Silberberg7e solution manual ch 06

6.3A Plan: Since heat is released in this reaction, the reaction is exothermic H < 0 and the reactants are above the products in an enthalpy diagram... 6.3B Plan: Since heat is absorbe

CHAPTER 6 THERMOCHEMISTRY: ENERGY FLOW AND CHEMICAL CHANGE

FOLLOW–UP PROBLEMS

6.1A Plan: The system is the liquid Since the system absorbs heat from the surroundings, the system gains heat and q is

positive Because the system does work, w is negative Use the equation E = q + w to calculate E Convert E

6.1B Plan: The system is the reactant and products of the reaction Since heat is absorbed by the surroundings, the

system releases heat and q is negative Because work is done on the system, w is positive Use the equation

E = q + w to calculate E Both kcal and Btu must be converted to kJ

6.2A Plan: Convert the pressure from torr to atm units Subtract the initial V from the final V to find ΔV Use w = -PΔV

to calculate w in atm•L Convert the answer from atm•L to J

6.3A Plan: Since heat is released in this reaction, the reaction is exothermic (H < 0) and the reactants are above the

products in an enthalpy diagram

6.3B Plan: Since heat is absorbed in this reaction, the reaction is endothermic (H > 0) and the reactants are below the

products in an enthalpy diagram

Solution:

6.4A Plan: Heat is added to the aluminum foil, so q will be positive The heat is calculated using the equation

q = c x mass x T Table 6.2 lists the specific heat of aluminum as 0.900 J/g•K

6.4B Plan: Heat is transferred away from the ethylene glycol as it cools so q will be negative The heat released is

calculated using the equation q = c x mass x T Table 6.2 lists the specific heat of ethylene glycol as 2.42 J/gK

The volume of ethylene glycol is converted to mass in grams by using the density

Solution:

T = 25.0°C – 37.0°C = (–12.0°C) change in 1 K

change in 1 degree C = –12.0 K Mass (g) of ethylene glycol = 5.50 L 1 mL3 1.11 g

6.5A Plan: The heat absorbed by the water can be calculated with the equation c x mass x T; the heat absorbed by the

water equals the heat lost by the hot metal Since the mass and temperature change of the metal is known, the specific heat capacity can be calculated and used to identify the metal

T = Tfinal – Tinitial = 27.25°C – 65.00°C = (–37.75°C) change in 1 K

change in 1 degree C = –37.75 K 2





 = 0.386738 = 0.387 J/gK

From Table 6.2, the metal with this value of specific heat is copper

6.5B Plan: The heat absorbed by the titanium metal can be calculated with the equation c x mass x T; the heat

absorbed by the metal equals the heat lost by the water Since the mass and temperature change of the water is known, along with the specific heat and final temperature of the titanium, initial temperature of the metal can be calculated

ΔTmetal = –cH2O x massH2O x ∆TH2O

c metal x mass metal = –(4.184 J/gK) (75.0 g) (–0.70 K)

(0.228 J/gK) (33.2 g) = 29.0187 = 29.0 K

ΔTmetal =29.0187 K = 322.45 K – Tinitial (using unrounded numbers to avoid rounding errors)

Tinitial = 293.4313 K – 273.15 = 20.2813 = 20.3 o C

6.6A Plan: First write the balanced molecular, total ionic and net ionic equations for the acid-base reaction To find

specific heat capacity We know the solutions’ volumes (25.0 mL and 50.0 mL), so we find their masses with the

given density (1.00 g/mL) Then, to find qsoln, we multiply the total mass by the given c (4.184 J/g•K) and the change in T, which we find from Tfinal – Tinitial The heat of reaction (qrxn) is the negative of the heat of solution (qsoln)

Solution:

a) The balanced molecular equation is: HNO 3(aq) + KOH(aq)  KNO3(aq) + H2O(l)

The total ionic equation is: H+(aq) + NO3– (aq) + K+(aq) + OH– (aq)  K+(aq) + NO3– (aq) + H2O(l)

The net ionic equation is: H +

(aq) + OH–(aq)  H2O(l)

b) Total mass (g) of solution = (25.0 mL + 50.0 mL) x 1.00 g/mL = 75.0 g

6.6B Plan: Write a balanced equation Multiply the volume by the molarity of each reactant solution to find moles of each

reactant Use the molar ratios in the balanced reaction to find the moles of water produced from each reactant; the smaller amount gives the limiting reactant and the actual moles of water produced Divide the heat evolved by the

moles of water produced to obtain the enthalpy in kJ/mol Since qsoln is positive, the solution absorbed heat that was

released by the reaction; qrxn and ΔH are negative

Solution:

Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)

Moles of Ba(OH)2 =  10 L3 0.500 mol Ba(OH)2

1 mol Ba(OH)

  = 0.0500 mol H2O

Moles of H2O from HCl =  2 mol H O2

ΔH (kJ/mol) =

2

1.386 kJ =

moles of H O produced 0.0250 mol



q

= –55.44 = – 55.4 kJ/mol

6.7A Plan: The bomb calorimeter gains heat from the combustion of graphite, so –qgraphite = qcalorimeter Convert the mass

of graphite from grams to moles and use the given kJ/mol to find qgraphite The heat lost by graphite equals the heat gained by the calorimeter, or T multiplied by Ccalorimeter

6.7B Plan: The bomb calorimeter gains heat from the combustion of acetylene, so –qrxn = qcalorimeter Use the given heat

capacity from Follow-up Problem 6.7A to find qrxn: the amount of heat lost by the reaction of acetylene equals the amount of heat gained by the calorimeter, or T multiplied by Ccalorimeter Divide the heat produced by the

combustion of acetylene (in kJ) by the moles of acetylene to obtain the enthalpy in kJ/mol

6.8A Plan: To find the heat required, write a balanced thermochemical equation and use appropriate molar ratios to

solve for the required heat

+180.58 kJ

2 mol NO = 1.05 x 10 7 kJ

6.9A Plan: Manipulate the two equations so that their sum will result in the overall equation Reverse the first equation

(and change the sign of H); reverse the second equation and multiply the coefficients (and H) by two

Solution:

2NO(g) + 3/2O2(g)  N2O5(s) H = –(223.7 kJ)= –223.7 kJ

2NO2(g)  2NO(g) + O2(g) H = –2(–57.1 kJ) = 114.2 kJ

Total: 2NO2(g) + 1/2O2(g)  N2O5(s) H = –109.5 kJ

6.9B Plan: Manipulate the three equations so that their sum will result in the overall equation Reverse the first equation

(and change the sign of H) and multiply the coefficients (and H) by 1/2 Multiply the coefficients of the second equation (and H) by 1/2 Reverse the third equation (and change the sign of H)

6.10A Plan: Write the elements as reactants (each in its standard state), and place one mole of the substance formed on

the product side Balance the equation with the following differences from “normal” balancing — only one mole

of the desired product can be on the right hand side of the arrow (and nothing else), and fractional coefficients are allowed on the reactant side The values for the standard heats of formation (Hf) may be found in the appendix Solution:

a) C(graphite) + 2H2(g) + 1/2O2(g)  CH3OH(l) Hf= –238.6 kJ/mol

b) Ca(s) + 1/2O2(g)  CaO(s) Hf= –635.1 kJ/mol

c) C(graphite) + 1/4S8(rhombic)  CS2(l) Hf= 87.9 kJ/mol

6.10B Plan: Write the elements as reactants (each in its standard state), and place one mole of the substance formed on

the product side Balance the equation with the following differences from “normal” balancing — only one mole

of the desired product can be on the right hand side of the arrow (and nothing else), and fractional coefficients are allowed on the reactant side The values for the standard heats of formation (Hf) may be found in the appendix Solution:

a) C(graphite) + 1/2H2(g) + 3/2Cl2(g)  CHCl3(l) Hf= –132 kJ/mol

b) 1/2N2(g) + 2H2(g) + 1/2Cl2(g)  NH4Cl(s) Hf= –314.4 kJ/mol

c) Pb(s) + 1/8S8(rhombic) + 2O2(g)  PbSO4(s) Hf= –918.39 kJ/mol

6.11A Plan: Look up values from the appendix and use the equation Hrxn = mHf(products) – nHf(reactants)

  = {1Hf[CO2(g)] + 2Hf[H2O(g)]} – {1Hf[CH3OH(l)] + 3/2Hf [O2(g)]}

–638.6 kJ = [(1 mol)(–393.5 kJ/mol) + (2 mol)(–241.826 kJ/mol)]

– [Hf[CH3OH(l)] + (3/2 mol)(0 kJ/mol)]

f

H

–638.6 kJ = (–877.152 kJ) –Hf[CH3OH(l)]

f

H[CH3OH(l)] = –238.552 = –238.6 kJ

CHEMICAL CONNECTIONS BOXED READING PROBLEMS

B6.1 Plan: Convert the given mass in kg to g, divide by the molar mass to obtain moles, and convert moles to kJ of

energy Sodium sulfate decahydrate will transfer 354 kJ/mol

B6.2 Plan: Three reactions are given Equation 1) must be multiplied by 2, and then the reactions

can be added, canceling substances that appear on both sides of the arrow Add theHrxn

values for the three reactions to get the Hrxn for the overall gasification reaction of 2 moles of

coal Use the relationshipHrxn = mHf (products) – nHf (reactants) to find the heat of combustion of 1 mole

of methane Then find the Hrxn for the gasification of 1.00 kg of coal and Hrxn for the combustion of the methane produced from 1.00 kg of coal and sum these values

  = [(1 mol)(–395.5 kJ/mol) + (2 mol)(–241.826 kJ/mol)]

rxn

H

  = –804.282 kJ/mol CH4

Total heat for gasification of 1.00 kg coal:

H° = 1.00 kg coal 10 g3 1 mol coal 12.4 kJ

1 kg 12.00 g coal 2 mol coal

6.1 The sign of the energy transfer is defined from the perspective of the system Entering the system is positive, and

leaving the system is negative

6.2 No, an increase in temperature means that heat has been transferred to the surroundings, which makes q negative

6.3 E = q + w = w, since q = 0

Thus, the change in work equals the change in internal energy

6.4 Plan: Remember that an increase in internal energy is a result of the system (body) gaining heat or having work

done on it and a decrease in internal energy is a result of the system (body) losing heat or doing work

Solution:

The internal energy of the body is the sum of the cellular and molecular activities occurring from skin level inward The body’s internal energy can be increased by adding food, which adds energy to the body through the breaking of bonds in the food The body’s internal energy can also be increased through addition of work and heat, like the rubbing of another person’s warm hands on the body’s cold hands The body can lose energy if it performs work, like pushing a lawnmower, and can lose energy by losing heat to a cold room

6.5 a) electric heater b) sound amplifier c) light bulb d) automobile alternator

e) battery (voltaic cell)

6.6 Plan: Use the law of conservation of energy

Solution:

The amount of the change in internal energy in the two cases is the same By the law of energy conservation, the change in energy of the universe is zero This requires that the change in energy of the system (heater or air conditioner) equals an opposite change in energy of the surroundings (room air) Since both systems consume the same amount of electrical energy, the change in energy of the heater equals that of the air conditioner

6.7 Heat energy; sound energy (impact)

6.8 Plan: The change in a system’s energy is E = q + w If the system receives heat, then its qfinal is greater than

qinitial so q is positive Since the system performs work, its wfinal < winitial so w is negative

Solution:

E = q + w

E = (+425 J) + (–425 J) = 0 J

6.9 q + w = –255 cal + (–428 cal) = –683 cal

6.10 Plan: The change in a system’s energy is E = q + w A system that releases thermal energy has a negative

value for q and a system that has work done on it has a positive value for work Convert work in calories to

6.16 The system does work and thus its internal energy is decreased This means the sign will be negative

6.17 Since many reactions are performed in an open flask, the reaction proceeds at constant pressure The

determination of H (constant pressure conditions) requires a measurement of heat only, whereas E requires

measurement of heat and PV work

6.18 The hot pack is releasing (producing) heat, thus H is negative, and the process is exothermic

6.19 Plan: An exothermic process releases heat and an endothermic process absorbs heat

Solution:

a) Exothermic, the system (water) is releasing heat in changing from liquid to solid

b) Endothermic, the system (water) is absorbing heat in changing from liquid to gas

c) Exothermic, the process of digestion breaks down food and releases energy

d) Exothermic, heat is released as a person runs and muscles perform work

e) Endothermic, heat is absorbed as food calories are converted to body tissue

f) Endothermic, the wood being chopped absorbs heat (and work)

g) Exothermic, the furnace releases heat from fuel combustion Alternatively, if the system is defined as the air

in the house, the change is endothermic since the air’s temperature is increasing by the input of heat energy from

6.20 The internal energy of a substance is the sum of kinetic (EK) and potential (EP) terms

EK (total) = EK (translational) + EK (rotational) + EK (vibrational)

EP = EP (atom) + EP (bonds)

EP (atom) has nuclear, electronic, positional, magnetic, electrical, etc., components

6.21 H = E + PV (constant P)

a) H < E, PV is negative

b) H = E, a fixed volume means PV = 0

c) H > E, PV is positive for the transformation of solid to gas

6.22 Plan: Convert the initial volume from mL to L Subtract the initial V from the final V to find ΔV Calculate w in

atm•L Convert the answer from atm•L to J

6.23 Plan: Convert the pressure from mmHg to atm Subtract the initial V from the final V to find ΔV Calculate w in

atm•L Convert the answer from atm•L to kJ

6.25

6.26 Plan: Combustion of hydrocarbons and related compounds require oxygen (and a heat catalyst) to yield

carbon dioxide gas, water vapor, and heat Combustion reactions are exothermic The freezing of liquid water is

an exothermic process as heat is removed from the water in the conversion from liquid to solid An exothermic

reaction or process releases heat, so the reactants have greater H (Hinitial) than the products (Hfinal)

b) Freezing of water: H2O(l)  H2O(s) + heat

6.27 a) Na(s) + 1/2Cl2(g)  NaCl(s) + heat

6.28 Plan: Combustion of hydrocarbons and related compounds require oxygen (and a heat catalyst) to yield carbon

dioxide gas, water vapor, and heat Combustion reactions are exothermic An exothermic reaction releases heat, so

the reactants have greater H (Hinitial) than the products (Hfinal) If heat is absorbed, the reaction is endothermic and

the products have greater H (Hfinal) than the reactants (Hinitial)

6.30 Plan: Recall that qsys is positive if heat is absorbed by the system (endothermic) and negative if heat is released

by the system (exothermic) Since E = q + w, the work must be considered in addition to qsys to find ΔEsys Solution:

a) This is a phase change from the solid phase to the gas phase Heat is absorbed by the system so qsys is positive (+)

b) The system is expanding in volume as more moles of gas exist after the phase change than were present before

the phase change So the system has done work of expansion and w is negative ΔEsys = q + w Since q is positive and w is negative, the sign of ΔEsys cannot be predicted It will be positive if q > w and negative if

q < w

c) ΔEuniv = 0 If the system loses energy, the surroundings gain an equal amount of energy The sum of the

energy of the system and the energy of the surroundings remains constant

6.31 a) There is a volume decrease; Vfinal < Vinitial so ΔV is negative Since wsys = –PΔV, w is positive, +

b) ∆Hsys is – as heat has been removed from the system to liquefy the gas

c) ∆Esys = q + w Since q is negative and w is positive, the sign of ΔEsys and ΔEsurr cannot be predicted ΔEsys

will be positive and ΔEsurr will be negative if w > q and ΔEsys will be negative and ΔEsurr will be positive if

w < q

6.32 The molar heat capacity of a substance is larger than its specific heat capacity The specific heat capacity of a

substance is the quantity of heat required to change the temperature of 1 g of a substance by 1 K while the molar heat capacity is the quantity of heat required to change the temperature of 1 mole of a substance by 1 K

The specific heat capacity of a substance is multiplied by its molar mass to obtain the molar heat capacity

6.33 To determine the specific heat capacity of a substance, you need its mass, the heat added (or lost), and the change

6.34 Specific heat capacity is an intensive property; it is defined on a per gram basis The specific heat capacity of

a particular substance has the same value, regardless of the amount of substance present

6.35 Specific heat capacity is the quantity of heat required to raise 1g of a substance by 1 K Molar heat

capacity is the quantity of heat required to raise 1 mole of substance by 1 K Heat capacity is also the quantity of heat required for a 1 K temperature change, but it applies to an object instead of a specified amount of a

substance Thus, specific heat capacity and molar heat capacity are used when talking about an element or

compound while heat capacity is used for a calorimeter or other object

6.36 In a coffee-cup calorimeter, reactions occur at constant pressure qp = H

In a bomb calorimeter, reactions occur at constant volume qv = E

6.37 Plan: The heat required to raise the temperature of water is found by using the equation

q = c x mass x T The specific heat capacity, cwater, is found in Table 6.2 Because the Celsius degree is the same size as the Kelvin degree, T = 100°C – 25°C = 75°C = 75 K

6.39 Plan: Use the relationship q = c x mass x T We know the heat (change kJ to J), the specific heat capacity, and

the mass, so T can be calculated Once T is known, that value is added to the initial temperature to find the final temperature

4

7.50x10 J0.900 J

T = 282.4859 K = 282.4859°C (Because the Celsius degree is the same size as the Kelvin degree, T is the

same in either temperature unit.)

6.41 Plan: Since the bolts have the same mass and same specific heat capacity, and one must cool as the other heats

(the heat lost by the “hot” bolt equals the heat gained by the “cold” bolt), the final temperature is an average of the two initial temperatures

6.43 Plan: The heat lost by the water originally at 85°C is gained by the water that is originally at 26°C Therefore

–qlost = qgained Both volumes are converted to mass using the density

6.45 Plan: Heat gained by water and the container equals the heat lost by the copper tubing so

qwater + qcalorimeter = –qcopper

= (159 g H2O)(4.184 J/g°C)(Tfinal – 22.8)°C + (10.0 J/°C)(Tfinal – 22.8)°C

– (176.085)(Tfinal – 89.5) = (665.256)(Tfinal – 22.8) + (10.0)(Tfinal – 22.8)

15759.6075 – 176.085Tfinal = 665.256Tfinal – 15167.8368 + 10.0Tfinal – 228

15759.6075 + 15167.8368 + 228 = 176.085Tfinal + 665.256Tfinal + 10.0Tfinal

31155.4443 = 851.341Tfinal

Tfinal = 31155.4443/(851.341) = 36.59573 = 36.6°C

6.46 –qlost = qgained = qwater + qcalorimeter

– (30.5 g alloy)(calloy)(31.1 – 93.0)°C = (50.0 g H2O)(4.184 J/g°C)(31.1 – 22.0)°C + (9.2 J/°C)(31.1 – 22.0)°C

– (30.5 g)(calloy)(–61.9°C) = (50.0 g)(4.184 J/g°C)(9.1°C) + (9.2 J/°C)(9.1°C)

1887.95(calloy) = 1903.72 + 83.72 = 1987.44

calloy = 1987.44/1887.95 = 1.052697 = 1.1 J/g°C

6.47 Benzoic acid is C6H5COOH, and will be symbolized as HBz

–qreaction = qwater + qcalorimeter

6.48 a) Energy will flow from Cu (at 100.0°C) to Fe (at 0.0°C)

b) To determine the final temperature, the heat capacity of the calorimeter must be known

c) – qCu = qFe + qcalorimeter assume qcalorimeter = 0