice => H2O (s)  H2O (l) => ∆Ssys >0 since
Sliquid>Ssolid => ∆Hsys>0 since bonds are broken =>
∆Ssurr<0
b. photosynthesis =>
6 CO2 (g) + 6 H2O (l)  C6H12O6 (s) + 6 O2 (g)
∆Ssys<0 since the moles of gas are equal and considering
the moles solids ↑ and liquids ↓ => ∆Hsys>0 because the
reaction is combustion flipped => ∆Ssurr<0
c. precipitation of AgCl => Ag+ (aq) + Cl-(aq)  AgCl(s)
∆Ssys<0 since Ssolid<Saqueous => ∆Hsys<0 since bonds
are forming => ∆Ssurr>0
Answer key – ch. 10
29. One mole of an ideal gas at 25°C is expanded isothermally
and reversibly from 125.0 L to 250.0 L. Which statement is
correct?
a. ∆Sgas = 0
b. ∆Ssurr = 0
c. ∆Suniv = 0
∆Sgas > 0
since expanding a gas causes
an increase in entropy
qrev > 0 => ∆Ssurr < 0
∆Suniv = 0
since reversible a.k.a. equilibrium
Answer key – ch. 10
30. One mole of an ideal gas is compressed isothermally and
reversibly at 607.4 K from 5.60 atm to 8.90 atm. Calculate ∆S
for the gas.
a. 2.34 J/K
b. -2.34 J/K
c. -3.85 J/K
∆S = nRlnV2/V1
d. 3.85 J/K
since P1V1=P2V2
e. 0 J/K
∆S = nRlnP1/P2
∆S = (1mol)(8.314J/molK)(ln(5.6atm/8.9atm))
∆S = -3.85J/K
Answer key – ch. 10
31. In a certain reversible expansion, a system at 300. K absorbs
exactly 6.00 × 102 J of heat. In the irreversible recompression
to the original state of the system, twice as much work is
done on the system as is performed on the surroundings in the
expansion. What is the entropy change of the system in the
recompression step?
a) -4.00 J/K
∆S = qrev/T
b) -2.00 J/K
for the expansion
c) 0.00 J/K
∆S = (600J)/(300K) = 2 J/K
d) 2.00 J/K
therefore for the recompression
e) 4.00 J/K
∆S = -2 J/K
Answer key – ch. 10
32. A system composed of a ideal gas expands spontaneously in
one step from an initial volume of 1.00 L to a final volume of
2.00 L at a constant temperature of 200 K. During the process
the gas does 200 J of work. What conclusion can be reached
about the value of the entropy change, ΔS, for this process?
a) ΔS = + 1.00 J/K
wirrev = -200J
b) ΔS = – 1.00 J/K
∆E = 0 since isothermal
c) ΔS is less than +1.00 J/K
qirrev = 200J (qirrev <qrev)
d) ΔS is greater than +1.00 J/K
∆S = qrev/T
e) None of the above
qrev > qirrev
∆S > qirrev/T
∆S > (200J)/(200K) > 1J/K
Answer key – ch. 10
33. Calculate the change in entropy for a process in which 3.00
moles of liquid water at 0°C is mixed with 1.00 mole of
liquid water at 100°C in a perfectly insulated container. The
molar heat capacity of liquid water is 75.3 Jmol-lK-1
∆Stotal = ∆Scold + ∆Shot
temperature is changing => ∆S = nClnT2/T1
∆Stotal = nClnT2/T1 + nClnT2/T1
T2 is unknown
-nC∆T = nC∆T where C cancels out
-(1mol)(T2-100°C) = (3mol)(T2-0°C)
=> T2 = 25 °C
∆Stotal = (3mol)(75.3J/molK)(ln(298K)/(273K)) +
2.89J/K
J/K
(1mol)(75.3J/molK)(ln(298K)/(373K)) = 2.89
Answer key – ch. 10
34. Calculate ∆S when 54 g of water is heated from -22°C to 156 °C at a
constant pressure of 1 atm. The heat capacities for solid, liquid and
gaseous water are 2.03 J/g°C, 4.18 J/g°C and 2.02 J/g°C respectively.
The enthalpies of fusion and vaporization are 6.01 kJ/mol and 40.7
kJ/mol respectively.
Water will undergo two phase changes within this temperature range –
so we need to break it down into steps
Step 1 => water will be a solid from -22°C to 0°C => ∆S1 = mClnT2/T1
∆S1 = (54g)(2.03J/g°C)(ln(273K/251K)) = 9.2 J/K
Step 2 => water will melt at 0°C => ∆S2 = qrev/T
∆S2 = (54g/18g/mol)(6010J/mol)/(273K) = 66 J/K
continue to next slide…
Answer key – ch. 10
34. …continued
Step 3 => water will be a liquid from 0°C to 100°C => ∆S3 = mClnT2/T1
∆S3 = (54g)(4.18J/g°C)(ln(373K/273K)) = 70.4 J/K
Step 4 => water will boil at 100°C => ∆S4 = qrev/T
∆S4 = (54g/18g/mol)(40,700J/mol)/(373K) = 327J/K
Step 5 => water will be a gas from 100°C to 156°C => ∆S5 =mClnT2/T1
∆S5 = (54g)(2.02J/g°C)(ln(429K/373K) = 15 J/K
∆Stotal = 9.2 J/K + 66 J/K + 70.4 J/K + 327 J/K + 15.3 J/K = 473 J/K
Answer key – ch. 10
35. Consider the process
A (l) at 75°C  A (g) at 155°C
which is carried out at constant pressure. The total ΔS for this
process is 75.0 Jmol-1K-1. For A (l) and A (g) the Cp values are 75.0
Jmol-1K-1 and 29.0 Jmol-1K-1 respectively. Calculate ΔHvap at 125°C
(its boiling point).
Since a phase change will occur within this temperature range => break
it into steps
Step 1 => Substance A is a liquid from 75°C to 125°C
∆S1 = ClnT2/T1
continue to next slide…
Answer key – ch. 10
Step 2 => Substance A will boil at 125°C
∆S2 = qrev/T
Step 3 => Substance A will be a gas from 125°C to 155°C
∆S3 = ClnT2/T1
∆Stotal = ∆S1 + ∆S2 + ∆S3
75J/molK = (75J/molK)(ln(398K/348K)) + qrev/398K +
(29J/molK)(ln(428K/398K))
qrev = 25 kJ/mol
Answer key – ch. 10
36. Indicate true or false for each of the following statements.
a. Spontaneous reactions must have a positive ΔSº for the
reaction. False - ΔSº can be either positive or negative in
spontaneous reactions
b. When the change in free energy is less than zero for a
chemical reaction, the reaction must be exothermic. False ΔHº can be either positive or negative in spontaneous
reactions
c. For a spontaneous reaction, if ΔSº < 0 then the reaction
must be exothermic. True
Answer key – ch. 10
37. A 100-mL sample of water is placed in a coffee cup
calorimeter. When 1.0 g of an ionic solid is added, the
temperature decreases from 21.5°C to 20.8°C as the solid
dissolves. Which of the following is true for the dissolving of
the solid?
a. ∆H < 0 false - it was observed that the T of the
surroundings cooled
b. ∆Suniv > 0 true – because the solid dissolved spontaneously
c. ∆Ssys < 0 false – aqueous ions have more S than in a solid
d. ∆Ssurr > 0 false – since the dissolving was endothermic
e. none of these
Answer key – ch. 10
38. At what temperatures (high or low) would the dissociation of
hydrogen be spontaneous?
H2 (g)  2 H (g)
∆H > 0 since bonds are broken
∆S > 0 since 2 moles of gas have more entropy than one
∆G <0 only at “high” temperatures
Answer key – ch. 10
39. For the freezing of water at -10 °C and 1 atm, predict whether
∆H, ∆S, and ∆G should be positive, negative or zero.
∆H < 0 since bonds are formed
∆S < 0 since liquids have more entropy than solids
∆G < 0 only at “low” temperatures => T < ∆H/∆S =>
We’re not given ∆H or ∆S values but we know the freezing
point of water is 0°C => ∆H/∆S = 0°C => so if the
temperature is below 0°C it’s spontaneous
Answer key – ch. 10
40. Given the following data, calculate the normal boiling point
for formic acid (HCOOH).
∆Hfo (kJ/mol)
S° (J/K·mol)
HCOOH (l)
– 410.
130.
HCOOH (g)
– 363
251
a. 2.57 K
Boiling points (and freezing points) are
b. 1730°C
temperatures at equilibrium (∆G=0)
c. 388°C
T = ∆H/∆S
d. 82°C
T = [(-363kJ/mol)-(-410kJ/mol)]/[(0.251J/molK) –
e. 115°C
(0.13J/molK)] = 388K or 115 °C
Answer key – ch. 10
41. Consider the following reaction at 25 oC.
CO (g) + H2 (g) → H2 (g) + CO (g)
For this reaction ΔHo = –5.36 kJ and ΔSo = –109.8 J /K. At
what temperatures will the reaction be spontaneous?
a. T > 48.8 K
Since both ΔHo and ΔS° are negative
b. T < 48.8 K
the reaction will be spontaneous at
T > ΔH°/ΔS°
c. T > 20.5 K
T > (-5.36 kJ)/(-0.1098 kJ/K)
d. T < 20.5 K
T > 48.8 K
e. Spontaneous at all temperatures.
Answer key – ch. 10
42. Consider the following reaction.
2 POCl3 (g) → 2 PCl3 (g) + O2 (g)
The free energies of formation at 25 °C are given below. ΔGf°
POCl3 (g) = –502 kJ/mol, ΔGf° PCl3 (g) = –270 kJ/mol
Indicate true or false.
a. The entropy change for the reaction is positive. True –
moles of gas increased
b. The reaction is not spontaneous at standard conditions and 25
°C but will eventually become spontaneous if the
temperature is increased. True –
ΔG° = Σ ∆Gf°(products) - Σ∆Gf°(reactants)
Continue to next slide…
Answer key – ch. 10
ΔG° = (2 mol PCl3)(-270 kJ/mol) –
(2 mol POCl3)(-502 kJ/mol) = 462 kJ =>
positive so nonspontaneous => how ever since ΔH° and ΔS° are
both positive it will be come spontaneous when the temperature
becomes high enough
c. The equilibrium constant for the reaction at 298 K is less
than 1. True – if ΔG° > 0 then reactants are favored or
K <1
d. The enthalpy change for the reaction is positive. True – this
reaction is a combustion flipped
Answer key – ch. 10
43. Consider the following reaction at 800 K.
2 NF3 (g) → N2 (g) + 3 F2 (g)
At equilibrium, the partial pressures are PN2= 0.040 atm, PF2=
0.063 atm and PNF3= 0.66 atm. Which of the following is true
about the value of ∆G°?
since
a. is a positive number
3/(0.66)2
K
=
(0.04)(0.063)
b. is a negative number
K = 2.3 x 10-5
c. is equal to zero
when K <1 the reaction
d. is independent of the temperature
favors the reactants or
e. can not be predicted from this data
∆G° > 0
Answer key – ch. 10
44. Use the following reaction to answer the following.
2 SO2 (g) + O2 (g)  2 SO3 (g)
a. Calculate ∆G°
b. What is the equilibrium constant at 25 °C?
c. At 25 °C the initial pressures for SO2, O2 and SO3 are
0.001 atm, 0.002 atm and 40 atm respectively. Will SO3
be consumed or will SO3 be formed?
Substance ∆Gf° (kJ/mol)
SO2 (g)
-300
SO3 (g)
-321
continue to next slide…
Answer key – ch. 10
44. …continued
a. ΔG° = Σ ∆Gf°(products) - Σ∆Gf°(reactants)
ΔG° = (2 mol SO3)(-321kJ/mol) – (2 mol SO2)(-300kJ/mol)
ΔG° = -42 kJ
b. ΔG° = -RTlnK
-42 kJ = -(8.314x10-3kJ/molK)(298K)(lnK)
K = 2.3x107
c. There’s 2 ways to do this =>
option 1 => ΔG = ΔG° + RTlnQ = -42kJ + (8.314x10-3kJ/molK)
(298K)(ln(40)2/(0.001)2(0.002)) = 26 kJ or nonspont.
option 2 => compare K to Q => Q = (40)2/(0.001)2(0.002) = 8x1011
=> since K < Q reaction wants to go backward
Answer key – ch. 10
45. For a certain reaction the Kp is 50 at 25 °C and Kp is
9 x 103 at 110 °C.
a. Is the reaction endothermic or exothermic? if K↑ and T↑
b. Calculate Kp at 275 °C? ln(K2/K1) = (ΔH/R)(1/T1 – 1/T2)
since we don’t have ΔH =>
ln(9x103/50) = (ΔH/(8.314 J/mol))(1/298K – 1/383K)
ΔH = 57,970 J/mol
now that we have ΔH lets make K2 be unknown
ln(K2/50) = ((57,970 J/mol)/(8.314 J/mol))(1/298K – 1/548K)
K2 = 2.16 x 106
Answer key – ch. 10
46. For a particular reaction a graph was made by plotting ln(Keq) as
a function of 1000/K. The x and y intercepts were found to be
3.0 and 40. respectively. Determine the values of ΔH° and ΔS°
for this reaction.
lnK = (-∆H °/R)(1/T) + ΔS°/R =>
equation of a line => y = mx + b
so y intercept = ΔS°/R => 40 = ΔS°/8.314 J/molK =>
ΔS°= 333 J/molK
x intercept means y = 0 or lnK = 0 and 1000/T = 3 =>
1/T = 0.003/K
0 = (- ∆H°/8.314J/molK)(0.003/K) + 40 => ∆H°= 111kJ/mol