Idea Transcript
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