Chemical Equilibrium Part II: Working with K Jespersen Chap. 15 Sec 2 thru 5 Dr. C. Yau Spring 2013

Mass Action Expression, Q and K H2(g) + I2(g) 2HI (g) The fraction is called the mass action expression (MAE). When we plug in actual molarity values into the expression, the numerical value is called Q. When the system has reached an equilibrium, the values we plug into expression are called “equilibrium concentrations” and Q becomes K.

Mass Action Expression, Q and K H2(g) + I2(g) 2HI (g) e.g. Suppose at 440oC, we place into a container, 0.022M H2, 0.035M I2 and 0.28M HI. When we plug these numbers into the mass action expression, we get the value 101. This is the mass action expression. This value is Q. At 440oC, it is known that Kc = 49.5. Q > Kc This tells us that the system will shift to the left (conc of HI will decrease and conc of H2 and I2 will increase) until Q becomes Kc.

Why are there no units to Q and K? In order to correct for the interaction of the species, we should be using an activity value, instead of molarity. This is done by dividing each term by the standard state of the substance. For solutions the standard state is 1 M. e.g. Instead of using [H2] = 0.022M, we should use 0.022M/1M = 0.022 (unit-less). As far as you are concerned, just use molar concentration. K will be close enough without that adjustment.

We are next going to look at what happens to the K expression when we “do things” to the chemical equation. For example, if we reverse the equation, what happens to K? If we double the equation, what happens to K?

Changing Direction of Equilibrium A + B C C A + B Note that When a reaction is reversed, K becomes its reciprocal.

Multiplying Coefficients by a Factor A + B C 2A + 2B 2C Kc" = Kc2

Adding Chemical Equilibria A + B C C + D B + F A + D F Note that Kc3 = Kc1 x Kc2 When eqns are added together, the resultant K is the product of the K’s involved.

Manipulating Equilibrium Laws: A Summary When we reverse an equation, the value of K for the new equation is its reciprocal (1/K) When we multiply an equation by a constant, the value of the equilibrium constant for the new equation is raised to the exponent of the multiplier n • Rxn → Kn When we add reactions, we multiply their K values for the net reaction. ChemFAQ: How do changes in the way a chemical equation is written change mass action expressions and equilibrium constants?

For the reaction N2(g) + 3H2(g) 2NH3(g), Kc=500. for a particular temperature. What would be Kc for the following: 2NH3(g) N2(g) + 3H2(g) N2(g) +  H2(g) NH3(g) 0.00200 22.3 Do Pract Exer 3 & 4 p.703

Kp for Gaseous Rxns When reactants and products are all gases, we can write either Kc or Kp. For example: N2 (g) + 3H2 (g) 2NH3 (g) where p stands for partial pressure. Do Pract. Exer. 15.5 and 15.6 on p.704

Remember PV=nRT. When rearranged to solve for P, we get Note: n/V is the molar concentration Does this remind you of anything? This equation tells us that at a given temperature T, P is proportional to M. (Partial pressure of a gas is proportional to its molar concentration.)

Using Partial Pressures in Kp Solving P=MRT for M gives us... A + B C where all are gases GENERAL EQUATION: Kp = Kc(RT)Δn where Δn = change in #mol gas

Kc versus Kp Kp = Kc(RT)Δn Δ n = n(products) – n(reactants) for gases only Example 15.3 p.705 At 500oC, N2(g) + 3H2(g) 2NH3(g) has Kc = 6.0x10-2. What is the numerical value of Kp for this reaction? R = 0.08206 atm.L.mol-1K-1 Δn = -2 Kp = ? Note which R value we are using! (NOT 8.314 J.mol-1K-1) Kp=1.5x10-5 Do Pract Exer 7 & 8 p.706

Heterogeneous Equilibria These are systems that contain more than one phase (physical state). 2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g) Note that 1 mol of solid occupies a certain space. 2 mol would occupy twice as much space. This means (mol: volume) ratio is constant. Concentration of a pure solid cannot change. The same goes for a pure liquid. Thus, MAE for heterogeneous systems does not include terms for solids and liquid.

Consider 1 mol of Solid NaHCO3. X 2 = 2 moles would occupy twice the volume. X 2 = Molarity of solids is constant regardless of sample size. Similarly, molarity of liquids is constant also.

Heterogeneous Equilibria 2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g) These are constants.

Example 15.4 p.708 The air pollutant sulfur dioxide can be removed from a gas mixture by passing the gases over calcium oxide. CaO(s) + SO2(g) CaSO3(s) Write the equilibrium law for the reaction. Do Pract Exer 9 & 10 p.708