# Lecture #19 (a) Advanced Calculations using the Equilibrium Lecture #19 (a) Advanced Calculations using the Equilibrium Constant; (b) Le Chateliers Principle; (c) Equilibria of Real Gases Chemistry 142 B Autumn Quarter 2004 J. B. Callis, Instructor Solving Equilibrium Problems Write the balanced equation for the reaction.

Write the equilibrium expression. List the initial concentrations. Calculate Q and determine the direction of shift to equilibrium. Define the change needed to reach equilibrium and define the equilibrium concentrations. Substitute the equilibrium concentrations into the equilibrium expression and solve for the unknown. Check the solution by calculating K and making sure it is identical to the original K. Problem 19-1 Graphical Solution to the Quadratic Equation Let us begin by considering the equilibrium between NO2, a red-brown gaseous pollutant formed by automobiles, and its dimer, N2O4. This equilibrium can be expressed by the chemical equation:

2 NO2(g) = N2O4(g) The equilibrium constant for this reaction is where PN2O4 and PNO2 are the equilibrium partial pressures of the two gases. KP has the numerical value of 8.8 at T = 25 oC when the partial pressures are expressed in atmospheres. Problem 19-1 Graphical Solution cont. Let us now consider the problem of finding the equilibrium partial pressures of NO2 and N2O4, given the value of KP, and the initial pressures of NO2 and N2O4 ( (PNO2)0 and (PN2O4)0 ). Then if x atm is the additional amount of N2O4 formed by the equilibrium, the partial pressure of NO2 must decrease by 2x atm. Inserting the equilibrium partial pressures into the equilibrium expression results in the equation: K which can be rationalized to yield :

Finding the Solution to an Equilibrium Problem Using a Graphical Method =4*K*x^2-(4*K*PNO2o+1)*x+K*PNO2o^2-PN2O4o x= Graphical Solution to the Equilibrium Problem and x= 35.00 30.00

The later is chosen because 25.00 20.00 f(x) x, atm f(x) 4.0 31.20 4.1 24.41 4.2 18.33 4.3 12.95

4.4 8.27 4.5 4.30 4.6 1.03 4.7 -1.53 4.8 -3.39 4.9 -4.55 5.0 -5.00 5.1 -4.75 5.2

-3.79 5.3 -2.13 5.4 0.23 5.5 3.30 5.6 7.07 5.7 11.55 5.8 16.73 5.9 22.61 6.0 29.20

From Graph: 15.00 10.00 5.00 0.00 -5.00 4.0 4.5 5.0 5.5

-10.00 x, atm 6.0 6.5 it gives all positive concentrations. (PNO2)eq = 0.726 atm (PN2O4)eq = 4.637 atm

Problem 19-2: Using Exact Solution of Quadratic Equation The dissociation of isopropyl alcohol to acetone and hydorgen gas is written as follows : CH 3 2 CHOH ( g ) CH 3 2 CO ( g ) H 2 ( g ) (K 0.444 at 179 K) For 10 g of isopropyl alcohol in 10 L initially, calculate the partial pressure of all the species at equilibrium at 179 K. Problem 19-2: Solution(a) 0 Initial partial pressure of the isopropanol PIso = 0.617 atm, 0

0 Initial partial pressure of the products PAc PH2 0.00 atm Q is not calculated - it is obvious that the reaction proceeds to the right Construct the reaction table : Pressure Isopropyl alcohol, IPA Acetone, Ac CH 3 2 CHOH(g) CH 3 2 CO(g) (atm) Init. Change Equil. Hydrogen

H2(g) Problem 19-2: Continued (b) Substituting the equilibrium concentrations from the table into the equilibrium expression : K To solve for x, we perform the indicated multiplication, and rearrange in descending powers of x on the left side with zero on the right : 0 This expression is a quadratic of the general form where the (two) roots can be obtained from the quadratic formula : Problem 19-2: Continued (c) From the quadratic, we obtain two candidate solutions x and x

Since only the positive root leads to all positive concentrations, we ignore the negative root : x atm Calculating equilibrium concentrations : PAc PH 2 x atm PIPA Check : PAC PH 2 K PIPA atm 0.444

Problem 19-3: Problems Involving Higher Order Polynomials Consider the following reaction : CH 4 ( g ) H 2 O CO ( g ) 3 H 2 ( g ) (K P 1.8 10 7 at 600 K) Gaseous CH 4 , H 2 O and CO are introduced into an evacuated chamber at 600 K, and their initial partial pressures (before reaction) are 1 .40 atm, 2.30 atm and 1.60 atm, respectively. Determine the partial pressure of H 2 g that will result at equilibrium. Problem 19-3: Solution(a) Q is not calculated - it is obvious that the reaction proceeds to the right Construct the reaction table : Pressure CH4(g) H2O(g) (atm)

Init. Change Equil. CO(g) 3 H2(g) Problem 19-3: Continued (b) Substituti ng the equilibrium concentrations from the table into the equilibrium expression : KP 1.8 10 7 To solve for x, we perform the indicated multiplication, and rearrange in descending powers of

y on the left side with zero on the right : This expression is a fourth order polynomial. Problem 3- Continued (c) Fortunately since K is small, the reaction does not shift that much to the right, i.e. we imagine that y is small. Thus we use the approximate equation 1.8 10 7 Solving for y yields y3 Taking the cube root of both sides gives : y This value is indeed small compared with the starting pressures. At equilibrium, then PH2 atm

Le Chateliers Principle If a change in conditions (a stress) is imposed on a system at equilibrium, the equilibrium position will shift in a direction that tends to reduce that change in conditions. Henri Le Chatelier, 1884 The Effect of a Change in Concentration If a gaseous reactant or product is added to a system at equilibrium, the system will shift in a direction to to reduce the concentration of the added component. If a gaseous reactant or product is removed from a system at equilibrium, the system will shift in a direction to to increase the concentration of the removed component.

Problem 19-4: The Effect of a Change in Concentration Consider the following reaction: 2 H2S(g) + O2(g) = 2 S(s) + 2 H2O(g) What happens to: (a) [H2O] if O2 is added? Ans: (b) [H2S] if O2 is added? Ans: (c) [O2] if H2S is removed? Ans: (d) [H2S] if S is added? Ans: Effect of a Change in Pressure There are three ways to change the pressure of

a reaction system involving gaseous components at a given temperature: 1. Add or remove a gaseous product at constant volume. 2. Add an inert gas (one not involved in the reaction) at constant volume. 3. Change the volume of the container. Problem 19-5: The Effect of a Change in Pressure How would you change the total pressure of each of the following reactions to increase the yield of the products: (a) CaCO3 (s) = CaO(s) + CO2 (g) Ans: (b) S(s) + 3 F2 (g) =

SF6 (g) Ans: (c) Cl2(g) + I2(g) = Ans: 2 ICl (g) The Effect of a Change in Temperature Changes in concentration or pressure alter the equilibrium position. In contrast, changes in temperature alter the value of the equilibrium constant. Exothermic Reactions

Releases heat upon reaction. H is negative. Addition of heat to an exothermic reaction shifts the equilibrium to the left. The value of K decreases in consequence. Endothermic Reactions Absorbs heat upon reaction. H is positive. Addition of heat to an endothermic reaction shifts the equilibrium to the right. The value of K increases in consequence. Using Le Chateliers Principle to Describe the Effect of a Temperature Change on a System in Equilibrium Treat the energy as a reactant (in an exothermic process) or as a product (in

an endothermic process). Predict the direction of the shift as if an actual reactant or product has been added or removed. Problem 19-6: The Effect of a Change in Temperature on the Position of Equilibrium How does an increase in temperature affect the equilibrium concentration of the indicated substance and K for the following reactions: (a) CaO(s) + H2O (l) = Ca(OH)2 (aq) H0 =-82 kJ Ans: (b) (a) CaCO3 (s) = + CaO(s) + CO2 (g)

Ans: (c) SO2 (g) = S(s) + O2(g) H0 = 297 kJ Ans: H0 = 178 kJ Equilibria Involving Real Gases We have thus far assumed that gas phase equilibria involve gases that behave as ideal gases. The effect of non-ideal behavior is to cause real equilibrium constants to become nonconstant under differing conditions of total pressure. The Activity Coefficient The activity of the ith gaseous component of the

equilibrium system is represented as: obs i Pi ai Pref Where i is called the activity coefficient for correcting Piobs to the ideal value. For equilibrium pressures of 1 atm or less, the value of Kp calculated from the observed pressures is expected to be within about 1% of the true value Answers to Problems in Lecture 19 1. From graph: x = 5.391 and x = 4.637. The later is chosen because it gives all positive concentrations. 2. PAc PH 2 0.347 atm PIPA 0.270 atm

3. PH 2 7.1 10 3 atm 4. (a) The reaction proceeds to the right so H2O increases. (b) Some H2S reacts with the added O2 to move the reaction to the right, so [H2S] decreases. (c) The reaction proceeds to the left to re-form H2S, more O2 is formed as well, O2 increases. (d) S is a solid, so its concentration does not change. Thus, [H2S] is unchanged. 5. (a) The only gas is the product CO2. To move the reaction to the right decrease the pressure. (b)With 3 moles of gas on the left and only one on the right, we increase the pressure to form more SF6. (c) The number of moles of gas is the same on both sides of the equation, so a change in pressure will have no effect.

6. (a) Add heat to the right side. Adding heat shifts the system to the left. [Ca(OH)2] and K will decrease. (b) Add heat to the left side. Adding heat shifts the system to the right. [CO2] and K will increase. (c) Add heat to the left side. Adding heat shifts the system to the right. [SO2] will decrease and K will increase.