Only in choice (a)does rxnf.Δ= ΔHHooIn choice (b), C(diamond) is notthe most stable form of elemental carbon under standard conditions; C(graphite) is the most stable form.6.93w=−PΔV=−(1.0 atm)(0.0196 −0.0180)L =−1.6 ×10−3L⋅atm Using the conversion factor 1 L⋅atm =101.3 J: 3101.3 J= ( 1.610L atm)=L atm−−×⋅×⋅0.16 J−w0.16 J of work are done by water as it expands on freezing. 6.94 (a)No work is done by a gas expanding in a vacuum, because the pressure exerted on the gas is zero. (b)w=−PΔVw=−(0.20 atm)(0.50 −0.050)L =−0.090 L⋅atm Converting to units of joules: 101.3 J0.090 L atmL atm= −⋅×=⋅9.1 J−w(c)The gas will expand until the pressure is the same as the applied pressure of 0.20 atm. We can calculate its final volume using the ideal gas equation. L atm(0.020 mol) 0.0821(27320)Kmol K0.20 atm⋅⎛⎞+⎜⎟⋅⎝⎠===2.4 LnRTPVThe amount of work done is: w=−PΔV=(0.20 atm)(2.4 −0.050)L =−0.47 L⋅atm Converting to units of joules: 101.3 J0.47 L atmL atm= −⋅×=⋅48 J−w6.95The equation corresponding to the standard enthalpy of formation of diamond is: C(graphite) ⎯⎯→C(diamond) Adding the equations: C(graphite) +O2(g) ⎯⎯→CO2(g) ΔH°=−393.5 kJ/mol CO2(g) ⎯⎯→C(diamond) +O2(g) ΔH°=395.4 kJ/mol C(graphite) ⎯⎯→C(diamond) ΔH°=1.9 kJ/molSince the reverse reaction of changing diamond to graphite is exothermic, need you worry about any diamonds that you might have changing to graphite?
CHAPTER 6: THERMOCHEMISTRY 1796.96 (a)The more closely packed, the greater the mass of food. Heat capacity depends on both the mass and specific heat. C=msThe heat capacity of the food is greater than the heat capacity of air; hence, the cold in the freezer will be retained longer. (b) Tea and coffee are mostly water; whereas, soup might contain vegetables and meat. Water has a higher heat capacity than the other ingredients in soup; therefore, coffee and tea retain heat longer than soup.6.97The balanced equation is: C6H12O6(s) ⎯⎯→2C2H5OH(l) +2CO2(g) rxnf25f2f6126[2(C H OH)2(CO )](C HO )Δ=Δ+Δ− ΔHHHHoooorxnoΔH=[(2)(−276.98 kJ/mol) +(2)(−393.5 kJ/mol)] −(1)(−1274.5 kJ/mol) =−66.5 kJ/mol6.98 4Fe(s) +3O2(g) →2Fe2O3(s). This equation represents twice the standard enthalpy of formation of Fe2O3. From Appendix 3, the standard enthalpy of formation of Fe2O3=−822.2 kJ/mol. So, ΔH°for the given reaction is: rxn(2)( 822.2 kJ/mol)1644 kJ/molΔ=−= −HoLooking at the balanced equation, this is the amount of heat released when four moles of Fe react. But, we are reacting 250 g of Fe, not 4 moles. We can convert from grams of Fe to moles of Fe, then use ΔH°as a conversion factor to convert to kJ. 1 mol Fe1644 kJ250 g Fe55.85 g Fe4 mol Fe−××=31.8410 kJ−×6.99One conversion factor needed to solve this problem is the molar mass of water. The other conversion factor is given in the problem. It takes 44.0 kJ of energy to vaporize 1 mole of water.
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