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 ...