thermochemistry solutions

8/18/2019 Thermochemistry Solutions

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Chem 131 Solutions to Thermochemistry Problems Page 1

Thermochemistry Problems

Problem 1. Enthalpy and Phase Change

a) If a bathtub contains 160. L of water at 37ºC (body temperature), what mass of ice (at 0ºC)would you need to add in order to end up with water at a uniform temperature of 25ºC?

Since the bathtub water has a higher temperature than the ice, it will transfer its heat into the

ice. This will first melt the ice to water at 0ºC, and then warm this melted ice from 0ºC up to

25ºC. As the bath water gives up its heat to the ice, it will cool from 37ºC to 25ºC. We

assume that all of the heat lost by the bath water is gained by the ice to (1) melt the ice and

(2) warm the melted ice:

( ) 02 =++ warmmelt O H qqq or ( )warmmelt O H qqq +−=2

The amount of heat given up by the bath water would be:

( ) J K gK

J

mL

g

L

mL LT C mq O H O H O H O H

6

2222 1003.83725184.4

1

00.1

1

1000160 ×−=−⎟⎟

⎠

⎞⎜⎜⎝

⎛

⎭⎬⎫

⎩⎨⎧

⎟ ⎠

⎞⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ =Δ=

For the ice, the first part involves a change of state (use enthalpy of fusion), while the

second involves a change of temperature (use specific heat capacity) of the melted ice:

Heat to melt ice: ( )g J m H mq icemice fusicemelt 333., =Δ=

Heat to warm: ( ) ( )g J mK gK

J mT C mq icemicemicemicemicemwarm 6.401025

184.4..... =−⎟⎟

⎠

⎞⎜⎜⎝

⎛ =Δ=

Plugging these three q expressions into the original energy balance equation above gives:

( ) ( )g J mg J m J iceice 3336.4011003.8 6 +−=×−

We can then solve for the mass of ice:

( ) J g J g J mice 61003.83336.401 ×=+ or kggg J

J m

ice 18350,816.734

1030.86

==×

=


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b) Repeat the analysis, but this time for a coffee cup that contains 250. mL of water at 95ºC.

same process as previous, but different numbers for qH2O

( ) J K gK J

mL

g

mLT C mq O H O H O H O H 4

2222 1023.79525

184.4

1

00.1

250 ×−=−⎟⎟ ⎠

⎞

⎜⎜⎝

⎛

⎭⎬

⎫

⎩⎨

⎧

⎟ ⎠

⎞

⎜⎝

⎛

=Δ=

gg J

J mice 170

6.734

1023.7 4

=×

=

c) When you finish taking a hot shower, the air in the 1.0m x 2.0m x 3.0m shower enclosure is

saturated with water vapor (i.e., 100% relative humidity) at 45ºC. How much total heat is

released from the condensation of this water vapor as the air cools to 25ºC? (Note: ignore

any energy changes associated with the temperature change; focus on the energy from the

phase change.)

In this problem, the heat comes from one particular process—the condensation of water

vapor: H2O (g) H2O (l) ΔH = −ΔHvap = −40.7 kJ/mol

The challenge in this problem is to figure out how many moles of water condense. The wayto do this is to determine the number of moles of water vapor in the air initially, and thenumber in the air at the end. The difference between the two gives the number that are nolonger in the air (i.e., the number of moles that condensed).

Note that the air in the shower stall is a gas mixture—it contains (primarily) N 2, O2, and H2Ogases. We can determine the number of moles of water in the air in each case using theideal gas law, if we know the partial pressure of water PH2O in each case.

RT

V Pn O H O H

2

2 =

We can determine PH2O in each case by recognizing that the air is “saturated” with watervapor, meaning that the partial pressure of water corresponds to the vapor pressure of water at that temperature (you can look these up in Appendix G):

PH2O (@45C) = 71.9 torr and PH2O (@25C) = 23.8 torr

The volume V of the gas (the moist air in the shower enclosure) is 6.0m3 = 6.0 x 106 cm3 =6.0 x 106 mL = 6.0 x 103 L. Applying the ideal gas equation:


3/8


( ) ( )( )

( )O H mol

torr

atm

K K mol

atm L

Ltorr

RT

V PC n O H O H 2

3

2

2 8.12760

1

31808206.0

100.69.7145 =⎟⎟

⎠

⎞⎜⎜⎝

⎛

⎟ ⎠

⎞⎜⎝

⎛

⋅

⋅

×==°

( ) ( )( )( )

O H moltorr

atm

K K mol

atm L Ltorr

RT V PC n O H O H 2

32

2 7.7760

1

29808206.0

100.68.2325 =⎟⎟ ⎠ ⎞⎜⎜

⎝ ⎛

⎟ ⎠

⎞⎜⎝

⎛

⋅

⋅×==°

So, the amount of water that condenses is 21.8 – 7.7 mol H2O = 14.1 mol H2O. Thecorresponding amount of heat released is therefore:

( ) kJ molkJ mol H nq O H vapO H cond 0757.401.412,2 −=−=Δ−=

Or, in other words, 570 kJ of heat is released from the condensation of the water vapor.

Problem 2. Energy Conversion

A typical incandescent light bulb emits approximately 10kJ of useful (i.e., visible) light energy

over a 1-hour period of use.

a) List all of the conversions between different forms of energy needed to provide this light,

assuming that the electricity is supplied from a natural-gas fired power plant.

If we start with the natural gas, the chemical potential energy of the natural gas is converted

during combustion into the thermal energy of the products (CO2 and water vapor). This

thermal energy is then converted into the mechanical (kinetic) energy of the spinning

turbine. This is converted in the generator into electrical energy, which then travels over the

transmission and distribution wires to your home. At the light bulb, the electrical energy is

converted into light (electromagnetic radiation), as well as into a lot of thermal energy in

heating the light bulb filament to high temperatures. (Note that we could start “all the way at

the beginning”—nuclear potential energy is converted into thermal energy and light in the

fusion reactions of the sun, the light is then converted into chemical potential energy by

plants during photosynthesis, this chemical potential energy is then altered slightly during

decomposition chemical reactions that result in the formation of natural gas, then …)

b) If we assume natural gas to be 100% methane (CH4), what mass of methane is needed to

provide the light energy above? Assume the following efficiencies: 33% for generation of

electricity from the methane, 90% for electricity transmission and distribution, 2% for the light

fixture and bulb. Note that the enthalpy of combustion of methane is:

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) ∆Hrxn° = –802.30kJ

mol


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Working backwards, 10kJ of light requires kJ kJ

50002.0

10 = of electrical energy entering

the light fixture. This requires kJ kJ

55690.0

500 = of electrical energy leaving the power

plant. This requires kJ kJ

168033.0

556 = of chemical potential energy in the methane. We

can see from the combustion reaction of methane above that we get about 800 kJ of energy

for each mole of methane burned, so roughly speaking, we would need about 2 moles

(~32g). More precisely:

4

4

44 341

0.16

30.802

11680 CH g

CH mol

CH g

kJ

CH molkJ =⎟⎟

⎠

⎞⎜⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛

Problem 3. Coal Gasifi cation

The “water gas” reaction can be used to produce hydrogen by treating carbon (from coal) with

high temperature steam:

C (s) + H2O (g) CO (g) + H2 (g)

a) Use heats of formation to determine the enthalpy change for this reaction. Is the reaction

endothermic or exothermic?

ΔHºrxn = ∑ p ΔHf º(P) – ∑ r ΔHf º(R)

ΔHºrxn = ΔHf º[CO (g)] + ΔHf º[H2 (g)] – ΔHf º[C (g)] – ΔHf º[H2O (g)]

ΔHºrxn = (–110.525 kJ mol-1

) + 0 – 0 – (–241.83 kJ mol-1

)

Hºrxn = + 131.31 kJ mol-1

Since ΔHºrxn is positive, the reaction is endothermic.

b) In practice, the energy needed to drive this reaction is usually supplied by burning (i.e.,

combusting) some additional carbon (to produce CO2). What mass of carbon needs to be

combusted in order to provide the energy needed to convert 1.0kg of carbon in the water-

gas reaction?

Since the water-gas reaction is endothermic, an input of energy is needed in order toconvert C in this reaction. From the ΔHrxn value calculated above, we can determine the

amount of energy needed for 1.0kg C. The problem statement indicates that all of this

energy will be provided from a second reaction, the combustion of carbon, which is

exothermic. We can write a reaction equation for this combustion, and then use heats of

formation to find the ΔHrxn of this reaction. Finally, we can determine how much C needs to

be combusted in this second reaction to provide the total amount of energy needed.


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Energy needed to convert 1.0 kg C in water-gas reaction:

kJ rxnmol

kJ

C mol

rxnmol

C g

C mol

C kg

C gC kg 41090.1

1

31.131

1

1

01.12

1

1

10000.1 ×=⎟⎟

⎠

⎞⎜⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛

Balanced reaction equation for combustion of carbon:

C (s) + O2 (g) CO2 (g)

ΔHºrxn = ΔHf º[CO2 (g)] – ΔHf º[C (s)] – ΔHf º[O2 (g)]

ΔHºrxn = (–393.5 kJ mol-1

) – 0 – 0 = –393.5 kJ mol-1

The heat from this combustion reaction will provide all of the energy to supply the energy

needs of the water-gas reaction, or in other words, qcomb = – qwg rxn = – 1.09 x 104 kJ. We can

then determine how many moles of combustion reaction are needed for this amount ofenergy, and in turn, the corresponding moles and mass of carbon in the reaction:

C kgC g

C kg

C mol

C g

rxnmol

C mol

kJ

rxnmolkJ 33.0

1000

1

1

01.12

1

1

5.393

11090.1 4 =⎟⎟

⎠

⎞⎜⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛

−×−

Problem 4. Heats of Formation/Calorimetry

Some camping stoves use ethanol (C2H5OH) as fuel.

a) Write a balanced equation for the combustion of ethanol.

C2H5OH (l) + 3 O2 (g) ⎯→ 2 CO2 (g) + 3 H2O (l)

b) Use these heats of formation to calculate ∆H° for the combustion of ethanol.

∆Hf ° C2H5OH (l) = –277.7kJ

mol , ∆Hf ° CO2 (g) = –393.5kJ

mol , ∆Hf ° H2O (l) = –285.8kJ

mol

∆Hrxn° = 2 (–393.5kJ

mol ) + 3 (–285.8kJ

mol ) – 1 (–277.7kJ

mol ) – 3 (0) = –1366.7 kJ

c) If 66% of the heat released by the burning ethanol is absorbed by a pot of water on thestove, how is qwater mathematically related to qreaction?

qwater = – 0.66 qreaction


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d) How many grams of ethanol must burn to melt 500. g of ice (at 0°C) and raise the

temperature of the liquid water to 100°C?

For the ice (water), the first part involves a change of state (use enthalpy of fusion), while

the second involves a change of temperature (use specific heat capacity) of the melted ice:

Heat to melt ice: ( )( ) kJ J g J g H mq ice fusicemelt 5.616500,166333.500, ===Δ=

Heat to warm: ( ) ( ) kJ J K gK

J gT C mq icemicemicemwarm 2.920200,2090100

184.4.500... ==−⎟⎟

⎠

⎞⎜⎜⎝

⎛ =Δ=

Total heat input to water: qwater = qmelt + qwarm = 375.7 kJ

qreaction = –1

0.66 qwater = –1

0.66 (375.7 kJ) = –569 kJ

–569 kJ × 1 mol C2H5OH

–1366.7 kJ ×

46.07 g C2H5OH

1 mol C2H5OH = 19 g C2H5OH

Problem 5. Enthalpy of Reaction/Hess’s Law

Because the following chemical reactions occur in solution, their enthalpies of reaction can be

easily measured in a calorimetry experiment without loss of heat to the surroundings:

Rxn 1: Mg (s) + 2 HCl (aq) MgCl2 (aq) + H2 (g)

Rxn 2: MgO (s) + 2 HCl (aq) MgCl2 (aq) + H2O (l)

a) Using heats of formation, calculate the reaction enthalpy that should result for each of

Reactions 1 and 2 above. (The heat of formation of aqueous MgCl2 is –801.15 kJ mol-1; all

others can be found in your textbook.)

ΔHºrxn1 = ΔHf º[MgCl2 (aq)] + ΔHf º[H2 (g)] – ΔHf º[Mg (s)] – 2 ΔHf º[HCl (aq)]

ΔHºrxn1 = (–801.15 kJ mol-1

) + 0 – 0 – 2 (–167.159 kJ mol-1

) = –466.83 kJ mol-1

ΔHºrxn2 = ΔHf º[MgCl2 (aq)] + ΔHf º[H2O (l)] – ΔHf º[MgO (s)] – 2 ΔHf º[HCl (aq)]

ΔHºrxn2 = (–801.15 kJ mol-1

) + (–285.83 kJ mol-1

) – (–601.24 kJ mol-1

) – 2 (–167.159 kJ mol-1

)

= –151.42 kJ mol-1


7/8


b) The formation of water from its elements can be represented by the reaction:

Rxn 3: 2 H2(g) + O2(g) 2H2O(l) ΔH = -572 kJ

Using only the reaction enthalpies for Reactions 1, 2, and 3 above, use Hess’s Law to

calculate the enthalpy of the following reaction (the burning of a strip of magnesium metal):

2 Mg(s) + O2(g) 2 MgO(s)

The strategy here is to rearrange reaction equations 1, 2, and 3 in such a way that they can

be combined together to produce the reaction above. First, look for substances in the

combined reaction that only appear in one of the 3 reactions; this will automatically dictate

how that particular reaction must be written. In this case, all 3 substances in the combined

reaction fall into this category.

Since Mg(s) only appears in Rxn 1, we know that Rxn 1 must be written in its current

direction, but multiplied by 2 to get matching stoichiometric coefficients.

Since O2(g) only appears in Rxn 3, this reaction must be written as is.

Since MgO(s) only appears in Rxn 2, we know that Rxn 2 must be reversed (so that MgO

will appear on the product side of the reaction), and it must be multiplied by 2.

In each case, we can manipulate the ΔHrxn values appropriately:

2 x Rxn1: 2 Mg(s) + 4 HCl(aq) 2 MgCl2(aq) + 2 H2(g) 2 x ΔH1 = –933.66 kJ

-2 x Rxn2: 2 MgCl2(aq) + 2 H2O(l) 2 MgO(s) + 4 HCl(aq) –2 x ΔH2 = 302.84 kJ

1 x Rxn3: 2 H2(g) + O2(g) 2H2O(l) 1 x ΔH3 = –572 kJ

When we combine these reactions together, we confirm that all of the necessary substances

cancel out in order to give the correct overall reaction, and we calculate the overall ΔH by

combining the 3 values above to get – 1203 kJ mo l-1.

c) How much heat could theoretically be released from the burning of 2.00g of magnesium

metal in 1.75L of air at a pressure of 760 torr and 25ºC?

Here, we are in essence given amounts of both reactants (Mg and O 2), so this must be a limiting

reactant problem. In order to determine the total amount of heat released, we will need to know

how many “moles of reaction” will occur based on the amount of limiting reactant. Since Mg is a

solid, we are given the amount as a mass, which we will need to convert as usual to moles Mg,

and then to moles of reaction.

rxnmol Mgmol

rxnmol

Mgg

Mgmol Mgg 41041.0

2

1

305.24

100.2 =⎟⎟

⎠

⎞⎜⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛


8/8


Since O2 is a gas, we will need to use the ideal gas equation. Moreover, O2 is present in a gas

mixture (air), so we will need to use partial pressures. Note that air is 21% O2 on a per mole

basis, and thus the partial pressure of O2 in air will be 21% of the total pressure (alternately, you

could calculate the total number of moles of gas in the air sample, and then take 21% of that).

( )( ) atmtorr

atmtorr PO 21.0760

176021.02

=⎟⎟ ⎠ ⎞⎜⎜

⎝ ⎛ =

( )( )

( )rxnmol

Omol

rxnmolOmol

K K mol

atm L

Latm

RT

V Pn

O

O 0501.01

10150.0

29808206.0

75.121.0

2

22

2=⎟⎟

⎠

⎞⎜⎜⎝

⎛ ×=

⎟ ⎠

⎞⎜⎝

⎛

⋅

⋅==

Thus, oxygen is the limiting reactant, and only 0.015 moles of reaction will occur. Finally, since

we know the number of moles of reaction, we can determine the amount of heat released:

( ) kJ rxnmolkJ

rxnmolqrxn 181

1203

015.0 −=⎟⎟ ⎠

⎞

⎜⎜⎝

⎛ −

= (or, 18 kJ of heat will be released)

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