Ch.9 - Thermochemistry: Chemical Energy
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- Assume that the nutritional content of an apple-say, 50 Cal (1 Cal = 1000 cal) - could be used to light a light bulb. For how many minutes would there be light from each of the following? (a) A 100-watt incandescent bulb (1 W = 1 J/s)
Problem 52
Problem 54a
A reaction inside a cylindrical container with a movable RAN piston causes the volume to change from 12.0 L to 18.0 L while the pressure outside the container remains constant at 0.975 atm. (The volume of a cylinder is V = pr2h, where h is the height; 1 L # atm = 101.325 J.) (a) What is the value in joules of the work w done during the reaction?
Problem 54b
A reaction inside a cylindrical container with a movable RAN piston causes the volume to change from 12.0 L to 18.0 L while the pressure outside the container remains constant at 0.975 atm. (The volume of a cylinder is V = pr2h, where h is the height; 1 L # atm = 101.325 J.) (b) The diameter of the piston is 17.0 cm. How far does the piston move?
Problem 55a
At a constant pressure of 0.905 atm, a chemical reaction takes place in a cylindrical container with a movable piston having a diameter of 40.0 cm. During the reaction, the height of the piston drops by 65.0 cm. (The volume of a cylinder is V=pr2h,wherehistheheight;1Latm=101.3J.) (a) What is the change in volume in liters during the reaction?
Problem 55b
At a constant pressure of 0.905 atm, a chemical reaction takes place in a cylindrical container with a movable piston having a diameter of 40.0 cm. During the reaction, the height of the piston drops by 65.0 cm. (The volume of a cylinder is V=pr2h,wherehistheheight;1Latm=101.3J.) (b) What is the value in joules of the work w done during the reaction?
- When a sample of a hydrocarbon fuel is ignited and burned in oxygen, the internal energy decreases by 7.20 kJ. If 5670 J of heat were transferred to the surroundings, what is the sign and magnitude of work? If the reaction took place in an environ- ment with a pressure of 1 atm, what was the volume change?
Problem 56
- What is the difference between the internal energy change ∆E and the enthalpy change ∆H? Which of the two is mea- sured at constant pressure and which at constant volume?
Problem 58
- Under what circumstances are ΔE and ΔH essentially equal?
Problem 59
- The enthalpy change for the reaction of 50.0 mL of ethylene with 50.0 mL of H2 at 1.5 atm pressure (Problem 9.51) is ∆H = -0.31 kJ. What is the value of ∆E?
Problem 62
- Assume that a particular reaction evolves 244 kJ of heat and that 35 kJ of PV work is gained by the system. What are the values of ∆E and ∆H for the system? For the surroundings?
Problem 63
- What is the enthalpy change (ΔH) for a reaction at a constant pressure of 1.00 atm fi the internal energy chagne (ΔE) is 44.0 kJ and the volume increase is 14.0 L? (1 L-atm = 101.325 J)
Problem 64
- A reactiont akes place at a constant pressure of 1.10 atm with an internal energy change (ΔE) of 71.5 kJ and a volume decrease of 13.6 L. What is the enthalpy change (ΔH) for the reaction? (1 L-atm = 101.325 J)
Problem 65
- What is the difference between heat capacity and specific heat?
Problem 76
- Does a measurement carried out in a bomb calorimeter give a value for ∆H or ∆E? Explain.
Problem 77
- Sodium metal is sometimes used as a cooling agent in heat-exchange units because of its releatively high molar heat capacity fo 28.2 J/(mol·°C). What is the specific heat and molar heat capacity of sodium in J/g·°C?
Problem 78
- Titanium metal is used as a structural material in many high-tech applications, such as in jet engines. what is the specific heat of titanium in J/(g·°C) if it takes 89.7 J to raise the temeprature of a 33.0 g block of 5.20 °C? What is the molar heat capacity of titanium J/(mol·°C)?
Problem 79
- Assuming that Coca-Cola has the saem specific heat as water [4.18 J/(g C)], calculate the amount of heat in kilojoules transferred when one can (about 350 g) is cooled from 25 C to 3 C.
Problem 80
- Calculate the amount of heat required to raise the tempera- ture of 250.0 g (approximately 1 cup) of hot chocolate from 25.0 °C to 80.0 °C. Assume hot chocolate has the same spe-cific heat as water 34.18 J>1g °C24.
Problem 81
- Instant cold packs used to treat athletic injuries contain solid NH4NO3 and a pouch of water. When the pack is squeezed, the pouch breaks and the solid dissolves, lowering the tem-perature because of the endothermic reaction NH4NO31s2 ¡ NH4NO31aq2 ∆H = +25.7 kJ What is the final temperature in a squeezed cold pack that contains 50.0 g of NH4NO3 dissolved in 125 mL of water? Assume a specific heat of 4.18 J/(g C) for the solution, an initial temperature of 25.0 °C, and no heat transfer between the cold pack and the environment.
Problem 82
- Instant hot packs contain a solid and a pouch of water. When the pack is squeezed, the pouch breaks and the solid dis- solves, increasing the temperature because of the exothermic reaciton. The following reaction is used to make a hot pack: H2O LiCl1s2 ¡ Li 1aq2 + Cl 1aq2 ∆H = -36.9 kJ. What is the final temperature in a squeezed hot pack that contains 25.0 g of LiCl dissolved# in 125 mL of water? Assume a specific heat of 4.18 J>1g °C2 for the solution, an initial temperature of 25.0 °C, and no heat transfer between the hot pack and the environment.
Problem 83
- When 0.187 g of benzene, C6H6, is burned in a bomb calorimeter the temperature rises by 3.45 °C. If the heat capacity of the calorimeter is 2.46 kJ>°C, calculate the combustion energy 1∆E2 for benzene in units of kJ/g and kJ/mol.
Problem 86
- When 1.50 g of magnesium metal is allowed to react with 200 mL of 6.00 M aqueous HCl, the temperature rises from 25.0 °C to 42.9 °C. Calculate ΔH in kilojoules for the reaction, assumign that the heat capacity of the calorimeter is 776 J/°C, that the specific heat of the final soltuion is the same as that of water [4.18 J(g·°C)] and that the density of the solution is 1.00 g/mL
Problem 88
- A 110.0 g piece of molybdenum metal is heated to 100.0 °C and placed in a calorimeter that contains 150.0 g of water at 24.6 °C. The system reaches equilibirum at a final temeprature of 28.0 °C. Calcualte the specific heat of molybdenum metal in J/g·°C. The specific heat of water is 4.18 J/g·°C
Problem 89
- Citric acid has three dissociable hydrogens. When 5.00 mL of 0.64 M citric acid and 45.00 mL of 0.77 M NaOH are mixed at an initial temperature of 26.0 °C, the temperature rises to 27.9 °C as the citric acid is neutralized. The combined mixture ahs a mass of 51.6 g and a specific heat of 4.0 J/(g·°C). Assuming that no heat is transferred to the surroundings, cal- culate the enthalpy change for the reaction of 1.00 mol of cit- ric acid in kJ. Is the reaction exothermic or endothermic?
Problem 90
- What is Hess's law, and why does it 'work'?
Problem 92
- The following steps occur in the reaction of ethyl alcohol (CH3CH2OH) wiht oxygen to yield acetic acid (CH3CO2H). Show that equations 1 and 2 sum to give the net equation and calculate ΔH° for the net equation. (1) CH3CH2OH(l) + 1/2 O2(g) → CH3CHO (g) + H2O(l) ΔH° = -174.2 kJ (2) CH3CHO(g) + 1/2 O2(g) → CH3CO2H(l) ΔH° = -318.4 kJ (Net) CH3CH2OH(l) + O2(g) → CH3CO2H(l) + H2O(l) ΔH° = ?
Problem 93
- The industrial degreasing solvent methylene chloride, CH2Cl2, is prepared from methane by reaction with chlorine: CH4(g) + 2 Cl2(g) → CH2Cl2(g) + 2 HCl(g) Use the following data to calcualte ΔH° in kilojoules for the reaction: CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g) ΔH° = -98.3 kJ CH3Cl(g) + Cl2(g) → CH2Cl2(g) + HCl(g) ΔH° = -104 kJ
Problem 94
- Hess's law can be used to calculate reaction enthalpies for hypothetical processes that can't be carried out in the labo- ratory. Set up a Hess's law cycle that will let you calculate ∆H° for the conversion of methane to ethylene: 2 CH4(g) → C2H4(g) + 2 H2(g) You can use the following information: 2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(l) ∆H° = -3120.8 kJ CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) ∆H° = -890.3 kJ C2H4(g) + H2(g) → C2H6(g) ∆H° = -136.3 kJ H2O(l) ∆H°f = -285.8 kJ/mol
Problem 95
- Find ∆H° in kilojoules for the reaciton of nitric oxide with oxygen, 2 NO(g) + O2(g) → N2O4(g), given the following data: N2O4(g) → 2 NO2(g) ∆H° = 55.3 kJ NO(g) + 1/2 O2(g) → NO2(g) ∆H° = -58.1 kJ
Problem 96
- Set up a Hess's law cycle, and use the following information to calculate ΔH°f for aqueous nitiric acid, HNO3(aq). You will need to use fractional coefficients for some equations. 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g) ΔH° = -137.3 kJ 2 NO(g) + O2(g) → 2 NO2(g) ΔH° = -116.2 kJ 4 NH3(g) + 5 O2(g) → 4 NO (g) + 6 H2O(l) ΔH° = -1165.2 kJ NH3(g) ΔH°f = -46.1 kJ/mol H2O(l) ΔH°f = -285.8 kJ/mol
Problem 97