Alberta Inquiry into Chemistry February 12, 2007

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1 Unit 5 Thermochemical Changes Chapter 0 Theories of Energy and Chemical Changes Solutions to Practice s. Ethene, C 2 H 4 (g), reacts with water to form ethanol, C 2 H 5 OH(l), as shown below: C 2 H 4 (g) + H 2 O(l) C 2 H 5 OH(l) Determine the enthalpy change of this reaction, given the following thermochemical equations. () C 2 H 5 OH(l) + 3O 2 (g) 3H 2 O(l) + 2CO 2 (g) Δ c H o = kj/mol (2) C 2 H 4 (g) + 3O 2 (g) 2H 2 O(l) + 2CO 2 (g) Δ c H o = 4.0 kj/mol You must manipulate the given equations to calculate the enthalpy change for the overall reaction. The thermochemical equations: () C 2 H 5 OH(l) + 3O 2 (g) 3H 2 O(l) + 2CO 2 (g) Δ c H o = kj/mol (2) C 2 H 4 (g) + 3O 2 (g) 2H 2 O(l) + 2CO 2 (g) Δ c H o = 4.0 kj/mol Applying Hess s Law, manipulate the given thermochemical equations to obtain an algebraic sum equivalent to the overall reaction. The overall enthalpy change, Δ r H o, is the algebraic sum of the enthalpy changes for the reactions used. () 3H 2 O(l) + 2CO 2 (g) C 2 H 5 OH(l) + 3O 2 (g) Δ c H o = kj/mol (2) C 2 H 4 (g) + 3O 2 (g) 2H 2 O(l) + 2CO 2 (g) Δ c H o = 4.0 kj/mol C 2 H 4 (g) + H 2 O(l) C 2 H 5 OH(l) Δ r H o, = 44.2 kj/mol The answer is reasonable for this reaction. The unit is correct (kj/mol of ethanol) and the answer has the correct number of significant digits after the decimal (). 2. A typical automobile uses a lead-acid battery. During discharge, the following chemical reaction takes place: Pb(s) + PbO 2 (s) + 2H 2 SO 4 (l) 2PbSO 4 (aq) + 2H 2 O(l) Determine the enthalpy change of this reaction, given the following equations: () Pb(s) + PbO 2 (s) + 2SO 3 (g) PbSO 4 (aq) ΔH o = 773 kj (2) SO 3 (g) + H 2 O(l) H 2 SO 4 (l) ΔH o = 33 kj Page of 6

2 You must manipulate the given equations to calculate the enthalpy change for the overall reaction. The thermochemical equations: () Pb(s) + PbO 2 (s) + 2SO 3 (g) PbSO 4 (aq) ΔH o = 773 kj (2) SO 3 (g) + H 2 O(l) H 2 SO 4 (l) ΔH o = 33 kj Applying Hess s Law, manipulate the given thermochemical equations to obtain an algebraic sum equivalent to the overall reaction. The overall enthalpy change, Δ r H o, is the algebraic sum of the enthalpy changes for the reactions used. () Pb(s) + PbO 2 (s) + 2SO 3 (g) PbSO 4 (aq) ΔH o = 773 kj (2) 2 2H 2 SO 4 (l) 2SO 3 (g) + 2H 2 O(l) ΔH o = +266 kj Pb(s) + PbO 2 (s) + 2H 2 SO 4 (l) 2PbSO 4 (aq) + 2H 2 O(l) Δ r H o = 507 kj The reaction has a negative enthalpy change and therefore it is an exothermic reaction, as expected. The unit is correct (kj) and the answer has the correct number of digits after the decimal (0). 3. Mixing household cleansers can result in the production of hydrogen chloride gas, HCl(g). Not only is this gas toxic and corrosive, but it also reacts with oxygen to form poisonous chlorine gas according to the following equation: 4HCl(g) + O 2 (g) 2Cl 2 (g) + 2H 2 O(g) Determine the enthalpy change of this reaction, given the following equations: () H 2 (g) + Cl 2 (g) 2HCl(g) ΔH = 84.6 kj (2) H 2 (g) + 2 O2 (g) H 2 O(l) Δ f H = kj/mol (3) H 2 O(l) H 2 O(g) Δ vaph = 40.7 kj/mol You must manipulate the given equations to calculate the enthalpy change for the overall reaction. The thermochemical equations: () H 2 (g) + Cl 2 (g) 2HCl(g) ΔH = 84.6 kj Page 2 of 6

3 (2) H 2 (g) + 2 O2 (g) H 2 O(l) Δ f H = kj/mol (3) H 2 O(l) H 2 O(g) Δ vap H = 40.7 kj/mol Applying Hess s Law, manipulate the given thermochemical equations to obtain an algebraic sum equivalent to the overall reaction. The overall enthalpy change, Δ r H o, is the algebraic sum of the enthalpy changes for the reactions used. () 2 4HCl(g) 2H 2 (g) + 2Cl 2 (g) ΔH = kj (2) 2 2H 2 (g) + O 2 (g) 2H 2 O(l) ΔH = 57.6 kj (3) 2 2H 2 O(l) 2H 2 O(g) ΔH = +8.4 kj 4HCl(g) + O 2 (g) 2Cl 2 (g) + 2H 2 O(g) Δ r H o = 2.0 kj The reaction has a negative enthalpy change and it is exothermic, as expected. The unit is correct (kj) and the answer has the correct number of digits after the decimal (). 4. Calculate the enthalpy change of the following reaction between nitrogen gas and oxygen gas, given thermochemical equations (), (2), and (3): 2N 2 (g) + 5O 2 (g) 2N 2 O 5 (g) () 2H 2 (g) + O 2 (g) 2H 2 O(l) ΔH = 572 kj (2) N 2 O 5 (g) + H 2 O(l) 2HNO 3 (aq) ΔH = 77 kj (3) 2 N2 (g) O2 (g) + 2 H2 (g) HNO 3 (l) Δ f H = 74 kj/mol You must manipulate the given equations to calculate the enthalpy change for the overall reaction. The thermochemical equations: () 2H 2 (g) + O 2 (g) 2H 2 O(l) ΔH = 572 kj (2) N 2 O 5 (g) + H 2 O(l) 2HNO 3 (aq) ΔH = 77 kj (3) 2 N2 (g) O2 (g) + 2 H2 (g) HNO 3 (l) Δ f H = 74 kj/mol Applying Hess s Law, manipulate the given thermochemical equations to obtain an algebraic sum equivalent to the overall reaction. The overall enthalpy change, Δ r H o, is the algebraic sum of the enthalpy changes for the reactions used. Page 3 of 6

4 () 2H 2 O(l) 2H 2 (g) + O 2 (g) ΔH = +572 kj (2) 2 4HNO 3 (aq) 2N 2 O 5 (g) + 2H 2 O(l) ΔH = +54 kj (3) 4 2N 2 (g) + 6O 2 (g) + 2H 2 (g) 4HNO 3 (l) ΔH = 696 kj 2N 2 (g) + 5O 2 (g) 2N 2 O 5 (g) Δ r H o = 30 kj The reaction has a positive enthalpy change and it is endothermic, as expected. The unit is correct (kj) and the answer has the correct number of digits after the decimal (0). 5. Ethene, C 2 H 4 (g), is used in the manufacturing of many polymers, including polyethylene terephthalate (PET), which is used to make pop bottles. Determine the molar enthalpy of formation for ethene, as shown by 2C(s) + 2H 2 (g) C 2 H 4 (g), given the following equations: () C(s) + O 2 (g) CO 2 (g) ΔH = kj (2) H 2 (g) + 2 O2 (g) H 2 O(l) ΔH = kj (3) C 2 H 4 (g) + 3O 2 (g) 2CO 2 (g) + 2H 2 O(l) ΔH = 4.2 kj You must manipulate the given equations to calculate the enthalpy of formation for ethene, C 2 H 4 (g). The thermochemical equations: () C(s) + O 2 (g) CO 2 (g) ΔH = kj (2) H 2 (g) + 2 O2 (g) H 2 O(l) ΔH = kj (3) C 2 H 4 (g) + 3O 2 (g) 2CO 2 (g) + 2H 2 O(l) ΔH = 4.2 kj Applying Hess s Law, manipulate the given thermochemical equations to obtain an algebraic sum equivalent to the overall reaction. The overall enthalpy change is the algebraic sum of the enthalpy changes for the reactions used and represents the enthalpy of formation, Δ f H o. () 2 2C(s) + 2O 2 (g) 2CO 2 (g) ΔH = kj (2) 2 2H 2 (g) + O 2 (g) 2H 2 O(l) ΔH = 57.6 kj (3) 2CO 2 (g) + 2H 2 O(l) C 2 H 4 (g) + 3O 2 (g) ΔH = +4.2 kj 2C(s) + 2H 2 (g) C 2 H 4 (g) Δ f H o = kj/mol Page 4 of 6

5 The unit is correct (kj/mol) and the answer has the correct number of digits after the decimal (). 6. From the following equations, determine the molar enthalpy of formation for HNO 2 (aq), as shown below in the overall equation: H2 (g) + N2 (g) + O 2 (g) HNO 2 (aq) 2 2 () NH 4 NO 2 (aq) N 2 (g) + 2H 2 O(l) ΔH = 320. kj (2) NH 3 (aq) + HNO 2 (aq) NH 4 NO 2 (aq) ΔH = 37.7 kj (3) 2NH 3 (aq) N 2 (g) + 3H 2 (g) ΔH = kj (4) H 2 (g) + 2 O2 (g) H 2 O(l) ΔH = kj You must manipulate the given equations to calculate the enthalpy of formation for HNO 2 (aq). The thermochemical equations: () NH 4 NO 2 (aq) N 2 (g) + 2H 2 O(l) ΔH = 320. kj (2) NH 3 (aq) + HNO 2 (aq) NH 4 NO 2 (aq) ΔH = 37.7 kj (3) 2NH 3 (aq) N 2 (g) + 3H 2 (g) ΔH = kj (4) H 2 (g) + 2 O2 (g) H 2 O(l) ΔH = kj Applying Hess s Law, manipulate the given thermochemical equations to obtain an algebraic sum equivalent to the overall reaction. The overall enthalpy change is the algebraic sum of the enthalpy changes for the reactions used and represents the enthalpy of formation, Δ f H o for HNO 2 (aq). () N 2 (g) + 2H 2 O(l) NH 4 NO 2 (aq) ΔH = kj (2) NH 4 NO 2 (aq) NH 3 (aq) + HNO 2 (aq) ΔH = kj (3) /2 NH 3 (aq) 2 N2 (g) H2 (g) ΔH = kj (4) 2 2H 2 (g) + O 2 (g) 2H 2 O(l) ΔH = 57.6 kj H2 (g) + N2 (g) + O 2 (g) HNO 2 (aq) Δ f H o = 28.8 kj/mol 2 2 Page 5 of 6

6 The unit is correct (kj/mol) and the answer has the correct number of digits after the decimal (). 7. Hydrogen can be added to ethene, C 2 H 4 (g), to obtain ethane, C 2 H 6 (g): C 2 H 4 (g) + H 2 (g) C 2 H 6 (g) Write thermochemical equations for the formation of both ethene, C 2 H 4 (g), and ethane, C 2 H 6 (g). Show that these equations can be algebraically combined to obtain the equation for the addition of hydrogen to ethene. Determine the enthalpy of reaction. You must write the thermochemical equations representing the formation of ethene, C 2 H 4 (g), and ethane, C 2 H 6 (g) and demonstrate that these equations can be combined algebraically to give the overall reaction between ethene and hydrogen to produce ethane. From this answer calculate the enthalpy of reaction. The balanced equation for the reaction between ethene and hydrogen to produce ethane is given. The heats of formation of ethene, C 2 H 4 (g), and ethane, C 2 H 6 (g) can be found in the Chemistry Data Booklet.. Write the equation for the formation of ethene, C 2 H 4 (g), and ethane, C 2 H 6 (g). Apply Hess s Law to obtain the overall reaction between ethene, C 2 H 4 (g), and ethane, C 2 H 6 (g). The overall enthalpy change, Δ r H o is the algebraic sum of the enthalpy changes for the reactions used. () formation of C 2 H 4 (g): 2C(s) + 2H 2 (g) C 2 H 4 (g) Δ f H o = kj/mol (2) formation of C 2 H 6 (g): 2C(s) + 3H 2 (g) C 2 H 6 (g) Δ f H o = 84.0 kj/mol () C 2 H 4 (g) 2C(s) + 2H 2 (g) Δ f H o = 52.4 kj/mol (2) 2C(s) + 3H 2 (g) C 2 H 6 (g) Δ f H o = 84.0 kj/mol C 2 H 4 (g) + 3H 2 (g) C 2 H 6 (g) Δ r H o = 36.4 kj The enthalpy of reaction is reasonable. The answer has the correct unit (kj) and the correct number of digits after the decimal () 8. Zinc sulfide reacts with oxygen gas to produce zinc oxide and sulfur dioxide: 2ZnS(s) + 3O 2 (g) 2ZnO(s) + 2SO 2 (g) Page 6 of 6

7 Calculate the enthalpy change of this reaction by using enthalpies of formation from Appendix G or your Chemistry Data Booklet. You must calculate the enthalpy change for the given reaction using standard enthalpies of formation of the reactants and products. The balanced equation for the reaction is given and the standard enthalpies of formation are in Appendix J. Look up the standard enthalpies of formation, Δ f H o, for each reactant and product in this reaction and calculate the enthalpy of reaction by substituting these values into the expression Δ r Hº = Σ(nΔ f Hº products) Σ(nΔ f Hº reactants) 2ZnS(s) + 3O 2 (g) 2ZnO(s) + 2SO 2 (g) Δ r Hº = Σ(nΔ f Hº products) Σ(nΔ f Hº reactants) Δ r Hº = [(2 mol) (Δ f Hº ZnO(s)) + (2 mol)(δ f Hº SO 2 (g))] [(2 mol) (Δ f Hº ZnS(s)) + (3 mol) (Δ f Hº O 2 (g))] Δ r Hº = [(2 mol) ( kj/mol) + (2 mol) ( kj/mol)] [(2 mol) ( kj/mol) + (3 mol) (0)] Δ r Hº = [ 70.0 kj + ( 593.6)] [ 42.0 kj] Δ r Hº = kj The answer has the correct unit (kj) and the correct number of digits after the decimal (). 9. Small amounts of oxygen gas can be produced in a laboratory by heating potassium chlorate, KClO 3 (s): 2KClO 3 (s) 2KCl(s) + 3O 2 (g) Calculate the enthalpy change of this reaction by using enthalpies of formation. You must calculate the enthalpy change for the given reaction using standard enthalpies of formation of the reactants and products. The balanced equation for the reaction is given and the standard enthalpies of formation are in the Chemistry Data Booklet. Page 7 of 6

8 Look up the standard enthalpies of formation, Δ f H o, for each reactant and product in this reaction and calculate the enthalpy of reaction by substituting these values into the expression Δ r Hº = Σ(nΔ f Hº products) Σ(nΔ f Hº reactants) 2KClO 3 (s) 2KCl(s) + 3O 2 (g) Δ r Hº = Σ(nΔ f Hº products) Σ(nΔ f Hº reactants) Δ r Hº = [(2 mol) (Δ f Hº KCl(s)) + (3 mol) (Δ f Hº O 2 (g))] [(2 mol) (Δ f Hº KClO 3 (s))] Δ r Hº = [(2 mol) ( kj/mol) + (3 mol) (0)] [(2 mol) ( kj/mol)] Δ r Hº = [ kj] [ kj] = 77.6 kj The answer has the correct unit (kj) and the correct number of digits after the decimal (). 0. Use the following equation to answer the questions below: CH 3 OH(l) + O 2 (g) CO 2 (g) + 2H 2 O(g) a) Calculate the molar enthalpy of combustion of methanol by using enthalpies of formation. b) Use your answer from a) to determine how much energy is released when 25 g of methanol undergoes complete combustion. a) You must calculate the enthalpy change for the given reaction using standard enthalpies of formation of the reactants and products. b) You must use the answer from a) and the balanced equation to calculate the energy released when 25 g of methanol is burned. a) The balanced equation for the reaction is given and the standard enthalpies of formation are in the Chemistry Data Booklet. b) You know that 25 g of methanol is burned. a) Look up the standard enthalpies of formation, Δ f H o, for each reactant and product in this reaction and calculate the enthalpy of reaction by substituting these values into the expression Δ r Hº = Σ(nΔ f Hº products) Σ(nΔ f Hº reactants) b) Determine the molar mass (M) of methanol, CH 3 OH(l), and convert 25 g to moles using n = m. Use the mole ratio in the balanced equation to calculate the energy released from the M burning of this number of moles of methanol. Page 8 of 6

9 a) CH 3 OH(l) O2 (g) CO 2 (g) + 2H 2 O(g) Δ r Hº = Σ(nΔ f Hº products) Σ(nΔ f Hº reactants) ΔHº = [( mol) (Δ f Hº CO 2 (g)) + (2 mol) (Δ f Hº H 2 O(g))] [( mol) (Δ f Hº CH 3 OH(l)) Δf Hº O 2 (g))] Δ r Hº = [( mol) ( kj/mol) + (2 mol) ( 24.8 kj/mol)] [( mol) ( kj/mol) mol (0)] Δ r Hº = [ 877. kj] [ kj] Δ r Hº = kj per mol of CH 3 OH(l) b) MCH 3 OH(l) = g/mol nch 3 OH(l) = M m = 25g 32.05g/mol = 3.90 mol energy released = 3.90 mol CH 3 OH(l) ( kj/mol) = kj a) The enthalpy of reaction is negative and the reaction is exothermic as expected. The answer has the correct unit (kj per mol of CH 3 OH(l)), and the correct number of digits after the decimal (). b) The heat given off has the correct unit (kj) and it has the correct number of significant digits (3).. Consider the following equation representing the reaction of methane and chlorine to form chloroform, CHCl 3 (g): CH 4 (g) + 3Cl 2 (g) CHCl 3 (g) + 3HCl(g) ΔH = kj Use standard molar enthalpies of formation to determine the molar enthalpy of formation for chloroform, CHCl 3 (g). You must calculate the enthalpy of formation, Δ f Hº, of chloroform, CHCl 3 (g). The balanced equation for the reaction is given. The enthalpy of reaction, Δ r Hº, is known and the standard enthalpies of formation are in the Chemistry Data Booklet. Look up the standard enthalpies of formation, Δ f H o, for CH 4 (g) and HCl(g). Use these values and the given enthalpy of reaction, Δ r Hº, in the expression Page 9 of 6

10 Δ r Hº = Σ(nΔ f Hº products) Σ(nΔ f Hº reactants). Solve for Δ f H of CHCl 3 (g). CH 4 (g) + 3Cl 2 (g) CHCl 3 (g) + 3HCl(g) ΔH = kj Δ r Hº = Σ (nδ f Hº products) Σ(nΔ f Hº reactants) Δ r Hº = [( mol) (Δ f Hº CHCl 3 (g)) + (3 mol) (Δ f Hº HCl(g)] [( mol) (Δ f Hº CH 4 (g)) + (3 mol) Δ f Hº Cl 2 (g))] kj = [( mol) (Δ f Hº CHCl 3 (g)) + (3 mol) ( 92.3 kj/mol)] [( mol) ( 74.6 kj/mol) + (3 mol) (0)] ( mol)δ f Hº CHCl 3 (g) = kj kj 74.6 kj Δ r Hº = 02.7 kj/mol This answer is reasonable, has the correct unit (kj/mol) and the correct number of digits after the decimal (). 2. The molar enthalpy of combustion of heptane, C 7 H 6 (l), in a bomb calorimeter is kj/mol of heptane. Using this and Δ fh data, determine the molar enthalpy of formation of heptane. You must calculate the enthalpy of formation, Δ f Hº, of heptane, C 7 H 6 (l). The balanced equation for the reaction is given. The enthalpy of reaction, Δ r Hº, is known and the standard enthalpies of formation are in the Chemistry Data Booklet. Write the balanced equation for the combustion of heptane, C 7 H 6 (l), to CO 2 (g) and H 2 O(l). Look up the standard enthalpies of formation, Δ f H o, for CO 2 (g) and H 2 O(l). Use these values and the given enthalpy of reaction, Δ r Hº, in the expression Δ c H o = Σ(nΔ f Hº products) Σ(nΔ f Hº reactants). Solve for Δ f H C 7 H 6 (l). C 7 H 6 (l) + O 2 (g) 7CO 2 (g) + 8H 2 O(l) Δ c H o = kj/mol Δ c H o = Σ(nΔ f Hº products) Σ(nΔ f Hº reactants) Δ c H o = [(7 mol)(δ f Hº CO 2 (g)) + (8 mol)(δ f Hº H 2 O(l)] [( mol) (Δ f Hº C 7 H 6 (l)) + (( mol) Δ f Hº O 2 (g))] kj = [(7 mol) ( kj)] + (8 mol) ( kj/mol)] [( mol) (Δ f Hº C 7 H 6 (l)) + ( mol) (0)] kj = [( kj.mol) + ( kj)] [( mol) (Δ f Hº C 7 H 6 (l))] ( mol) (Δ f Hº C 7 H 6 (l)) = kj kj Δ r Hº = kj/mol Page 0 of 6

11 This answer is reasonable, has the correct unit (kj/mol) and the correct number of digits after the decimal (). 3. Using the data for the molar enthalpy of combustion of butane from Table 9. (page 347), determine the efficiency of a lighter as a heating device if 0.70 g of butane is required to raise the temperature of a 250 g stainless steel spoon (c = J/g C) by 45 C. You must determine the efficiency of a butane lighter by comparing the heat released by the butane as it burns with the heat gained by a stainless steel pot. Energy Input mc 4 H 0 (g) = 0.70 g Δ c H o = kj/mol (from Table 9.) Energy output m spoon = 250 g c spoon = J/g C Δt = 45.0 o C Determine the molar mass, M, for C 4 H 0 (g) and calculate the number of moles of nc 4 H 0 (g) = M m. Calculate the energy input. Energy input = Q = nδ c H o Calculate the energy output. Energy output = Q = mcδt Calculate the efficiency. Energyoutput Efficiency = 00% Energyinput MC 4 H 0 (g) = 58.4 g/mol nc 4 H 0 (g) = M m = 0.70g = mol 58.4g/mol Page of 6

12 Energy input = mol C 4 H 0 (g) ( kj/mol) = 3.99 kj Energy output = mcδt = 250 g J/g o C 45.0 o C = J = kj Energyoutput kj Efficiency = 00% = 00% = 8% Energyinput 3.99 kj The efficiency is low, as expected when heating a spoon in this manner. The answer has the correct unit (%) and the correct number of significant digits (2). 4. A camping fuel has a posted energy content of 50 kj/g. Determine its efficiency if a 2.50 g piece was required to raise the temperature of 500 g of soup (c = 3.77 J/g C) in a 50 g aluminium pot by 45 C. You must determine the efficiency of a fuel by comparing the heat released by the fuel as it burns with the heat gained by an aluminium pot and the soup in the pot. Energy Input m fuel = 2.50 g energy content of fuel = 50 kj/g Energy output m soup = 500 g c soup = 3.77 J/g o C m Al pot = 20 g c Al pot = J/g C Δt = 45.0 o C Calculate the heat output of the fuel using the energy content and the mass of fuel. Energy input = Q = m(energy content) Calculate the energy output. Energy output = Q soup + Q aluminium = (mcδt) soup + (mcδt) aluminium Calculate the efficiency. Energyoutput Efficiency = 00% Energyinput Page 2 of 6

13 Energy input = 2.50 g 50 kj/g = 25 kj Energy output = (mcδt) soup + (mcδt) aluminium = 500 g 3.77 J/g o C 45 o C + 50 g J/g o C 45 o C = J J = J = 86.8 kj Energyoutput 86.8kJ Efficiency = 00% = 00% = 69% Energyinput 25 kj Considering heat loss to the surroundings, this is a reasonable answer. The answer has the correct unit (%) and the correct number of significant digits (2). 5. What mass of pentane, C 5 H 2 (g), would have to be burned in an open system to heat 250 g of hot chocolate (c = 3.59 J/g C) from 20.0 ºC to 39.8 ºC if the energy conversion is 45% efficient? You must calculate the mass of pentane, C 5 H 2 (g), needed to heat a cup of hot chocolate. m hot chocolate = 250 g c hot chocolate = 3.59 J/g o C t i = 20.0 o C t f = 39.8 o C Δt = t f t i = 39.8 o C 20.0 o C = 9.8 o C Δ c H o C 5 H 2 (g) = kj/mol (from Table 9.) Efficiency = 45 % MC 5 H 2 (g) = 72.7 g/mol Calculate energy output. Energy output = Heat absorbed = Q hot chocolate = (mcδt) hot chocolate Use the expression for efficiency to calculate energy input. Energyoutput Efficiency = 00% Energyinput Page 3 of 6

14 Use the expression for energy input to calculate the number of moles of C 5 H 2 (g). n = Q ΔcH o mc 5 H 2 (g) = n M Energy output = (mcδt) hot chocolate = 250 g 3.59 J/g o C 9.8 o C = 7770 J = 7.77 kj Energyoutput 7.77 kj Energy input = 00% = 00% = kj Efficiency 45% nc 5 H 2 (g) = Q ΔcH o = kj kj = mol mc 5 H 2 (g) = n M = mol 72.7 g/mol = 0.88 g The answer is reasonable for this efficiency. The answer has the correct unit (g) and number of significant digits (2). 6. Determine the efficiency of a heating device that burns methanol, CH 3 OH(l), given the following information: Data for Determining the Efficiency of a Methanol-Burning Heater Quantity being measured Data Initial mass of burner g Final mass of burner g Mass of aluminium can g Mass of aluminium can and water g Initial temperature of water 0.45 C Final temperature of water C c H(CH 3 OH) kj 726. mol You must determine the efficiency of a methanol burner by comparing the heat released by the fuel as it burns with the heat gained by an aluminium can and the water in the can. Page 4 of 6

15 Energy Input initial mass of methanol burner = g final mass of methanol burner = g mch 3 OH(l) = g g =.45 g Δ c H o CH 3 OH(l) = kj/mol (from Chemistry Data Booklet) MCH 3 OH(l) = g/mol Energy output mh 2 O(l) = [m(al can + water)] [m(al can)] = g g = 79.9 g ch 2 O (l) = 4.9 J/g o C m Al can = g c Al can = J/g C t i = 0.45 o C t f = o C Δt = t f t i = o C 0.45 o C = 2.9 o C nch 3 OH(l) = M m Calculate the energy input. Energy Input = Q = nδ c H o Calculate the energy output. Energy output = Heat absorbed = Q water + Q aluminium = (mcδt) water + (mcδt) aluminium Calculate the efficiency. Energyoutput Efficiency = 00% Energyinput m.45g nch 3 OH(l) = = M g mol = mol Energy input = mol kj/mol = kj Energy output = (mcδt) water + (mcδt) alumium = 79.9 g 4.9 J/g o C 2.9 o C g J/g o C 2.9 o C = J J = 27.6 J = 2.7 kj Page 5 of 6

16 Energyoutput 2.7kJ Efficiency = 00% = 00% = 38.7% Energyinput kj Considering heat loss to the surroundings, this is a reasonable answer. The answer has the correct unit (%) and the correct number of significant digits (3). Page 6 of 6