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### Ch6 part 4 NOTES corr

Course: CHEM 245, Fall 2009
School: Virgin Islands
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Word Count: 739

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Chapter 6 Part 4 LECTURE NOTES ' 6.10 The variation of KP with temperature In paragraph 6.3 we showed that /T(G/T)P = -H/T2. This result can be used to calculate KP at a different temperature than standard temperature (=298.15 K). If we integrate left and right over dT of the above equation we get: Remark: There is a mistake in Example Problem 6.10 on page 133: the number 210.6 in the second line should read...

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Chapter 6 Part 4 LECTURE NOTES ' 6.10 The variation of KP with temperature In paragraph 6.3 we showed that /T(G/T)P = -H/T2. This result can be used to calculate KP at a different temperature than standard temperature (=298.15 K). If we integrate left and right over dT of the above equation we get: Remark: There is a mistake in Example Problem 6.10 on page 133: the number 210.6 in the second line should read 211.4. ' 6.11 Equilibria involving ideal gases and solid and liquid phases In this ' we will show how to deal with equilibria if there are ideal gases in the presence of a solid or solids that do not mix. This means that the solids are present in this equilibrium as pure compounds Let's look at dG = -SdT + VdP For ideal gases this gave: For liquids and solids, knowing that they are both hardly compressible, we can treat Vm as a constant, so we get: Chem 245 Fall 2008 Lecture Notes / Copyright F.C.J.M. van Veggel 1 The term VP is small compared to the term RTlnPeq/Po, so we often neglect it. Let's look at the following reaction with two solids that do not mix: ' 6.12 Expressing the equilibrium constant in terms of <a href="/keyword/mole-fraction/" >mole fraction</a> or molarity We go back to eq. 6.44 (remember that this is only slightly rewritten and derived for ideal gases): With Pi = xiP (P is the total pressure, not Po), this can be rewritten as: We know have the chemical potential i in a mixture in terms of the chemical potential of pure compound I at T and P and a term with the <a href="/keyword/mole-fraction/" >mole fraction</a> : We will first deal with ideal gases only. The general reaction is: We will derive Kx similarly to KP by summation of eq. 1 over the reaction equation: Chem 245 Fall 2008 Lecture Notes / Copyright F.C.J.M. van Veggel 2 And Kx is similarly to KP, with products in the numerator and reactants in the denonimator: Chem 245 Fall 2008 Lecture Notes / Copyright F.C.J.M. van Veggel 3 The same can be done with concentrations, i.e. with molarities, which gives: ' 6.13 The dependence of eq on T and P In summary we have: We also have: Applying eq (5) to eq (3) gives: Remark: KP is not a function of the pressure, because of the superscript in Ho which means at Po (= 1 bar). Chem 245 Fall 2008 Lecture Notes / Copyright F.C.J.M. van Veggel 4 Applying eq (5) to eq (4) gives: Remark: Kx is a function of the pressure because Hreaction is a function of the pressure. The question thus arises why is Kx = f(P) while KP f(P). The answer lies in the relation we derived on page 4: Problem 6.12b (page 146 of the book). The given reaction is: FeO(s) + CO(g) = Fe(s) + CO2(g) in (a) we calculated Goreaction (600 oC) = 765 J/mole. in (b) they ask us to calculate xCO2 at 600 oC. The equilibrium constant KP (600 oC) = 0.900. With the following relation we get our answer in a straightforward way: Chem 245 Fall 2008 Lecture Notes / Copyright F.C.J.M. van Veggel 5 ' 6.14 A case study: the synthesis of ammonia The ammonia is a classic case that illustrates nicely our theory. First some standard calculations: This gives KP = 778, so the equilibrium, if achieved, lies on the product side (= NH3) which is what we want. However, at room temperature the reaction rate is negligible. The reason is that the activation barrier, is too high to overcome. The solution to this is to raise the temperature. What does this do to KP? With Horeaction &lt; 0, an increase in the temperature will decrease the KP, which is not desired. In ...

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Virgin Islands - CHEM - 245
Chapter 6 Part 5 LECTURE NOTES ' 6.14 A case study: the synthesis of ammonia CONTINUED Eq. 1 Eq. 2 The following data are given: N2(g) + 1H2(g) = NH3(g) N2(g) + 3H2(g) = 2NH3(g)1st a check on the data with Go = Ho - TSo. For eq. 1 we get for So = 1
Virgin Islands - CHEM - 245
Chapter 8 Part 1 LECTURE NOTES ' 8.1 What determines the relative stability of the solid, liquid, and gas phases? From our experience we know that most (pure) substances exist in three phases: solid, liquid, and gas. We also know that solids normally
Virgin Islands - CHEM - 245
Chapter 8 Part 2 LECTURE NOTES ' 8.6 The vapor pressure of a pure substance depends on the applied pressure If we have a pure compound in a cylinder with piston, we can have condition where we have both the liquid and the vapor (the gas) in equilibri
Virgin Islands - CHEM - 245
Chapter 9 Part 1 LECTURE NOTES ' 9.1 Defining the ideal solution Look at the following experiment in which we start with a mixture of pure benzene vapour and liquid. After the addition of some toluene, we measure the composition of the vapour and liq
Virgin Islands - CHEM - 245
Chapter 9 Part 2 LECTURE NOTES ' 9.5 The Gibbs-Duhem equation The Gibb-Duhem equation, derived below, shows that the component i's in a mixture are not independent. And because of this, the chemical potential of a non-volatile solute, such as sugars,
Virgin Islands - CHEM - 245
Chapter 9 Part 3 LECTURE NOTES ' 9.9 Real solutions exhibit deviations from Raoult's law Many solutions do not follow Raoult's law, i.e. Pi = xiPi* with xi the mole fraction in the liquid mixture and the * means the vapour pressure of the pure liquid
Virgin Islands - CHEM - 245
Virgin Islands - CHEM - 245
One derivation of the following five will be asked on the final. Equation numbers refer to the book. Equations: 6.78 (page 135) 6.37 (page 121) 6.27 (page 118) 3.19 (page 46) 3.44 (page 53)
Virgin Islands - CHEM - 245
Virgin Islands - CHEM - 245
Chemistry 245 Introductory Physical ChemistryMidterm 1A October 2008 (7 questions in total)Name: Student number:Q1-5 each 2 pnt; Q6 3 pnt; Q7 7 pnt Conversions: 1 bar = 105 Pa (1 Pa = 1 N/m2) 1 atm = 1.013 bar Equations and laws: 0th law of therm
Virgin Islands - CHEM - 245
Chemistry 245 Introductory Physical ChemistryMidterm 1B October 2008 (7 questions in total)Name: Student number:Q1-5 each 2 pnt; Q6 3 pnt; Q7 7 pnt Conversions: 1 bar = 105 Pa (1 Pa = 1 N/m2) 1 atm = 1.013 bar Equations and laws: 0th law of therm
Virgin Islands - CHEM - 245
Chemistry 245 Introductory Physical ChemistryMidterm 2A Friday November 7th 2008 (5 questions in total)Please do not remove this first page Name . Student number .Formulae and constants R = 8.314 J / mole K A wexpansion + wnon-expansion A = U T
Virgin Islands - CHEM - 245
Virgin Islands - CHEM - 245
Chemistry 245 Introductory Physical ChemistryQUIZ 3A Wednesday November 7th 2007 (5 questions in total)Name . Student number .Formulae and constants A wexpansion + wnon-expansion A = U TS dU = TdS PdV dA = SdT PdV G wnon-expansion G = H TS
Virgin Islands - CHEM - 245
Virgin Islands - CHEM - 245
Virgin Islands - CHEM - 245
Virgin Islands - CHEM - 245
Virgin Islands - CHEM - 245
Virgin Islands - CHEM - 245
Virgin Islands - CHEM - 245
Virgin Islands - CHEM - 245
Virgin Islands - CHEM - 245
Virgin Islands - CHEM - 245
Virgin Islands - CHEM - 245
Virgin Islands - CHEM - 245
Virgin Islands - CHEM - 245
Virgin Islands - CHEM - 245
Chapter 2 part 1 Figure 2.3Chapter 2 part 1 Figure 2.4Chapter 2 part 1 Figure 2.5State Functions Another example than the internal energy UY (Xfinal, Yfinal)(Xinitial, Yinitial) XIs the position (X,Y) a state function ?State FunctionsY
Virgin Islands - CHEM - 245
Chapter 2, part 2 State FunctionY (Xfinal, Yfinal)(Xinitial, Yinitial) XWe concluded: The position (X,Y) a state function Who knows other state functions from every day life?Reversible Work Figure 2.111A new State Function: H, the enthalpy
Virgin Islands - CHEM - 245
Chapter 3 part 1 Figure 3.1Chapter 3 part 1 Figure 3.31Chapter 3 part 1 Figure 3.42
Virgin Islands - CHEM - 245
Chapter 3 part 2Remark about the large number of equationsChapter 3 part 2 The Joule-Thomson experiment Figure 3.5Chapter 3 part 2 Figure 3.6+ 0Assignment 3Page 59 and on: P3.2 P3.4 P3.9 P3.11 P3.16 P3.25