Homework+3+Solutions - BME 221 Homework#3 Due Beginning of...

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BME 221 Homework #3 Due: Beginning of class on October 23 rd 1. The enthalpy of combustion of benzoic acid (C 6 H 5 COOH) is commonly used as the standard for calibrating calorimeters; its value has been accurately determined to be -3226.7 kJ mol -1 . The following experiments are done in a constant-volume calorimeter. (a) When 0.9862g of benzoic acid was oxidized, the temperature rose from 21.84˚C to 25.67˚C. What is the heat capacity of the calorimeter? (b) In a separate experiment, 0.4654 g of α-D-glucose (C 6 H 12 O 6 ) was oxidized in the same calorimeter, and the temperature rose from 21.22˚ C to 22.28˚C. Calculate the enthalpy of combustion of glucose, the value of Δ r U for the combustion, and the molar enthalpy of formation of glucose. (a) Balanced Equation: C 6 H 5 COOH (s)+ 7.5 O 2 (g) 7CO 2 (g)+3H 2 O (l) ΔH = ΔU + Δ(PV) = ΔU + RTΔn (change in moles of gas) -3226700 J mol -1 = ΔU + (8.314 J mol -1 K -1 )(294.99 K)(7-7.5) ΔU= -3225.5 kJ mol -1 = q v 0.9862 g of benzoic acid x x -3225.5 kJ mol -1 = -26.07 kJ (Note that -qv lost by reaction = +qv gained by calorimeter) @ Constant volume C v ΔT = q v (C v )(298.82-294.99)=(26070 J) C v =6808 J K -1 . (b) Balanced Equation: C 6 H 12 O 6 (s)+ 6O 2 (g) 6CO 2 (g)+ 6H 2 O (l), thus change in moles of gas Δn = 0, so ΔH=ΔU=q v Enthalpy of combustion = q v = C v ΔT q v = (6808J K -1 )(1.06K)=7216.5 J 0.4654g glucose x = 0.00259 moles enthalpy combustion = = - 2791.1 kJ mol -1 . (Once again, heat gained by calorimeter = heat lost by reaction—sign change needed) Since ΔH=ΔU=q v , Δ r U = - 2791.1 kJ mol -1 To calculate molar enthalpy of formation: Δ r H combustion = (6Δ f H CO2 (g) +6Δ f H H2O (l))-(6Δ f H O2 + 1Δ f H glucose ) plugging in known values -2791.1 kJ mol -1 = (6(-393 kJ/mol) + 6(-285.8 kJ/mol)) – ( 0 + Δ f Glucose)) Δ f Glucose = -1284.7 kJ mol -1 . (Actual value at STP is -1274.6 kJ/mol). 2. The dissolution of large amounts of most salts in water results in a warming of the solution. There are, however, a few kinds of salts, which when dissolved in water cool down the solution. For example, instant cold packs use ammonium nitrate, which dissolves in water and makes the cold packs cold. (a) What are the signs of the values for ∆G, ∆H, and ∆S for the dissolution of ammonium nitrate in water?
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ΔG = - ΔH = + ΔS = + (b) What is the driving force for this dissolving process? Explain. Though the ΔH is positive, which is unfavorable, the process is spontaneous because of the large, positive ΔS. Entropy increases because the solid salt dissolves into ions in the water, increasing disorder of the solution. 3.
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Homework+3+Solutions - BME 221 Homework#3 Due Beginning of...

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