Chemical Equilibrium Examples, Study Guides, Projects, Research of Chemistry

Download Chemical Equilibrium Examples and more Study Guides, Projects, Research Chemistry in PDF only on Docsity! 1 WRITE OUT THE EQUILIBRIUM CONSTANT 1 Chemical Equilibrium Examples These are the examples to be used along with the powerpoint lecture slides. The problems are numbered to match the tags in the the lower left hand corner of the powerpoint slides. The numbers of the examples are # the in the CEQ EX# tags on the slides. 1 Write out the equilibrium constant (This is homework problem 12-2 Gold or 26-2 Red in McQuarrie and Simon.) Write out and compare the equilibrium constant for the following two reactions (which are the same with a different choice of the balanced stoichiometric coefficients), 2 SO2(g) + O2(g) 2 SO3(g) SO2(g) + 1 2 O2(g) SO3(g) For the first reaction we have, KP (T ) = P 2 SO3 PO2P 2 SO2 and for the second reaction, KP (T ) = PSO3 P 1/2 O2 PSO2 This means that the equilibrium constant for reaction two is the square root of the equilib- rium constant for reaction one,√ KP (T )[rxn 1] = KP (T )[rxn 2] Note that both are the same chemical reaction, only the stoichiometry has been written differently. This should make clear that a particular equilibrium constant only has meaning when referenced to a specific balanced chemical equation. Again note that K is unitless. From Zumdalh’s freshman chemistry text, “. . .equilibrium constants are given without units” [and this is beyond the scope of this freshman text] “. . .involves corrections for nonideal behavior.” So now you know why! 2 From KP to Kc (This is example 12-3 Gold or 26-3 Red in McQuarrie and Simon) Given the value of KP (T ) for the following reaction calculate the corresponding value of 3 GIBBS VS EXTENT TO FIND K 2 Kc(T ). NH3(g) 3 2 H2(g) + 1 2 N2(g) KP = 1.36× 10−3 at 298.15 K The general expression is, KP = Kc ( c◦RT P ◦ )νH2 +νN2 −νNH3 When converting from KP (T ) to Kc(T ) we must know what standard states we are using. In this case we will use one mole · L−1 for c◦ and one bar for P ◦. The choice of standard states, and the fact that KP (T ) and Kc(T ) are unitless (why is this... remember to divide by standard states) determines the units for R, in this case R = 0.083145 L · bar ·K−1 ·mol−1. The stoichiometric coefficients are, νNH3 = 1 νH2 = 3 2 νN2 = 1 2 leaving us with, KP = Kc ( c◦RT P ◦ )1 and the conversion factor at 298.15 K is,( c◦RT P ◦ ) = (1 mol · L−1) ( 0.083145 L · bar ·K−1 ·mol−1 ) (298.15 K) 1 bar = 24.79 And finally we have, Kc = KP 24.79 = 4.59× 10−5 3 Gibbs vs extent to find K (Largely taken from pages 486-487 Gold or 1058-1059 Red in McQuarrie and Simon) We are considering the reaction, N2O4(g) 2 NO2(g) • First use the extent of the reaction at the minimum of G(ξ) to calculate KP . Starting with the expression shown on the slide for G(ξ) (no need to write this on the board), G(ξ) = (1− ξ)∆fG ◦ N2O4 + 2ξ∆fG ◦ NO2 + (1− ξ)RT ln 1− ξ 1 + ξ + 2ξRT ln 2ξ 1 + ξ 5 CHANGE IN K WITH T 5 5 Change in K with T (This is example 12-6 Gold or 26-6 Red in McQuarrie and Simon) For the reaction, PCl3(g) + Cl2(g) PCl5(g) Estimate KP at 700 K given, • KP = 0.0408 at 500 K • ∆rH ◦ has an average value of -69.8 kJ · mol−1 over the range of 500 K ≥ T ≥ 700 K Assume that ∆rH ◦ is independent of T over this range and use, ln KP (T2) KP (T1) = −∆rH ◦ R ( 1 T2 − 1 T1 ) Plug in the numbers, ln KP (700 K) 0.0408 = − −69.8 kJ ·mol−1 8.314 J ·K−1 ·mol−1 ( 1 700 K − 1 500 K ) = −4.80 KP (700 K) = (0.0408)e−4.80 = 3.36× 10−4 *So is this reaction endothermic or exothermic? Since KP is smaller at a higher temperature, then the reactants must be higher in energy and the reaction is exothermic. 6 Extra Problems 6.1 Example of ξeq in terms of KP and P (This is problem 12-7 Gold or 26-7 Red in MS ) Consider the reaction, 2NOCl(g) 2NO(g) + Cl2(g) and you start with n0 moles of NOCl(g) and zero moles of the two products. 1. Derive and expression for KP in terms of ξeq and P . Write the partial pressures in terms of ξeq and P , Total number of moles: nNOCl + nNO + nCl2 = (n0 − 2ξeq) + (2ξeq) + (ξeq) = n0 + ξeq 6 EXTRA PROBLEMS 6 Partial pressure, Pi = xiP , PNOCl = n0 − 2ξeq n0 + ξeq P PNO = 2ξeq n0 + ξeq P PCl2 = ξeq n0 + ξeq P Now write out the equilibrium constant, KP = PCl2P 2 NO P 2 NOCl = 4ξ3 eqP (n0 + ξeq)(n0 − 2ξeq)2 And you can express this as a function of the initial amount of reactant, in terms of ξeq/n0 by multiplying through by n0/n0, KP = 4(ξeq/n0) 3P (1 + (ξeq/n0))(1− 2(ξeq/n0))2 2. Plug in KP = 2.0×10−4 for 2 different pressures (remember the KP is NOT dependent on total pressure P ) and see if this follows Le Chatelier’s principle, For P = 0.080 bar ξeq/n0 = 0.078 For P = 0.160 bar ξeq/n0 = 0.063 So we can see that as the total pressure goes up equilibrium moves back to the left, as expected from Le Chatelier. Sketch a qualitative plot of P vs ξeq P (e xt en t) e q / n 0 6 EXTRA PROBLEMS 7 6.2 Consider the effect of the standard state (This is problem 12-8 Gold or 26-8 Red MS.) Consider the reaction, COCl2(g) CO(g) + Cl2(g) given that KP = 34.8 at 1000◦C IF THE STANDARD STATE IS TAKEN TO BE 1 BAR. But what if you had selected 0.5 bar as the standard state? KP (1 bar) = (PCO/1 bar)(PCl2/1 bar) (PCOCl2/1 bar) = 34.8 KP (0.5 bar) = (PCO/0.5 bar)(PCl2/0.5 bar) (PCOCl2/0.5 bar) = 2KP (1 bar) = 69.6 Demonstrating the value of the equilibrium constant only has meaning if you know the standard state that was selected. Notice that this also means that if you use KP to get ∆rG ◦ then the Gibb energy is also for the standard state used for KP , ∆rG ◦ = −RT ln KP