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Chem 1B – CLAS - Chapter 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.
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?
a. 17.8 J
b. –17.8 J
c. 8.2 J
d. –8.2 J
e. 0 J
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.
4. Which of the following are always exothermic?
a. Breaking bonds b. Melting c. Combustion
d. Condensation
e. Sublimation
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
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
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)
7. 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
8. 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. Endothermic reactions have stronger bonds or a greater number of bonds in the products.
d. If qp for a process is negative, the process is exothermic.
e. The freezing of water is an example of an exothermic reaction.
9. For the following reactions at constant pressure, predict if ∆H > ∆E, ∆H < ∆E, or ∆H = ∆E.
ΔH = ΔE + Δ(PV) or ΔH = ΔE + Δ(nRT)
a. H2 (g) + Cl2 (g) → 2 HCl (g)
since the moles of gas stay constant ∆H = ∆E
b. N2 (g) + 3 F2 (g) → 2 NF3 (g)
since the moles of gas decrease ∆H < ∆E
10. Consider a sample containing 2 moles of xenon that undergoes the following changes:
Pa = 10 atm,Va = 10L
Pb = 10 atm, Vb = 5L
Pc = 15 atm, Vc = 8L
For each step, assume that the external pressure is constant and equals the final pressure of the gas for that
step. Calculate ΔE and ΔH (in kJ) to go from state a to state c.
ΔE = 3/2Δ(PV) = 3/2 (15 atm x 8 L – 10 atm x 10 L) = 30 atmL or 3.04 kJ
ΔH = 5/2Δ(PV) = 5/2 (15 atm x 8 L – 10 atm x 10 L) = 50 atm or 5.06 kJ
w = -PΔV ⇒ w1 = 50 atmL or 5.06 kJ, w2 = -45 atmL or -4.56 kJ, w (total) = 0.51 kJ
q = ΔE – w = 3.04 kJ – 0.51 kJ = 2.53 kJ
11. Calculate ∆E (in kJ) when 3.2 mol of liquid is vaporized at 1 atm and its boiling point (73°C). ∆Hvap = 28.6
kJ/mol at 73°C. ΔE = q + w
for vaporization ⇒ q = n∆H and w = -ΔnRT ⇒
ΔE = n∆Hvap –ΔnRT
ΔE = (3.2mol)(28.6 kJ/mol) – (3.2mol)(0.0083145 kJ/molK)(73 + 273 K) = 82.3 kJ
12. 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 combined with 28.1 g of oxygen
(32 g/mol), how much energy is released as heat?
A) 87.8 kJ B) 2.14 × 104 kJ C) 4.64 × 102 kJ D) 4.01 × 102 kJ E) 4.75 kJ
for a reaction q =
moles C2H5OH =
moles O2 =
𝑛∆𝐻
𝑐𝑜𝑒𝑓
(if use moles of limiting reagent or a product)
15.6 g
46.1 g/mol
28.1 g
32 g/mol
= 0.338 moles
= 0.878 moles
even though there are more moles of O2 it is the LR because it has a coefficient of 3
q=
(0.878 𝑚𝑜𝑙)(−1.37 𝑥 103 kJ
3 𝑚𝑜𝑙
= -401kJ
13. A 25.0 g piece of unknown metal was transferred from an oven at 115 °C into a coffee cup calorimeter
containing 150. g 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) When you have multiple systems the heat lost equals the heat gained between the systems and
therefore q total = 0. In this case you have 3 systems: metal, water and the coffee cup.
q total = q metal + q water + q coffee cup
when the temperature is changing ⇒ q = nC∆T or q = mC∆T or q = C∆T (look at units for C)
q total = mC∆T metal + mC∆T water + C∆T coffee cup
0 = (25 g)(C)(115 °C – 28 °C) + (150 g)(4.18 J/g°C)(28 °C – 24 °C) + (25 J/K)(28 °C – 24 °C)
C metal = 1.2 J/g°C
14. In a coffee cup calorimeter 20.0 mL of 0.30 M HCl and 50.0 mL of 0.10 M NaOH 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
mass of the resulting solution is 70.0 g and the specific heat capacity for the solution is 4.18 J/g°C.
Here you have two systems: the reaction and the solution
q total = q reaction + q solution
q total =
𝑛∆𝐻
𝑐𝑜𝑒𝑓
+ mCΔT
The reaction ⇒ HCl + NaOH → H2O + NaCl
moles of HCl = (0.02 L)(0.3 mol/L) = 0.006 mol
moles of NaOH = (0.05 L)(0.1 mol/L) = 0.005 mol
NaOH is the LR
0=
(0.005 𝑚𝑜𝑙)(∆𝐻)
1 𝑚𝑜𝑙
+ (70 g)( 4.18 J/g°C)(31°C – 23°C)
ΔH = - 468 kJ
15. In a coffee cup calorimeter, 27 grams of NaNO3 (85.00 g/mol) is dissolved in enough water to make 150
grams of solution. The following reaction occurs: NaNO3 (s) → Na+ (aq) + NO3– (aq) H = 20.5 kJ/mol
The initial temperature of the water and the NaNO3 is 24.4 °C. What is the final temperature of the
solution? The heat capacity of the solution is 4.18 J/g°C. Assume no heat is lost to the calorimeter.
a) 11.7 °C b) 34.8 °C c) 14.0 °C d) 37.1 °C e) 82.1 °C
Here you have two systems: the reaction and the solution
q total = q reaction + q solution
q total =
𝑛∆𝐻
𝑐𝑜𝑒𝑓
+ mCΔT
There’s only one reactant so it is the LR by default
27 𝑔
moles of NaNO3 (s)=
=0.318 mol
g
0=
(85.00
)
mol
20,500𝐽
(0.318 𝑚𝑜)(
)
𝑚𝑜𝑙
1 𝑚𝑜𝑙
+ (150 g)(4.18 J/g °C)(Tf – 24.4 °C)
Tf = 14.0 °C
16. Given the following data calculate the ∆H° for the reaction: NO + O → NO2
Rxn 1: 2 O3 → 3 O2
∆H° = – 427 kJ
Rxn 2: O2 → 2 O
∆H° = +495 kJ
Rxn 3: NO + O3 → NO2 + O2
∆H° = – 195 kJ
Rxn 1 ⇒ flip and divide by 2 ⇒ 3/2 O2 → O3
∆H° = ½(+427 kJ)
Rxn 2 ⇒ flip and divide by 2 ⇒ O → O2 ∆H° = ½(–495 kJ)
Rxn 3 ⇒ keep as is ⇒ NO + O3 → NO2 + O2 ∆H° = –195 kJ
When you add the 3 reactions together the O2 and O3 cancel out
∆H°total = ½(+427 kJ) + ½(–495 kJ) + (-195 kJ) = -229 kJ
17. Using the following data, calculate the standard heat of formation of ICl (g) in kJ/mol:
Rxn 1: Cl2 (g) → 2 Cl (g)
ΔH° = 242.3 kJ
Rxn 2: I2 (g) → 2 I (g)
ΔH° = 151 kJ
Rxn 3: ICl (g) → I (g) + Cl (g)
ΔH° = 211.3 kJ
Rxn 4: I2 (s) → I2 (g)
ΔH° = 62.8 kJ
Heats of formation reactions have to be written in a specific manner ⇒ only 1 mol of one substance is
made from its elements in their standard states ⇒ so for ICl the formation reaction is
½ I2 (s) + ½ Cl2 (g) → ICl (g)
Rxn 1 ⇒ divide by 2 ⇒ ½ Cl2 (g) → Cl (g)
ΔH° = ½ (242.3 kJ)
Rxn 2 ⇒ divide by 2 ⇒ ½ I2 (g) → I (g)
ΔH° = ½(151 kJ)
Rxn 3 ⇒ flip ⇒ I (g) + Cl (g) → ICl (g)
ΔH° = - 211.3 kJ
Rxn 4 ⇒ divide by 2 ⇒ ½ I2 (s) → ½ I2 (g)
ΔH° = ½ (62.8 kJ)
When you add the four reactions Cl (g), I (g) and I2 (g) cancel out
∆H°total =16.75 kJ
18. 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 kJ/mol 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
Combustion of acetylene ⇒ C2H2 (g) + 5/2 O2 (g) → 2 CO2 (g) + H2O (l)
∆H°rxn = Σ∆H°(products) - Σ∆H°(reactants)
–1299 kJ = [(2 moles CO2 (g))( –393 kJ/mol) + (1 mol H2O (l))( –286 kJ/mol)] – [(1 mol C2H2 (g))(Δ𝐻𝑓° )
Δ𝐻𝑓° = 227 kJ/mol