Transcript Document

Example 6.1
A gas does 135 J of work while expanding, and at the same time it absorbs 156 J of heat.
What is the change in internal energy?
Strategy
The key to getting the correct value of ∆U in Equation (6.7) is to use the correct signs for
the given quantities, w and q. Note that heat is absorbed by the system; this means q is a
positive quantity: q = +156 J. Work is done by the system; this means w is a negative
quantity: w = –135 J.
Solution
We add the values for q and w, each with its correct sign. Note that because more heat is
absorbed than work done, the internal energy increases.
∆U = q + w = +156 J + (–135 J) = +21 J
Exercise 6.1A
In a process in which 89 J of work is done on a system, 567 J of heat is given off. What is
DU of the system?
Exercise 6.1B
In a particular process, the internal energy of a system increases by 41.4 J and the
quantity of work the system does on its surroundings is 81.2 J. Is heat absorbed or given
off by the system? What is the value of q?
Example 6.2
The internal energy of a fixed quantity of an ideal gas depends only on its temperature. If
a sample of an ideal gas is allowed to expand against a constant pressure at a constant
temperature, (a) what is ∆U for the gas? (b) Does the gas do work? (c) Is any heat
exchanged with the surroundings?
Analysis and Conclusions
(a) Because the expansion occurs at a constant temperature, the expanded gas (state 2)
is at a lower pressure than the compressed gas (state 1) but the temperature is
unchanged. Because the internal energy of the ideal gas depends only on the
temperature, U2 = U1 and ∆U = U2 – U1 = 0.
(b) The gas does work in expanding against the confining pressure, P. The pressure–
volume work is w = –P∆V, as was illustrated in Figure 6.8. The work is negative
because it is done by the system.
(c) The work done by the gas represents energy leaving the system. If this were the only
energy exchange between the system and its surroundings, the internal energy of the
system would decrease, and so would the temperature. However, because the
temperature remains constant, the internal energy does not change. This means that
the gas must absorb enough heat from the surroundings to compensate for the work
that it does in expanding: q = –w. And, according to the first law of thermodynamics,
∆U = q + w = –w + w = 0.
Example 6.2 continued
Exercise 6.2A
In an adiabatic process, a system is thermally insulated from its surroundings so that
there is no exchange of heat (q = 0). If an ideal gas undergoes an adiabatic expansion
against a constant pressure, (a) does the gas do work? (b) Does the internal energy of the
gas increase, decrease, or remain unchanged? (c) What happens to the temperature?
Exercise 6.2B
A sample of an ideal gas does 152 kJ of work as it very slowly expands at constant
temperature. To compress the gas very quickly back to its original condition, 209 J of
work must be done. How is it possible for this expansion/compression process to occur?
What will be ∆U for the overall process? Explain.
Example 6.3
Given Equation (a), calculate ∆H for
Equation (b).
Strategy
Our first requirement is to determine how Equation (b) is related to Equation (a). Then
we must adjust the given enthalpy change to reflect the ways in which Equation (a) was
modified to get Equation (b).
Solution
Exercise 6.3A
Exercise 6.3B
Example 6.4
The complete combustion of liquid octane, C8H18, to produce gaseous carbon dioxide and
liquid water at 25 °C and at a constant pressure gives off 47.9 kJ of heat per gram of
octane. Write a chemical equation to represent this information.
Strategy
Because the reaction is carried out at constant pressure, the heat of reaction is a qP value
and therefore an enthalpy change ∆H. Moreover, the fact that heat is given off signifies
that the reaction is exothermic and that ∆H is a negative quantity. Finally, the enthalpy
change is given for 1 g of reactant; we must convert this to a molar basis because the
coefficients in the equation represent moles of substances.
Solution
Example 6.4 continued
Exercise 6.4A
Write a chemical equation to express the fact that, at 0 °C, ice melts by absorbing 334 J
of heat per gram of ice.
Exercise 6.4B
Express the following information as a chemical equation. At 25 °C and constant
pressure, dinitrogen trioxide gas decomposes to nitrogen monoxide and nitrogen dioxide
gases with the absorption of 0.533 kJ of heat for every gram of dinitrogen trioxide that
decomposes.
Example 6.5
What is the enthalpy change associated with the formation of 5.67 mol HCl(g) in this
reaction?
H2(g) + Cl2(g)  2 HCl(g)
∆H = –184.6 kJ
Strategy
The quantity we are seeking, ∆H, has the unit kilojoules. The quantity on which the
calculation is based is the amount of HCl(g) produced, and the only conversion factor
required relates kilojoules of heat to moles of HCl.
Solution
The conversion factor from the chemical equation is shown in red here:
? kJ = 5.67 mol HCl x
–184.6 kJ
= –523 kJ
2 mol HCl
The formation of 5.67 mol HCl in this exothermic reaction results in the release of 523 kJ
of heat to the surroundings.
Exercise 6.5A
What is the enthalpy change when 12.8 g H2(g) reacts with excess Cl2(g) to form HCl(g)?
H2(g) + Cl2(g)  2 HCl(g)
∆H = –184.6 kJ
Exercise 6.5B
What volume of CH4(g), measured at 25 °C and 745 Torr, must be burned in excess
oxygen to release 1.00 x 106 kJ of heat to the surroundings in the reaction
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
∆H = –890.3 kJ
Example 6.6
Calculate the heat capacity of an aluminum block that must absorb 629 J of heat from its
surroundings in order for its temperature to rise from 22 °C to 145 °C.
Solution
We need the ratio of q to ∆T. Because q = 629 J and ∆T is final temperature minus initial
temperature, we have
Exercise 6.6A
Calculate the heat capacity of a sample of brake fluid if the sample must absorb 911 J of
heat in order for its temperature to rise from 15 °C to 100 °C.
Exercise 6.6B
A burner on an electric range has a heat capacity of 345 J/K. What is the value of q, in
kilojoules, as the burner cools from 467 °C to 23 °C?
Example 6.7
How much heat, in joules and in kilojoules, does it take to raise the temperature of 225 g
of water from 25.0 to 100.0 °C?
Strategy
This calculation is a direct application of Equation (6.15) once we have replaced ∆T by
the difference in two temperatures. The conversion from joules to kilojoules requires
dividing the first answer by 1000.
Solution
The specific heat of water is 4.18 J g–1 °C –1. The temperature change is (100.0 – 25.0) °C
= 75.0 °C, and the quantity of water to be heated is 225 g. We substitute these data in
Equation (6.15):
Exercise 6.7A
How much heat, in calories and in kilocalories, does it take to raise the temperature of
814 g of water from 18.0 to 100.0 °C?
Exercise 6.7B
What mass of water, in kilograms, can be heated from 5.5 to 55.0 °C by 9.09 x 1010 J of
heat?
Example 6.8
What will be the final temperature if a 5.00-g silver ring at 37.0 °C gives off 25.0 J of
heat to its surroundings? Use the specific heat of silver listed in Table 6.1.
Strategy
The loss of heat from the system—the ring—means that q is negative: q = –25.0 J. This
loss is not compensated for by any energy input into the system, and so the temperature
must fall, meaning that ∆T < 0. We must solve the specific-heat Equation (6.15), first for
the temperature change (Tf – Ti ), and then for Tf .
Solution
Assessment
When using Equation (6.15), always check for the correct signs on quantities of heat and
temperature changes. Here, the temperature change must be negative because heat is lost
to the surroundings. The final temperature of the ring must be lower than the initial
temperature, and it is (15.7 °C < 37.0 °C).
Example 6.8 continued
Exercise 6.8A
A 454-g block of lead is at an initial temperature of 22.5 °C. What will be the temperature
of the lead after it absorbs 4.22 kJ of heat from its surroundings?
Exercise 6.8B
How many grams of copper can be heated from 22.5 to 35.0 °C by the same quantity of
heat that can raise the temperature of 145 g H2O from 22.5 to 35.0 °C?
Example 6.9
A 15.5-g sample of a metal alloy is heated to 98.9 °C and then dropped into 25.0 g of
water in a calorimeter. The temperature of the water rises from 22.5 to 25.7 °C. Calculate
the specific heat of the alloy.
Strategy
We need to use Equation (6.15) twice. First, we use it as written to find the heat absorbed
by the water, and then we rearrange it to solve for the unknown specific heat.
Solution
We begin by calculating the quantity of heat absorbed by the water (the surroundings):
As suggested in Figure 6.17, we assume that only the water (and not the container)
absorbs heat and that the only source of this heat is the alloy (the system) as it cools.
Because the alloy loses heat, we write
qalloy = –334 J
Now, we can calculate the specific heat of the alloy:
Example 6.9 continued
Assessment
We really are not justified in writing the intermediate result (334 J) to three significant
figures, because the ∆T term is precise only to two significant figures. However, we
ordinarily would not record this intermediate result and would simply round off to the
appropriate two significant figures in the final answer. We must be careful not to make
errors in signs when using Equation (6.15), but in the calculation here we can catch these
by noting that specific heats must always be positive. Also, the data in Table 6.1 give us a
good indication of the range of possible values for specific heats.
Exercise 6.9A
A 23.9-g sample of iridium is heated to 89.7 °C and then dropped into 20.0 g of water in
a calorimeter. The temperature of the water rises from 20.1 °C to 22.6 °C. Calculate the
specific heat of iridium.
Exercise 6.9B
A 135-g piece of iron is heated to 225 °C in an oven. It is then dropped into a calorimeter
containing 250.0 mL of glycerol (d = 1.261 g/mL) at 23.5 °C. The temperature of the
glycerol rises to a maximum value of 44.7 °C. Use these data to determine the specific
heat of glycerol.
Example 6.10 An Estimation Example
Without doing detailed calculations, determine which of the following is a likely
approximate final temperature when 100 g of iron at 100 °C is added to 100 g of 20 °C
water in a calorimeter: (a) 20 °C (b) 30 °C (c) 60 °C (d) 70 °C.
Analysis and Conclusions
Because we have the same mass of each substance, if the water and iron had the same
specific heat, the rise in temperature of the water would be the same as the drop in
temperature of the iron. The final temperature would be the average of 20 °C and 100 °C,
or 60 °C. However, from Table 6.1 we see that the specific heat of water (4.18 J g –1 °C–1)
is much greater than that of iron (0.45 J g–1 °C–1). It takes nearly ten times more heat to
change the temperature of a given mass of water than it takes to change the temperature
of the same mass of iron. The final water temperature must be below 60 °C, but it also
has to be above 20 °C, the initial temperature. The only possible approximate temperature
of the four values we are given is 30 °C.
Exercise 6.10A
Without doing detailed calculations, determine the final temperature if 200.0 mL of water
at 80 °C is added to 100.0 mL of water at 20 °C.
Exercise 6.10B
Calculate the exact final temperature in Example 6.10.
Example 6.11
A 50.0-mL sample of 0.250 M HCl at 19.50 °C is added to 50.0 mL of 0.250 M NaOH,
also at 19.50 °C, in a calorimeter. After mixing, the solution temperature rises to 21.21
°C. Calculate the heat of this reaction.
Strategy
To identify the quantities that enter into the calculation and to keep it simple, we can
make four assumptions:
• The solution volumes are additive. The volume of NaCl(aq) that forms is equal to
50.0 mL + 50.0 mL = 100.0 mL.
• The NaCl(aq) is sufficiently dilute that its density and specific heat are about the same
as those values for pure water: 1.00 g/mL and 4.18 J g–1 °C–1, respectively.
• The system is completely isolated: No heat escapes from the calorimeter.
• The heat required to warm any part of the calorimeter other than the NaCl(aq) is
negligible.
Solution
The heat produced by the reaction and retained in the calorimeter is
Example 6.11 continued
Exercise 6.11A
A 100.0-mL portion of 0.500 M HBr at 20.29 °C is added to 100.0 mL of 0.500 M KOH,
also at 20.29 °C, in a calorimeter. After mixing, the temperature rises to 23.65 °C.
Calculate the heat of this reaction. You may make the same assumptions we made in
Example 6.11.
Exercise 6.11B
Suppose the acid used in Example 6.11 was HI(aq) instead of HCl(aq). Based on the
same assumptions used in Example 6.11, how would you expect the result of the
hypothetical experiment using HI(aq) to compare with that found in Example 6.11?
Explain.
Example 6.12
Express the result of Example 6.11 for molar amounts of the reactants and products. That
is, determine the value of ∆H that should be written in the equation for the neutralization
reaction:
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
∆H = ?
Strategy
With data from Example 6.11, we can first determine the number of moles of each
reactant. Next, we can determine the number of moles formed of one of the products (for
example, moles of H2O). As implied by the above equation, the value of ∆H we are
seeking is that for the formation of 1 mol H2O. This we can obtain by dividing the heat of
reaction in Example 6.11 by the number of moles of H2O found here.
Solution
Example 6.12 continued
Finally, we write the reaction equation based on the formation of 1 mol H 2O with the
corresponding value of ∆H.
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
∆H = –57.2 kJ
Exercise 6.12A
Express the result of Exercise 6.11A for molar amounts of reactants and products.
HBr(aq) + KOH(aq)  KBr(aq) + H2O(l)
∆H = ?
Exercise 6.12B
A 125-mL sample of 1.33 M HCl and 225 mL of 0.625 M NaOH, both initially at
24.4 °C, are allowed to react in a calorimeter. What is the final temperature in the
calorimeter? (Use the type of assumptions stated in Example 6.11 and the value of ∆H
obtained in Example 6.12 to determine the final temperature.)
Example 6.13
In a preliminary experiment, the heat capacity of a bomb calorimeter assembly is found
to be 5.15 kJ/°C. In a second experiment, a 0.480-g sample of graphite (carbon) is placed
in the bomb with an excess of oxygen. The water, bomb, and other contents of the
calorimeter are in thermal equilibrium at 25.00 °C. The graphite is ignited and burned,
and the water temperature rises to 28.05 °C. Calculate ∆H for the reaction
C(graphite) + O2(g)  CO2(g)
∆H = ?
Strategy
First, we must use the heat capacity of the calorimeter and the temperature change to
calculate the quantity of heat absorbed by the calorimeter (qcalorim). The heat of the
reaction is the negative of qcalorim. Finally, we can convert qrxn for a 0.480-g sample of
graphite to a per-mole basis (similar to what we did in Example 6.12).
Solution
Example 6.13 continued
Exercise 6.13A
A 0.250-g sample of diamond (another form of carbon) is burned with an excess of O 2(g)
in a bomb calorimeter having a heat capacity of 6.52 kJ/°C. The calorimeter temperature
rises from 20.00 °C to 21.26 °C. Calculate ∆H for the reaction
C(diamond) + O2  CO2(g)
∆H = ?
Exercise 6.13B
A 0.8082-g sample of glucose (C6H12O6) is burned in a bomb calorimeter assembly, and
the temperature is noted to rise from 25.11 °C to 27.21 °C. Determine the heat capacity of
the assembly.
C6H12O6(s) + 6 O2  6 CO2(g) + 6 H2O(l)
∆H = –2803 kJ
Example 6.14
Calculate the enthalpy change for reaction (a), given the data in equations (b), (c), and (d).
Strategy and Solution
Example 6.14 continued
Assessment
It does not matter in what order we adjust the equations before combining them into the
required final equation, as long as each adjustment is a step toward our goal of getting the
correct terms on each side of the equation. In addition to achieving this correctly
balanced final equation, we must also carefully check all the adjustments made to ∆H
values in the individual equations to ensure that we get the correct ∆H for the reaction of
interest.
Exercise 6.14A
Exercise 6.14B
Example 6.15
Synthesis gas is a mixture of carbon monoxide and hydrogen that is used to synthesize a
variety of organic compounds. One reaction for producing synthesis gas is
3 CH4(g) + 2 H2O(l) + CO2(g)  4 CO(g) + 8 H2(g)
∆H° = ?
Use standard enthalpies of formation from Table 6.2 to calculate the standard enthalpy
change for this reaction.
Strategy
A good way to begin a calculation based on Equation (6.21) is to list, under the formula
of each substance in the balanced chemical equation, the value of its standard molar
enthalpy of formation. Next, we can substitute these ∆Hf° values into Equation (6.21),
being careful in each case to multiply the ∆Hf° value by the appropriate stoichiometric
coefficient (v) from the balanced equation. Throughout the calculation, we must pay
particular attention to the proper use of algebraic signs.
Example 6.15 continued
Solution
Exercise 6.15A
Ethylene, derived from petroleum, is used to make ethanol for use as a fuel or solvent.
The reaction is
C2H4(g) + H2O(l)  CH3CH2OH(l)
Use data from Table 6.2 to calculate ∆H° for this reaction.
Exercise 6.15B
Use data from Table 6.2 to calculate ∆H° for the combustion of butane gas, C4H10, to
produce gaseous carbon dioxide and liquid water.
Example 6.16
The combustion of isopropyl alcohol, common rubbing alcohol, is represented by the
equation
2 (CH3)2CHOH(l) + 9 O2(g)  6 CO2(g) + 8 H2O(l)
∆H° = –4011 kJ
Use this equation and data from Table 6.2 to establish the standard enthalpy of formation
for isopropyl alcohol.
Strategy
We need essentially the same strategy as in Example 6.15, except that we enter the
standard enthalpy change for the reaction on the left side of Equation (6.21) and the
unknown standard enthalpy of formation, ∆Hf°, with the appropriate other data, on the
right side. Then we solve for the desired ∆Hf°.
Solution
Example 6.16 continued
Exercise 6.16A
Tetrachloroethylene, a degreasing solvent, is produced by the reaction
C2H4(g) + 4 HCl(g) + 2 O2(g)  C2Cl4(l) + 4 H2O(l)
∆H° = –878.5 kJ
Use data from Table 6.2 to establish ∆Hf°for C2Cl4(l).
Exercise 6.16B
The combustion of thiophene, C4H4S(l), a compound used in the manufacture of
pharmaceuticals, produces carbon dioxide and sulfur dioxide gases and liquid water. The
enthalpy change in the combustion of 1 mol of C4H4S(l) is –2523 kJ. Use this information
and data from Table 6.2 to establish ∆Hf° for C4H4S(l).
Example 6.17
Without performing a calculation, determine which of these two substances should yield
the greater quantity of heat per mole upon complete combustion: ethane, C2H6(g), or
ethanol, CH3CH2OH(l).
Analysis and Conclusions
Each combustion has precisely the same final state: 2 mol CO 2(g) + 3 mol H2O(l).
The initial state is 1 mol of either ethane or ethanol and enough oxygen to complete the
combustion. Because ∆Hf°[O2(g)] = 0, the enthalpy of the initial state is simply the
standard enthalpy of formation of 1 mol of ethane in one case and 1 mol of ethanol in the
other. The combustion reaction that releases more heat is the one with the more negative
∆H°comb, and this will be the reaction that has in its initial state the compound with the
higher standard enthalpy of formation. The standard enthalpies of formation are –84.68
kJ/mol C2H6 and –277.7 kJ/mol CH3CH2OH. Ethane has the higher (that is, less negative)
standard enthalpy of formation, and its combustion will liberate more heat than the
combustion of ethanol. The sketch in Figure 6.20 should help you to visualize this
conclusion.
Example 6.17 continued
Exercise 6.17A
Without doing a detailed calculation, determine which alcohol gives off more heat upon
combustion on a per-mole basis: CH3OH(l) or C2H5OH(l)? Use data from Table 6.2.
Exercise 6.17B
Refer to Example 6.17 and Appendix C. Without doing a detailed calculation, determine
how the heat of combustion per mole of acetic acid should compare with those of ethane
and ethanol. Explain your reasoning.
Example 6.18
Use the net ionic equation just given, together with ∆Hf° = 0 for to obtain for H+(aq), to
obtain ∆Hf° for OH–(aq).
Solution
By proceeding exactly as we did in Example 6.16, we can write
Exercise 6.18A
Use data from Table 6.3 and the following data about the precipitation of BaSO 4(s) to
determine ∆Hf°[BaSO4(s)].
Ba2+(aq) + SO42–(aq)  BaSO4(s)
∆H° = 26 kJ
Exercise 6.18B
Given that ∆Hf°[Mg(OH)2(s)] = –924.5 kJ/mol, what is the standard enthalpy change,
∆H°, for the reaction between aqueous solutions of magnesium chloride and potassium
hydroxide?
Cumulative Example
The human body is about 67% water by mass. What mass of sucrose, C 12H22O11(s), for
which ∆Hf° = –2225 kJ/mol, must be metabolized by a 55-kg person with hypothermia
(low body temperature) to raise the temperature of the body water from 33.5 °C to the
normal body temperature of 37.0 °C? Assume the products of metabolism are in their
most stable states at 25 °C. What volume of air at 37 °C having a partial pressure of O2 of
151 Torr is required for the metabolism?
Strategy
We can treat the first question in two parts and then combine the parts. In the first part,
we write an equation for the metabolism of sucrose similar to the one for glucose in
Section 6.8, and we use standard enthalpies of formation to find the enthalpy change for
the reaction. In the second part, we determine the mass of water in the body, and we use
the specific heat of water and the temperature change to find the heat required to raise the
body temperature (Equation 6.15). At this point, we can combine the results of the first
two parts to find the mass of sucrose needed. To answer the second question, the volume
of air needed, we use the mass of sucrose in a stoichiometric calculation, together with
the ideal gas law and the concept of partial pressure.
Solution
Cumulative Example continued
Assessment
The mass of sucrose calculated—33 g, somewhat over an ounce—is not an unreasonable
amount of sucrose for a person to consume. The volume of gas may appear rather large at
first glance, but the balanced equation requires 12 mol O 2 per mole of sucrose,
corresponding to about 1.2 mol O2, which is about 30 L O2 at 1.0 atm pressure and 37 °C.
The partial pressure of oxygen is about 0.2 atm, and the volume of air would be about
five times greater than calculated for pure oxygen, that is, about 150 L.