Sample Problem
Sample Problem Find ΔH for H2O (s) 0 C → H2O (g) 100 C sublimation: H Happens all ll around: d snow, ice i in i f fridge, id d dry iice This is clearly not a constant T process. Since H is a state function, define any path to get to product. Make it simple to use. Calculate ΔH for that path. Here is one: H2O ((g)) H2O (s) fusion +40 6 kJ/mol +40.6 kJ/ l + 6 kJ/mol H2O () 0 ºC vaporization Hvap Hfusion Cp,l =33.6 J/mol K heating H2O () 100 ºC 100C H H fusion C p , dT H vap 0C CpT for H 2 O ConcepTest p #0 Find ∆H for the reaction: CH4(g) + H2O(l) H2CO(g) + 2H2(g) using the following standard heats of formation: CH4(g) H2CO(g) H2O(g) -74 74.8 8 kJ/mol -108.6 kJ/mol -241.8 kJ/mol a. a b. c. d. Less than 208.0 208 0 kJ/mol Equal to 208.0 kJ/mol G Greater t th than 208.0 208 0 kJ/mol kJ/ l Not possible to tell ConcepTest #0 Solution Find ∆H for the reaction: CH4(g) + H2O(l) H2CO(g) + 2H2(g) using us ng the follow following ng standard heats of formation: format on: CH4(g) H2CO(g) H2O(g) O( ) -74.8 kJ/mol -108.6 kJ/mol -241.8 241 8 kJ/mol We need this ΔH = [ΔHf(H2CO(g))+ ΔHf(2H2 (g)) ] –[ΔH [ΔHf(CH4(g)) + ΔHf(H2O (l)) ] Ho(H2O(l) ) = Ho(H2O(g) ) + Hfus(H2O(g) ) = -241.8 -241 8 + F (a negative number) ΔH =[-108.6 +2x0] –[-74.8 +(-241.8 +F)] = 208 –F Since F is negative, ΔH must be > 208 kJ/mol Thermochemistry Revisited Consider some chemical reaction is taking place at constant Pex, and the temperature changes from Ti to Tf as the reaction proceeds. We can write Hrxn Hrxn Ti C p p products T where Hrxn is the enthalpy change for the stated conditions, and ΔT = (Tf – Ti). STANDARD ENTHALPY CHANGE R f s to Refers t enthalpy th l change h under d set s t of f standard st d d conditions. diti s The Th standard st d d enthalpy change ΔHrxn for a process A → B is the change in enthalpy HB – HA when A and B are in their standard states. The components of A and B are also p from each other. separated That’s double talk! What is a standard state?? STANDARD STATE “The standard state of a substance at a specified temperature is its pure form at that temperature. temperature ” Moreover, for the very special case that T=298.14 K and P = 1 atm, we call Hrxn the standard reaction enthalpy and designate this condition with a superscript o. H process 298 Ho process at P = 1atm (or sometimes or even ) However, the standard enthalpy change of a Rxn can be defined at any temperature Rx Enthalpy as f(T) Consider the reaction 2A + 3B C + 2D at two temperatures T1 and T2. What is the difference in the standard enthalpies of formation at these temperatures? For constant Cp, r H2 r H1 r CP dT r CP T2 T1 o o T2 o o T1 r CP o CP,m products o CP,m o reactants The are the coefficients in the reaction above. m is molar. Ammonia Heat of Formation Thus, suppose we want to determine the heat of formation of NH3 at 400 K. ½ N2 + 3/2 H2 NH3 Hf 400 Hf CP T ? 400 298 kJ 46.11 mol How to obtain ΔCP? Cp(NH3) = 35.06 J/mol K Cp(N2) = 29.1 J/mol K Cp(H2) = 28.8 J/mol K all from tabulated values 1 J 3 J So CP 35.06 29.1 28.8 22.7 2 mol K 2 mol K and Hf 400 46.11 kJ mol 22.7 22 7 kJ kJ 400 298 48.43 1000 mol mo l get J to kJ This approach assumes that the CP are all constant over this temperature range. This assumption is certainly valid for ideal gases; it is certainly false if there is a phase change, and it is not exact for real gases! Propene Hydrogenation/Combustion What’s propene? H2C=CH-CH3 Does it matter where the double bond goes? What’s the hydrogenation Rxn? H H C H C C H + H2 H H H H C C C H H H H H H We know (from table) that ΔH for this reaction is -124 kJ/mol at standard conditions. All components are gases. Now we use Hess’ Law to find the standard heat of reaction for the combustion of propene, 9 C3 H 6 g O 2 g 3CO 2 g 3H 2 O 2 We need to find a series of reactions that have known ΔH and add up p to this one. Propene Hydrogenation/Combustion C3H6 9 O2 3CO2 3H2O 2 C3H6 H2 g C3H8 C3H8 5O2 3CO2 4H2O -124 -2220 3(-393.5)+ 3(-393 5)+ 4 285.8 285 8 103 103.8 8 1 O g 286 -Hf H2O 2 2 _____________________________________________ 9 kJ C3H6 O2 3CO2 3H2O -2058 2 mol H2O H2 g P Pressure ConcepTest p #1 1 4 3 2 An id A ideall gass goess through the four step process shown. In which p stages is work done? Vl Volume a. In 1 and 2 only d. In none of the stages b. In 1 and 3 only e. In all of the stages c. In 2 and 4 only ConcepTest p #2 Pressurre The curve shown at left r pr s nts th represents the expansion xp nsi n of an ideal gas. If this process is isothermal then we can also say: Volume a. There is no heat involved in this process. b There is no work involved in this process. b. process c. There is a transfer of thermal energy, or heat, into the gas. d. There is a transfer of thermal energy, or heat, out of the gas. ConcepTest p #3 Two 10 kg weights sit on a piston, compressing the air underneath. One of the weights g is removed, and the air underneath expands from 18.3 to 20.0L. Then the second weight is removed and the air expands from 20.0 to 22.0L. How does the amount of work done compare if instead both weights were removed at once? Assume the same total change in volume. a. More work is done removing one weight at a time than removing both weights at once. b Less b. L work k is i done d removing i one weight i h at a time i than h in i removing i both weights at once. c. The same amount of work is done removing one weight at a time, or if both are removed at once. once Adiabatic Expansion/Compression Consider the three paths, A, B and C P1, V1, T1 U A U B U C B P A P3, V3, T3 C P2, V2, T2 V This is why we call U or H state functions. and H A H B H C but q A qB qC and wA wB wC Suppose that path A in the 2 D slice in PV above characterizes the expansion, with P1, V1 and T1 being the initial condition. Further, let path A be a reversible adiabatic path. What can we say experimentally? P1, V1, T1 → P2, V2, T2 Is V1 < V2 or is V1 > V2 ? Is T1 < T2 , is T1 =T2, or is T1 > T2 ? Is P1 < P2 or is P1 > P2 ? Now we calculate the cooling in a the reversible adiabatic expansion of an ideal gas.
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