1. 14.4-14.7 Equilibrium Constant in terms of pressure, Heterogeneous Equilibria, and the Reaction Quotient

2. Equilibrium Constants: K c vs. K p So far, we have expressed equilibrium constants in terms of concentration, but we can also express equilibrium in terms of partial pressure for gases. We will now designate K c as the equilibrium constant with respect to concentrations in molarity. We will designate K p as the equilibrium constant with respect to partial pressures in atmospheres. – K p has the same form as K c except we use the partial pressures of gases instead of concentrations.

3. I Brief Walkthrough on How to Write K c and K p Expressions Since we already know how to write the equilibrium constant for concentration…write the K c expression for the following reaction: 2SO 3 (g) 2SO 2 (g) + O 2 (g) [SO 2 ] 2 [O 2 ] K c = ---------------- [SO 3 ] 2 Now this is how we write the K p expression for the same reaction (you must check for gases). (P SO 2 ) 2 P O 2 K p = --------------- (P SO 3 ) 2

4. The Relationship Between K c and K p Although the value for K c and K p are often not the same, we can derive a relationship between the two, so long as the gases are behaving ideally. K p = K c (RT) Δn where Δn = c + d – (a+b) If the total number of moles are the same before and after the reaction, and Δn = 0, then K c =K p

5. Let’s Try a Practice Problem! Consider the following reaction and corresponding value of K c : H 2 (g) + I 2 (g) 2HI(g) K c = 6.2X10 2 at 25 o C What is the value of K p at this temperature? K p = K c (RT) Δn K p = 6.2X10 2 (0.08206 L atm / mol K (298K)) 0 K p = 6.2X10 2

6. Let’s Try Another!!! Under which circumstances are Kp and Kc equal for the reaction aA(g) + bB(g) cC (g) + dD(g)? (a)If a + b = c + d (b)If the reverse reaction is reversible (c)If the equilibrium constant is small (a) If a + b = c + d, then the quantity Δn is zero so that K p = K c (RT) 0. Since (RT) 0 is equal to 1, K p =K c

7. Heterogeneous Equilibria: Reactions Involving Solids and Liquids Pure solids (substances in a chemical reaction with (s)), and pure liquids (substances written with an (l)) are not included in equilibrium expressions, because as you increase the amount of solid or liquid, you are not changing the concentration of it. Reactants and products that are gases and aqueous solutions are the only ones that are added to equilibrium expressions. Consider the following reaction: CO 2 (g) + H 2 O(l) H + (aq) + HCO 3 - (aq) Based on the information listed above, write the K c expression for this reaction. [H + ][HCO 3 - ] K c = ---------------- [CO 2 ]

8. Let’s Try Another! Write an equilibrium expression (K c ) for the equation: 4HCl(g) + O 2 (g) 2H 2 O(l) + 2Cl 2 (g) [Cl 2 ] 2 K c = ---------------- [HCl] 4 [O 2 ]

9. Let’s Try Another!!! For which reaction does K p = K c ? (a)2Na 2 O 2 (s) + 2CO 2 (g) 2Na 2 CO 3 (s) + O 2 (g) (b)Fe 2 O 3 (s) + 3CO(g) 2Fe(s) + 3CO 2 (g) (c)NH 4 NO 3 (s) N 2 O(g) + 2H 2 O(g) (b) Fe 2 O 3 (s) + 3CO(g) 2Fe(s) + 3CO 2 (g), since solids are not part of equilibrium expressions and there are three moles of gas on both sides of the reaction 3-3= 0

10. Calculating Equilibrium Constants from Measured Equilibrium Concentrations If we were given the concentrations of both the reactants and products of a reaction when that reaction reaches equilibrium, we could plug those concentrations directly into the K c expression, and use the balanced chemical equation to solve for K c But, if we weren’t given this information, and instead, only given the initial concentrations of the reactant(s) and the equilibrium concentration of one reactant or product, then we could deduce the other equilibrium concentrations from the stoichiometry of the reaction. To do this, we must set up an ICE table (I (initial), C (change), E(equilibrium)).

11. Here is an Example When An ICE Table Would Be Used Let’s consider the following reaction: A(g) 2B(g) The initial concentration of A is 1.00 M, the initial concentration of B is 0.00 M, and the equilibrium concentration of A is 0.75 M. We can set up an ICE table. Since we know that A’s concentration changes by -0.25, we can deduce that B changed by (+0.25 X 2) or +0.50. Let’s fill in the ICE table below: Now, since we have found equilibrium concentrations, we can plug them into the equilibrium expression and solve for K c [B] 2 [0.50] 2 K c = -------- = ---------- = 0.33 [A] [0.75] [A][B] Initial1.00 M0.00 M Change-0.25+0.50 Equilibrium0.75 M0.50 M

12. Let’s Try the Following Practice Problem! Consider the following reaction: CO(g) + 2H 2 (g) CH 3 OH(g) A reaction mixture at 780 o C initially contains [CO] = 0.500 M and [H 2 ] = 1.00 M. At equilibrium, the CO concentration is found to be 0.15 M. What is the value of the equilibrium constant? [CH 3 OH] [0.35] K c = ----------------= -------------------- = 26 [CO][H 2 ] 2 [0.15][0.30] 2 [CO][H 2 ][CH 3 OH] Initial0.500 M1.00 M0.00 M Change-0.35(-0.35x2)+0.35 Equilibrium0.15 M0.30 M0.35 M

13. The Reaction Quotient: Predicting the Direction of Change To gauge the progress of a chemical reaction relative to equilibrium (if the reversible reaction has produced both products, and reactants but is not at equilibrium yet), we use a quantity called the reaction quotient (Q). The reaction quotient takes the same form as the equilibrium constant, except the reaction does no need to be at equilibrium. Q c is the reaction quotient based on concentrations, and Q p is the reaction quotient based on pressure. The reaction quotient is useful because the value of Q relative to K is a measure of the progress of the reaction toward equilibrium. – At equilibrium, the reaction quotient is equal to the equilibrium constant.

14. Summarizing Direction of Change Predictions If Q < K Reaction goes to the right (Toward products. If Q > K Reaction goes to the left (Toward reactants) If Q = K Reaction is at equilibrium.

15. Let’s Try A Practice Problem! Consider the reaction and its equilibrium constant: N 2 O 4 (g) 2NO 2 (g) Kc = 5.85X10 -3 (at some temperature) A reaction mixture contains [NO 2 ] = 0.0255 M and [N 2 O 4 ] = 0.0331 M. Calculate Q c and determine the direction in which the reaction proceeds. [NO 2 ] 2 [0.0255] 2 Q c = ----------------= -------------------- = 1.96X10 -2 [N 2 O 4 ] [0.0331] Q c > K c so the reaction proceeds to the left (towards the reactants).

16. Let’s Try Another!!! For the reaction N 2 O 4 (g) 2NO 2 (g), a reaction mixture at a certain temperature initially contains both N 2 O 4 and NO 2 in their standard states (meaning they are gases with a pressure of 1 atm). If K p = 0.15, which statement is true of the reaction mixture before any reaction occurs? (a)Q = K; The reaction is at equilibrium. (b)Q < K; The reaction will proceed to the right. (c)Q > K; The reaction will proceed to the left. (c) Q > K; The reaction will proceed to the left, since Q p = 1, and 1 > 0.15.

17. 14.4-14.7 #’s 32 (a), 36, 38, 40, and 48 Read 14.8 pgs. 667-676 Quiz Chapter 14 on Fri. 02/05 (Exam 13-14 Tues. 02/09)