Determining the Enthalpy Change of a Reaction
(Student textbook page 305)
21. A pellet of potassium hydroxide, KOH(s), having a mass of 0.648 g, is dissolved in
40.0 mL of water in an insulated cup. The temperature of the water increases from
22.6ºC to 27.8ºC. What is the molar enthalpy of solution, ∆Hsolution, for KOH(s)?
Assume that the solution has a density and a specific heat capacity equal to that of
water.
What Is Required?
You need to calculate the molar enthalpy of solution, ∆Hsolution, for KOH(s).
What Is Given?
You know the mass of the pellet of KOH(s): mKOH(s) = 0.648 g
You know the mass of the solution, KOH(aq): msolution = 40.0 g
You know the initial temperature: Tinitial = 22.6ºC
You know the final temperature: Tfinal = 27.8ºC
You know the specific heat capacity of H2O(ℓ): 4.19 J/g•°C
Plan Your Strategy
Convert the mass of KOH(s) to an
amount in moles, n, using the molar
mass of potassium hydroxide and the
m
formula n =
.
M
Determine the temperature change of
the system.
Use the formula Q = mc∆T to
calculate the amount of heat absorbed
by the solution.
Act on Your Strategy
m
n=
M
0.648 g
nKOH(s) =
56.11 g /mol
= 0.011 548 7 mol
DT = Tfinal - Tinitial
= 27.8°C - 22.6°C
= 5.2°C
Q = msolultion csolution DTsolution
= (40.0 g )(4.19 J/ g•°C )(5.2 °C )
= 871.52 J
Since ∆Esystem = –∆Esurroundings, change
the sign of Q to find ∆H, the change
in the thermal energy of the system.
= 0.871 52 kJ
DH = -Q
= -0.871 52 kJ
Unit 3 Part B ● MHR 25
DH
to
n
determine the molar enthalpy of
solution.
Use the formula DH solution =
DH
n
-0.871 52 kJ
=
0.011 548 7 mol
= -75.5 kJ/mol
DH solution =
Check Your Solution
Since much less than 1 mol of KOH(s) was dissolved, it is reasonable that a heat of
solution per mol will be much more than the heat absorbed by the 40.0 g of solution.
You know the reaction is exothermic since the temperature increases. The calculated
value of ∆H is negative. The answer shows the correct number of significant digits.
26 MHR ● Chemistry 12 Solutions Manual 978-0-07-106042-4
22. When 5.022 g of sodium hydrogen carbonate, NaHCO3(s), reacts completely with
80.00 mL of acetic acid, CH3COOH(aq), the temperature increases from 18.6ºC to
28.4ºC.
CH3COOH(aq) + NaHCO3(s) ɦ CH3COONa(aq) + CO2(g) + H2O(ℓ)
Assume that the acid solution has the same density and specific heat capacity as water
and that the mass of the final solution is 80.00 g. Calculate the molar enthalpy of
reaction, ∆Hr.
What Is Required?
You need to calculate the molar enthalpy of reaction.
What Is Given?
You know the mass of NaHCO3(s): mNaHCO3 = 5.022 g
You know the mass of CH3COOH(aq) solution: msolution = 80.0 g
You know the initial temperature: Tinitial = 18.6ºC
You know the final temperature: Tfinal = 28.4ºC
You know the specific heat capacity of H2O(ℓ): cH O = 4.19 J/g•°C
2
Plan Your Strategy
Convert the mass of NaHCO3(s) to
amount in moles, n, using the molar
mass of sodium hydrogen carbonate
m
.
and the formula n =
M
Determine the temperature change of
the system.
Use the formula Q = mc∆T to
calculate the amount of heat absorbed
by the solution.
Since ∆Esystem = –∆Esurroundings, change
the sign of Q to find ∆H, the change in
the thermal energy of the system.
Act on Your Strategy
m
n=
M
5.022 g
nNaHCO3 (s ) =
84.01 g /mol
= 0.059 778 mol
DT = Tfinal - Tinitial
= 28.4°C - 18.6°C
= 9.8°C
Q = msolultion csolution DTsolution
= (80.0 g )(4.19 J/ g•°C )(9.8 °C )
= 3284.96 J
= 3.284 96 kJ
DH = -Q
= -3.284 96 kJ
Unit 3 Part B ● MHR 27
DH
to
n
determine the molar enthalpy of
reaction.
Use the formula DH r =
DH
n
-3.284 96 kJ
=
0.0597 78 mol
= 54.952 65 kJ/mol
DH r =
The enthalpy of reaction is –55.0 kJ/mol.
Check Your Solution
The answer is reasonable for this amount of NaHCO3(s) reacting. You know the
reaction is exothermic since the temperature increases. The calculated value of ∆H is
negative. The answer has the correct number of significant digits.
28 MHR ● Chemistry 12 Solutions Manual 978-0-07-106042-4
23. Sodium reacts violently to form sodium hydroxide when placed in water, as shown in
the following equation: 2Na(s) + 2H2O(ℓ) → 2NaOH(aq) + H2(g)
Determine an experimental value for the molar enthalpy of reaction for sodium given
the following data:
mass of sodium, Na(s): 0.37 g
mass of water in calorimeter: 175 g
initial temperature of water: 19.30°C
final temperature of mixture: 25.70ºC
What Is Required?
You need to calculate the molar enthalpy of reaction.
What Is Given?
You know the mass of sodium, Na(s): mNa = 0.37 g
You know the mass of water in the calorimeter: mcalorimeter = 175 g
You know the initial temperature: Tinitial = 19.30ºC
You know the final temperature: Tfinal = 25.70ºC
You know the specific heat capacity of H2O(ℓ): cH O = 4.19 J/g•°C
2
Plan Your Strategy
Convert the mass of Na(s) to amount
in moles, n, using the molar mass of
m
sodium and the formula n =
.
M
Determine the temperature change of
the system.
Use the formula Q = mc∆T to calculate
the amount of heat absorbed by the
solution.
Act on Your Strategy
n=
n Na =
m
M
0.37 g
22.99 g /mol
= 0.016 093 mol
DT = Tfinal - Tinitial
= 25.70°C - 19.30°C
= 6.40°C
Q = msolultion csolution DTsolution
= (175 g )(4.19 J/ g•°C )(6.40° C )
= 4692.8 J
Since ∆Esystem = –∆Esurroundings, change
the sign of Q to find ∆H, the change in
the thermal energy of the system.
= 4.692 8 kJ
DH = -Q
= -4.692 8 kJ
Unit 3 Part B ● MHR 29
DH
to
n
determine the molar enthalpy of
reaction.
Use the formula DH r =
DH
n
-4.692 8 kJ
=
0.016093 mol
= -291.605 kJ/mol
DH r =
= -2.9 ´102 mol
The molar enthalpy of reaction is –2.9 ×102
kJ/mol of sodium.
Check Your Solution
The answer is reasonable for this amount of Na(s) reacting. You know the reaction is
exothermic since the temperature increases. The calculated value of ∆H is negative.
The answer has the correct number of significant digits.
30 MHR ● Chemistry 12 Solutions Manual 978-0-07-106042-4
24. In a simple calorimeter, 250.0 mL of 0.120 mol/L barium chloride, BaCl2(aq), is
mixed with 150.0 mL of 0.200 mol/L sodium sulfate, Na2SO4(aq). A precipitate of
barium sulfate, BaSO4(s), forms. The initial temperature of the two solutions is
20.00ºC. After mixing, the final temperature of the solutions is 20.49ºC. Calculate the
enthalpy of reaction, in kJ/mol, of Na2SO4(aq). Assume that the solutions have
densities and specific heat capacities equivalent to those of water.
BaCl2(aq) + Na2SO4(aq) → 2NaCl(aq) + BaSO4(s)
What Is Required?
You need to calculate the enthalpy of reaction.
What Is Given?
You know the balanced chemical equation for the reaction that occurs in aqueous
solution.
You know the concentration of BaCl2(aq): cBaCl = 0.120 mol/L
2
You know the volume of BaCl2(aq): VBaCl = 250.0 mL
You know the mass of BaCl2(aq): mBaCl = 250.0 g
2
2
You know the concentration of Na2SO4(aq): cNa SO = 0.200 mol/L
You know the volume of Na2SO4(aq): VNa SO = 150.0 mL
2
2
4
4
You know the massof Na2SO4(aq): mNa SO = 150.0 g
You know the total mass of the solutions:mT = 400.0 g
You know the initial temperature: Tinitial = 20.0ºC
You know the final temperature: Tfinal = 20.49ºC
You know the specific heat capacity of each solution is the same as the specific heat
capacity of H2O(ℓ): cH O = 4.19 J/g•°C
2
4
2
Plan Your Strategy
Use the formula n = cV to calculate
the amount in moles of each reactant.
Determine if there is a limiting
reactant.
Determine the temperature change of
the system.
Act on Your Strategy
n = cV
nBaCl2 = ( 0.120 mol/ L ) (0.250 L )
= 0.300 mol
n = cV
nNa 2SO4 = ( 0.200 mol/ L ) (0.150 L )
= 0.300 mol
Neither reactant is limiting.
DT = Tfinal - Tinitial
= 20.49°C - 20.00°C
= 0.49°C
Unit 3 Part B ● MHR 31
Use the formula Q = mc∆T to
calculate the heat absorbed by the
mixture.
Since ∆Esystem = –∆Esurroundings, change
the sign of Q to find ∆H, the change in
the thermal energy of the system.
DH
Use the formula DH r =
to
n
determine the molar heat of reaction.
Q = mmixture cmixture DTmixture
= (400 g )(4.19 J/ g•°C )(0.49° C )
= 821.24 J
= 8.2124 kJ
DH = -Q
= -0.821 24 kJ
DH
n
- 0.821 124 kJ
=
0.0300 mol
= –27.370 8 kJ/mol
= -27.4 kJ/mol Na 2SO 4 (aq)
DH r =
The enthalpy of reaction is –27.4 kJ/mol.
Check Your Solution
You know the reaction is exothermic since the temperature increases. The calculated
value of ∆H is negative. The answer has the correct number of significant digits and
seems reasonable.
32 MHR ● Chemistry 12 Solutions Manual 978-0-07-106042-4
25. A neutralization reaction occurs when 100.0 mL of 0.200 mol/L aqueous ammonia,
NH3(aq), and 200.0 mL of 0.200 mol/L hydrochloric acid, HCl(aq), are mixed in an
insulated cup.
NH3(aq) + HCl(aq) → NH4Cl(aq) + 53.6 kJ
Assuming that the two solutions have the same density and specific heat capacity as
water, what temperature change is expected after mixing?
What Is Required?
You need to calculate the temperature change for the reaction.
What Is Given?
You know the balanced thermochemical equation for the reaction that occurs in
aqueous solution.
From this, you know the heat of reaction: DH r = –53.6 kJ/mol
You know the concentration of NH3(aq )solution: cNH = 0.200 mol/L
3
You know the volume of NH3(aq) solution: VNH = 100.0 mL
You know the mass of NH3(aq) solution: mNH = 100.0 g
3
3
You know the concentration of HCl(aq) solution: cHCl = 0.200 mol/L
You know the volume of HCl(aq) solution: VHCl = 250.0 mL
You know the mass of HCl(aq) solution: mHCl 250.0 g
You know the total mass of solution: mT = 350.0 g
You know the specific heat capacity of each solution is the same as the specific heat
capacity of H2O(ℓ): cH O = 4.19 J/g•°C
2
Plan Your Strategy
Use the formula n = cV to calculate
the amount in moles of each reactant.
Determine if there is a limiting
reactant.
Act on Your Strategy
n = cV
nNH3 = ( 0.200 mol/ L ) (0.100 L )
= 0.0200 mol
n = cV
nHCl = ( 0.200 mol/ L ) (0.200 L )
= 0.0400 mol
The balanced chemical equation indicates
that NH3(aq) and HCl(aq) react in a mole
ratio of 1:1. By inspection, since there is a
lesser amount in moles of NH3(aq) , this
reactant is limiting. The amount in moles of
NH3(aq) must be used to calculate ∆Hr.
Unit 3 Part B ● MHR 33
Use the formula ∆H = n∆Hr to find
∆Hr.
Since ∆Esystem = –∆Esurroundings, change
the sign of ∆H to find Q, the amount of
heat absorbed by the solutions.
Use the formula Q = mc∆T to calculate
the temperature change, ∆T.
DH = nDH r
n
DH r = NH3
DH
0.020 mol
=
-53.6 kJ/ mol
= -1.072 kJ
Q = -DH
= 1072 J
Q = mcDT
1072 J = (300 g)(4.19 J/g • °C)(DT )
(300 g )(4.19 J / g • °C)
DT =
1072 J
= 0.852 8°C
The temperature increases by 0.853°C.
Check Your Solution
You know the reaction is exothermic so the temperature must increase. This seems to
be a reasonable temperature change. The answer has the correct number of significant
digits.
34 MHR ● Chemistry 12 Solutions Manual 978-0-07-106042-4
26. In a simple calorimeter, 150.0 mL of 1.000 mol/L NaOH(aq) is mixed with 150.0 mL
of 1.000 mol/L HCl(aq). If both solutions were initially at 25.00°C and after mixing
the temperature increased to 30.00°C, what is the enthalpy of reaction as written?
Assume that the solutions have a density of 1.000 g/mL and a specific heat capacity of
4.19 J/g•°C.
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(ℓ)
What Is Required?
You need to calculate the enthalpy of reaction.
What Is Given?
You know the balanced chemical equation for the reaction that occurs in aqueous
solution.
You know the concentration of NaOH(aq) solution: cNaOH = 1.000 mol/L
You know the volume of NaOH(aq) solution = VNaOH = 150.0 mL
You know the mass of NaOH(aq) solution = mNaOH = 150.0 g
You know the concentration of HCl(aq) solution: cHCl = 1.000 mol/L
You know the volume of HCl(aq) solution: VHCl = 150.0 mL
You know the mass of HCl(aq) solution: mHCl = 150.0 g
You know the total mass of the solutions: 300.0 g
You know the initial temperature: Tinitial = 25.00ºC
You know the final temperature: Tfinal = 30.00ºC
You know the specific heat capacity of each solution is the same as the specific heat
capacity of H2O(ℓ): = cH O = 4.19 J/g•°C
2
Plan Your Strategy
Use the formula n = cV to calculate
the amount in moles of each reactant.
Determine the temperature change of
the system.
Use the formula Q = mc∆T to calculate
the heat absorbed by the mixture.
Since ∆Esystem = –∆Esurroundings, change
the sign of Q to find ∆H, the change in
the thermal energy of the system.
Act on Your Strategy
n = cV
nHCl = (1.000 mol/ L ) (0.1500 L )
= 0.1500 mol
DT = Tfinal - Tinitial
= 30.00°C - 25.00°C
= 5.00°C
Q = mmixture cmixture DTmixture
= (300 g )(4.19 J/ g•°C )(5.00° C )
= 6285 J
= 6.285 kJ
DH = -Q
= -6.285 kJ
Unit 3 Part B ● MHR 35
DH
to
n
determine the molar heat of reaction.
Use the formula ∆Hr =
-6.285 kJ
0.150 00 mol
= -41.9 kJ/mol
DH ro =
The enthalpy of reaction is –41.9 kJ/mol.
Check Your Solution
You know the reaction is exothermic since the temperature increases. The calculated
value of ∆H is negative. The answer has the correct number of significant digits and
seems reasonable.
36 MHR ● Chemistry 12 Solutions Manual 978-0-07-106042-4
27. The enthalpy of solution for sodium hydroxide, NaOH(s), is given as –55.0 kJ/mol. A
chemist prepares 250.0 mL of a 0.100 mol/L solution of NaOH(aq). Assuming that
this solution has the same specific heat capacity and density as water, by how much
should the water temperature increase as the NaOH(s) dissolves?
What Is Required?
You need to determine the temperature change as NaOH(s) dissolves.
What Is Given?
You know the enthalpy of solution: –55.0 kJ/mol
You know the concentration of NaOH(aq) solution: cNaOH = 0.100 mol/L
You know the volume NaOH(aq) solution: VNaOH = 100.0 mL
You know the mass of NaOH(aq) solution: mNaOH = 100.0 g
You know the specific heat capacity of the solution is the same as the specific heat
capacity of H2O(ℓ): cH O = 4.19 J/g•°C
2
Plan Your Strategy
Use the formula n = cV to
calculate the amount in moles, n,
of NaOH(aq).
DH
n
to determine the amount of heat
given off during dissolving.
Use the formula ∆Hsolution =
Since ∆Esystem = –∆Esurroundings,
change the sign of ∆H to find the
amount of heat, Q, absorbed by
the surroundings. .
Use the formula Q = mc∆T to
calculate the change in
temperature, ∆T.
Act on Your Strategy
n = cV
nNaOH = ( 0.100 mol/ L ) (0.2500 L )
= 0.02500 mol
DH solution =
DH
n
DH
0.025 00 mol
DH = (-55.0 kJ/ mol )(0.025 00 mol )
= -1.375 kJ
= -1375 J
= -DH
= 1375 J
-55.0 kJ/mol =
Qsurroundings
Q = mcDT
1375 J = (250 g)(4.19 J/g • °C)(DT )
1375 J
DT =
(250 g )(4.19 J / g • °C)
= 1.31°C
The temperature of the water increases by 1.31°C.
Check Your Solution
You know the reaction is exothermic, so the temperature must increase. The change in
temperature seems reasonable. The answer has the correct number of significant digits.
Unit 3 Part B ● MHR 37
28. A neutralization reaction occurs when 120.00 mL of 0.500 mol/L LiOH and 160.00
mL of 0.375 mol/L HNO3(aq) are mixed in an insulated cup. Initially, the solutions are
at the same temperature. If the highest temperature reached during mixing was 24.5°C,
what was the initial temperature of the solutions?
LiOH(aq) + HNO3(aq) → LiNO3(aq) + H2O(ℓ) + 53.1 kJ
Assume that both of these solutions have a density of 1.00 g/mL and a specific heat
capacity of 4.19 J/g •°C.
What Is Required?
You need to determine the initial temperature of solutions used in a neutralization
reaction.
What Is Given?
You know the concentration, c, of LiOH(aq) solution: cLiOH = 0.500 mol/L
You know the volume of LiOH(aq) solution: VLiOH = 120.0 mL
You know the mass of LiOH(aq) solution: mLiOH = 120.0 g
You know the concentration, c, of HNO3(aq) solution: cHNO = 0.3750 mol/L
3
You know the volume of HNO3(aq) solution: VHNO = 160.0 mL
You know the mass of HNO3(aq) solution: mHNO = 160.0 g
You know the total mass of the solutions: mT = 280.0 g
You know the final temperature: Tfinal = 24.5ºC
You know the specific heat capacity of each solution is the same as the specific heat
capacity of H2O(ℓ): cH O = 4.19 J/g•°C
3
3
2
Plan Your Strategy
Use the formula n = cV to
calculate the amount in moles, n, of
each reactant.
Determine if there is a limiting
reactant.
Act on Your Strategy
n = cV
nLiOH = ( 0.500 mol/ L ) (0.1200 L )
= 0.0600 mol
n = cV
nHNO3 = ( 0.375 mol/ L ) (0.1600 L )
= 0.0600 mol
Neither reactant is limiting.
38 MHR ● Chemistry 12 Solutions Manual 978-0-07-106042-4
DH
to
n
determine the amount of heat given
off in the neutralization.
Use the formula ∆Hsolution =
Since ∆Esystem = –∆Esurroundings,
change the sign of ∆H to find the
amount of heat, Q, absorbed by the
surroundings. .
Use the formula Q = mc∆T to
calculate the change in
temperature, ∆T.
Use the formula DT = Tfinal - Tinitial
to determine the initial temperature
of the solutions.
DH solution =
DH
n
DH
0.0600 mol
DH = (-53.1 kJ/ mol )(0.0600 mol )
= -3.186 kJ
= -3186 J
Qneutralization = -DH
= 3186 J
-53.1 kJ/mol =
Q = mcDT
3186 J = (280 g)(4.19 J/g • °C)(DT )
3186 J
DT =
(280 g )(4.19 J / g • °C)
= 2.7°C
DT = Tfinal - Tinitial
Tinitial = Tfinal - DT
= 24.5°C - 2.7°C
= 21.8°C
The initial temperature of the solutions was
21.8°C.
Check Your Solution
You know the reaction is exothermic. Since the temperature must increase, the initial
temperature must be lower than the final temperature. This seems to be a reasonable
initial temperature. The answer has the correct number of significant digits.
Unit 3 Part B ● MHR 39
29. Peroxides will react to release oxygen when added to water. By how much would the
water temperature change if 7.800 g of sodium peroxide, Na2O2(s), is added to 110.00
mL of water?
∆Ho = –285.0 kJ
2Na2O2(s) + 2H2O(ℓ) → 4NaOH(aq) + O2(g)
What Is Required?
You need to determine the change in temperature when sodium peroxide, Na2O2(s), is
dissolved in water.
What Is Given?
You know the balanced chemical equation for the reaction and the enthalpy of
reaction: ∆Ho = –285.0 kJ
You know the volume of water: VH O = 110.0 mL
2
You know the mass, m, of water: mH O = 110.0 g
You know the specific heat capacity of H2O(ℓ): cH O = 4.19 J/g•°C
2
2
You know the mass, m, of Na2O2(s): mNa O
2
Plan Your Strategy
Convert the mass of Na2O2(s) to
amount in moles using the molar
mass, M, of Na2O2(s) and the
m
formula n =
.
M
= 7.800 g
2 (s)
Act on Your Strategy
n=
nNa 2O2 =
m
M
7.800 g
77.98 g /mol
= 0.1000 mol
Let x represent ∆Hr. Use the mole
ratio in the balanced chemical
equation to calculate the enthalpy
change.
2 mol Na 2O 2 0.1000 mol Na 2O 2
=
x
-285.0 kJ
(0.1000 mol Na 2O 2 )(285.0 kJ)
x=
2 mol Na 2O 2
= -14.25 kJ
DH r = -14.25 kJ
= -14 250 J
Since ∆Esystem = –∆Esurroundings,
change the sign of ∆Hr to find the
amount of heat, Q, absorbed by the
surroundings. .
Q = -DH r
= 14 250 J
40 MHR ● Chemistry 12 Solutions Manual 978-0-07-106042-4
Use the formula Q = mc∆T to
calculate the change in
temperature, ∆T.
Q = mcDT
14 250 J = (110.0 g)(4.19 J/g • °C)(DT )
14 250 J
DT =
(110.0 g )(4.19 J / g • °C)
= 30.9°C
The temperature of the water increases by
30.9°C.
Check Your Solution
You know the reaction is exothermic so the temperature must increase. This seems to
be a reasonable temperature change. The answer has the correct number of significant
digits.
Unit 3 Part B ● MHR 41