Chemical Equilibrium Examples

Transcription

1 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), For the first reaction we have, and for the second reaction, 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) SO 2 (g) O 2(g) SO 3 (g) K (T ) = 2 SO 3 O2 2 SO 2 K (T ) = SO 3 1/2 O 2 SO2 This means that the equilibrium constant for reaction two is the square root of the equilibrium constant for reaction one, K (T )[rxn 1] = K (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 K to K c (This is example 12-3 Gold or 26-3 Red in McQuarrie and Simon) Given the value of K (T ) for the following reaction calculate the corresponding value of

2 3 GIBBS VS EXTENT TO FIND K 2 K c (T ). NH 3 (g) 3 2 H 2(g) N 2(g) K = at K The general expression is, K = K c ( c RT ) νh2 +ν N2 ν NH3 When converting from K (T ) to K c (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. The choice of standard states, and the fact that K (T ) and K c (T ) are unitless (why is this... remember to divide by standard states) determines the units for R, in this case R = L bar K 1 mol 1. The stoichiometric coefficients are, ν NH3 = 1 ν H2 = 3 2 ν N2 = 1 2 leaving us with, K = K c ( c RT and the conversion factor at K is, ( ) c RT = (1 mol L 1 ) ( L bar K 1 mol 1) ( K) 1 bar ) 1 = And finally we have, K c = K = Gibbs vs extent to find K (Largely taken from pages Gold or Red in McQuarrie and Simon) We are considering the reaction, N 2 O 4 (g) 2 NO 2 (g) First use the extent of the reaction at the minimum of G(ξ) to calculate K. Starting with the expression shown on the slide for G(ξ) (no need to write this on the board), G(ξ) = (1 ξ) f G N 2 O 4 + 2ξ f G NO 2 + (1 ξ)rt ln 1 ξ 1 + ξ + 2ξRT ln 2ξ 1 + ξ

3 3 GIBBS VS EXTENT TO FIND K 3 If you plug in the values for the heats of formation and R, the minimum (or equilibrium value) with respect to ξ can be found by setting the derivative equal to zero, dg(ξ) dξ = 0 (to find ξ min = ξ eq ) You will solve a similar problem in your homework. The result in this case is ξ eq = mol (shown with an arrow on the slide) Now we must express the equilibrium constant in terms of ξ eq, K = ( ) NO2 2 ( ) N2 O 4 Which leaves us to express the partial pressures in terms of ξ. First note the total number of moles in terms of ξ, And get the partial pressures, total moles = n NO2 + n N2 O 4 = ν NO2 ξ + (n N2 O 4 (0) ν N2 O 4 ξ) = 2ξ + (1 ξ) = 1 + ξ N2 O 4 = x N2 O 4 total = n N 2 O 4 (0) ν N2 O 4 ξ total = 1 ξ 1 + ξ 1 + ξ total NO2 = x NO2 total = ν NO 2 ξ 1 + ξ total = 2ξ 1 + ξ total Note that the units on the pressures are canceled since we take them relative to the standard state. For a total pressure of 1 bar, total = = 1 bar. lug it in, K = ( ) NO2 2 ( ) N2 = O 4 ( 2ξeq ) 2 total 1+ξ eq ( ) 1 ξeq total 1+ξ eq total = 1 bar 1 bar =1 ( ) 2 2ξeq 1+ξ eq ( ) = 1 ξeq 1+ξ eq 4ξ 2 eq 1 ξ 2 eq = 4(0.1892)2 1 (0.1892) 2 K = 0.148

4 4 THE REACTION QUOTIENT AND DIRECTION OF CHANGE 4 Now compare this to the calculation of K from r G, r G = RT ln K ln K = rg RT r G = 2 f G [NO 2 (g)] f G [N 2 O 4 (g)] = J mol 1 ( J mol 1 ) ln K = ( J K 1 mol 1 )( K) = K = The reaction quotient and direction of change (This is example 12-5 Gold or 26-5 Red in McQuarrie and Simon.) For the reaction, 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) K = 10 at 960 K Calculate r G and indicate which direction the reaction will spontaneously proceed starting from the following given condition, 2 SO 2 ( bar) + O 2 (0.20 bar) 2 SO 3 ( bar) First calculate the reaction quotient for the given conditions, (again note that the pressures are unitless since they are referenced to 1 bar) Q = 2 SO 3 O2 2 SO 2 = *redict the direction of the reaction: ( ) 2 ( ) 2 (0.20) = Since Q is smaller than K, Q must increase to proceed towards equilibrium (Q = K ). This means the reaction will move toward the products, or left to right as written. Now calculate r G, r G = RT ln Q K = (8.314 J K 1 mol 1 )(960 K) ln r G = 42.3 kj mol 1 Finding that r G < 0 indicates that that the reaction will proceed left to right as written, as we expected based on our value of Q.

5 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, Estimate K at 700 K given, K = at 500 K Cl 3 (g) + Cl 2 (g) Cl 5 (g) r H has an average value of kj mol 1 over the range of 500 K T 700 K Assume that r H is independent of T over this range and use, ln K ( (T 2 ) K (T 1 ) = rh 1 1 ) R T 2 T 1 lug in the numbers, ln K (700 K) kj mol 1 = J K 1 mol 1 ( K 1 ) = K K (700 K) = (0.0408)e 4.80 = *So is this reaction endothermic or exothermic? Since K is smaller at a higher temperature, then the reactants must be higher in energy and the reaction is exothermic. 6 Extra roblems 6.1 Example of ξ eq in terms of K and (This is problem 12-7 Gold or 26-7 Red in MS) Consider the reaction, 2NOCl(g) 2NO(g) + Cl 2 (g) and you start with n 0 moles of NOCl(g) and zero moles of the two products. 1. Derive and expression for K in terms of ξ eq and. Write the partial pressures in terms of ξ eq and, Total number of moles: n NOCl + n NO + n Cl2 = (n 0 2ξ eq ) + (2ξ eq ) + (ξ eq ) = n 0 + ξ eq

6 6 EXTRA ROBLEMS 6 artial pressure, i = x i, NOCl = n 0 2ξ eq n 0 + ξ eq NO = 2ξ eq n 0 + ξ eq Cl2 = ξ eq n 0 + ξ eq Now write out the equilibrium constant, K = Cl 2 2 NO 2 NOCl = 4ξ 3 eq (n 0 + ξ eq )(n 0 2ξ eq ) 2 And you can express this as a function of the initial amount of reactant, in terms of ξ eq /n 0 by multiplying through by n 0 /n 0, K = 4(ξ eq /n 0 ) 3 (1 + (ξ eq /n 0 ))(1 2(ξ eq /n 0 )) 2 2. lug in K = for 2 different pressures (remember the K is NOT dependent on total pressure ) and see if this follows Le Chatelier s principle, For = bar ξ eq /n 0 = For = bar ξ eq /n 0 = 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 vs ξ eq (extent)eq / n0

7 6 EXTRA ROBLEMS Consider the effect of the standard state (This is problem 12-8 Gold or 26-8 Red MS.) Consider the reaction, COCl 2 (g) CO(g) + Cl 2 (g) given that K = 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? K (1 bar) = ( CO/1 bar)( Cl2 /1 bar) ( COCl2 /1 bar) = 34.8 K (0.5 bar) = ( CO/0.5 bar)( Cl2 /0.5 bar) ( COCl2 /0.5 bar) = 2K (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 K to get r G then the Gibb energy is also for the standard state used for K, r G = RT ln K