Chemical Equilibrium Chapter 13

Not all reactions go to completion!

Chemical Equilibrium The state where the concentrations of all reactants and products remain constant with time. On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.

Equilibrium & Kinetics At equilibrium, Rf = Rr but kf is not necessarily equal to kr. kf[Reactant] = kr[Product] If equilibrium lies far to the right, [Products] will be large and kr will be small while[Reactants] will be small and kf will be large.

Reactions That Appear to Run to Completion 1. Formation of a precipitate. 2. Formation of a gas. 3. Formation of a molecular substance such as water. These reactions appear to run to completion, but actually the equilibrium lies very far to the right (favors products). All reactions in closed vessels reach equilibrium.

Molecular representation of the reaction 2NO2(g) ----> N2O4(g). c) & d) represent equilibrium.

Change in time and rates As the concentrations of reactants decrease, the forward reaction slows down and the reverse reaction speeds up UNTIL the system reaches equilibrium

The equilibrium position of a rxn- left, right, or in the middle depends on many factors: Initial Concentrations Relative energies of the reactants and products Organization of the reactants and products We will focus on #2&3 more in Chapter 16

Concentration profile for the Haber Process which begins with only H2(g) & N2(g).

Haber Process When gaseous nitrogen, hydrogen, and ammonia are mixed in a closed vessel at 25 C, concentrations don’t noticeably change over time Why? System is at equilbrium Forward and reverse rxn rates are SLOW N2 has a very strong triple bond, and unreactive H2 is a relatively strong single bond Under appropriate conditions, the system does reach equilibrium H2 disappears 3x as fast as N2 NH3 forms 2x as fast as N2 disappears

Shifting Gaseous Equilibrium If the size of a container is changed, the concentration of the gases change. A smaller container shifts the equilibrium to the right -- N2(g) + 3H2(g) ---> 2NH3(g). Four gaseous molecules produce two gaseous molecules. A larger container shifts to the left -- two gaseous molecules produce four gaseous molecules.

The Law of Mass Action For jA + kB  lC + mD The law of mass action (Cato Guldberg & Peter Waage) is represented by the equilibrium expression: Equilibrium Constant

Equilibrium Expression 4NH3(g) + 7O2(g)  4NO2(g) + 6H2O(g)

Notes on Equilibrium Expressions (EE) The Equilibrium Expression for a reaction is the reciprocal of that for the reaction written in reverse. When the equation for a reaction is multiplied by n, EEnew = (EEoriginal)n The units for K depend on the reaction being considered.

Equilibrium Expressions N2(g) + 3H2(g) ---> 2NH3(g) 2NH3(g) ---> N2(g) + 3H2(g) K´ = 1/K

Equilibrium Expressions 1/2N2(g) + 3/2H2(g) ---> NH3(g) K´ ´ = K1/2

Example: Calculate the values of K at 127°C for the reaction shown below if the [NH3]=3.1 x10-2 M, [N2]=8.5 x 10-1, [H2]=3.1 x 10-3 N2 + 3H2  2NH3 Work out for solution K=3.8 x 104 (NO UNITS)

Example: Calculate the value of the equilibrium constant for the reverse reation Work out for solution K=2.6 x 10-5(NO UNITS)

Example: Calculate the values of K at 127°C for the reaction shown below if the [NH3]=3.1 x10-2 M, [N2]=8.5 x 10-1, [H2]=3.1 x 10-3 (1/2)N2 + (3/2)H2  1NH3 Work out for solution K=1.9 x 102(NO UNITS)

What does K value mean? K > 1 Products Favored (equilibrium lies to right) K < 1 Reactants Favored (equilibrium lies to left) K = 1 System at equilibrium

K Values For the reverse reaction, Kreverse=1 /Koriginal For when the coefficients are multipled by a factor n Knew=(Koriginal)n K is written without units!

Equilibrium Position For a given reaction at a given temperature, there is only one equilibrium constant (K), (regardless of concentrations) but there are an infinite number of equilibrium positions. Where the equilibrium position lies is determined by the initial concentrations of the reactants and products. The initial concentrations do not affect the equilibrium constant.

Three equilibrium positions but only one equilibrium constant (K).

Kp K in terms of equilibrium partial pressures: Pressure and concentration are directly proportional when the temperature is held constant.

K versus Kp jA + kB  lC + mD For Kp = K(RT)n R= 0.08206 L·atm/mol·K n = sum of coefficients of gaseous products minus sum of coefficients of gaseous reactants. n = (l+m) – (j+k)

K versus Kp When n = 0 and Kp = K(RT)o, then Kp = K K and Kp are equal numerically but do not have the same units. Under what situations would n = 0 ?

Example N2(g) + 3H2(g) 2NH3(g) Using the value of Kp (3.9 × 104) from the previous example, calculate the value of K at 35°C.

Homogeneous Equilibria Homogeneous equilibria – involve the same phase: N2(g) + 3H2(g) 2NH3(g) HCN(aq) H+(aq) + CN-(aq)

Heterogeneous Equilibria . . . are equilibria that involve more than one phase. CaCO3(s)  CaO(s) + CO2(g) K = [CO2] The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present. DON’T INCLUDE SOLIDS OR LIQUIDS IN EQUILIBRIUM EXPRESSION!

Why? Concentrations of pure liquids and solids cannot change!

The position of the equilibrium CaCO3(s) ---> CaO(s) + CO2(g) does not depend upon the amounts of solid CaCO3 or CaO.

Magnitude of K A K value much larger than 1 means that the equilibrium system contains mostly products -- equilibrium lies far to the right. A very small K value means the system contains mostly reactants -- equilibrium lies far to the left. The size of K and the time required to reach equilibrium are not directly related!

A physical analogy illustrating the difference between thermodynamic and chemical stability. The magnitude of K depends on E, but reaction rate depends on Ea.

Reaction Quotient, Q . . . helps to determine the direction of the move toward equilibrium. The law of mass action is applied with initial concentrations.

Reaction Quotient, Q (continued) H2(g) + F2(g)  2HF(g)

Q versus K Q = K System is at equilibrium. Q > K System will shift to the left. Q < K System will shift to the right. See Sample Exercise 13.7 on page 625.

Reaction Quotient, Q Q = K; The system is at equilibrium. No shift will occur. Q > K; The system shifts to the left. Consuming products and forming reactants, until equilibrium is achieved. Q < K; The system shifts to the right. Consuming reactants and forming products, to attain equilibrium.

Solving Equilibrium Problems 1. Balance the equation. 2. Write the equilibrium expression. 3. List the initial concentrations. 4. Calculate Q and determine the shift to equilibrium.

Solving Equilibrium Problems (continued) 5. Define equilibrium concentrations. 6. Substitute equilibrium concentrations into equilibrium expression and solve. 7. Check calculated concentrations by calculating K.

To solve these equations…. We set up something called RICE or ICE tables R=reaction I=initial concentrations C=changed concentration E=Equilibrium

Example Consider the reaction represented by the equation: Fe3+(aq) + SCN-(aq) FeSCN2+(aq) Trial #1: 6.00 M Fe3+(aq) and 10.0 M SCN-(aq) are mixed at a certain temperature and at equilibrium the concentration of FeSCN2+(aq) is 4.00 M. What is the value for the equilibrium constant for this reaction? Note: Use the red box animation to assist in explaining how to solve the problem.

Set up RICE Table Fe3+(aq) + SCN–(aq) FeSCN2+(aq) Initial 6.00 10.00 0.00 Change – 4.00 – 4.00 +4.00 Equilibrium 2.00 6.00 4.00 K = 0.333 The value for K is 0.333.

Example #2 Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At a temperature of 700 K, the equilibrium constant is 5.10. Calculate the equilibrium concentrations of all species if 1.00 mol of each component is mixed in a 1.00 L flask. CO(g) + H2O (g)  CO2 (g) + H2(g)

Concentration of everything is 1 M Q=[CO2]o[H2]o [CO]o[H2O]o =[1][1] Step 1: Find Q. Concentration of everything is 1 M Q=[CO2]o[H2]o [CO]o[H2O]o =[1][1] [1][1] = 1 Step 2: Compare this to K. Q < K, so equilibrium shifts right If equilibrium shifts right, then [reactant] ↓, [product] ↑

Reaction: CO(g) + H2O(g)CO2(g) + H2(g) Initial 1.0 1.0 1.0 1.0 Step 3: Set up RICE TABLE Reaction: CO(g) + H2O(g)CO2(g) + H2(g) Initial 1.0 1.0 1.0 1.0 Change – x – x +x +x Equilib. 1.0-x 1.0-x 1.0+x 1.0+x

Step 4: Plug into K expression K= 5.10= [CO2][H2] = [1.0 +x][1.0+x] [CO][H2O] [1.0 -x][1.0-x] 5.10= (1.0 + x)2 (1.0 – x)2 √5.10= (1.0 + x) / (1.0 –x) 2.26=(1.0 + x) / (1.0 –x) Rearrange eqn: 0.387 M= x

Step 5: Plug #s in At equilibrium: [CO]=1-x=1-0.387=.0613 M [H2O]= 1-x=1-0.387=.0613 M [CO2]=1+x=1+0.387=1.387 M [H2]= 1+x=1+0.387=1.387 M

Step 6: CHECK Plug into original K expression to see if you get correct K value Original was 5.10 K=[1.387][1.387] =5.12 [0.387][0.387]

Example #3 Assume for the following reaction, K=1.15x102. In a particular experiment, 3.00 mol of each component was added to a 1.5 L flask. Calculate the equilibrium concentrations of all species. H2(g) + F2(g) <---> 2HF(g) [H2]o = [F2]o = [HF]o = 3.000 mol/1.500 L = 2.000M 1.000 Q < K  equilibrium shifts right.

H2(g) + F2(g) <---> 2HF(g) RICE H2(g) + F2(g) <---> 2HF(g) [H2] [F2] [HF] Initial (mol/L) 2.000 2.000 2.000 Change (mol/L) - x - x + 2x Equil. (mol/L) 2.000 - x 2.000 - x 2.000 + 2x

RICE Continued x = 1.528 [H2] = [F2] = 2.000 M - 1.528 M = 0.472 M Substitute into K expression to check validity!

RICE & the Quadratic Formula See Sample Exercise 13.12 on pages 633-634 Use the quadratic formula to solve for K.

Assume that gases hydrogen iodide is synthesized from hydrogen gas and iodine vapor at a temperature where the equilibrium constant is 1.0 x 102. Suppose HI at 5.000 x 10-1 atm, H2 is 1.0 x 10-2 atm, and I2 at 5.000 x 10-3 atm are mixed in a 5.000 L flask. Calculate the equilibrium pressures of all species. H2(g) + I2(g)  2HI(g)

Step 1: Find Q. Use initial pressures Q=[PHI]o2 [PH2]o[PI2]o =(5.000 x 10-1 atm)2 (1.000 x 10-2 atm) (5.000 x 10-3 atm) = 5.000 x 103 Step 2: Compare this to K. Q > K, so equilibrium shifts left If equilibrium shifts left, then [reactant] ↑, [product] ↓

Reaction: H2(g) + I2(g) 2HI(g) Initial 0.01 .005 .5 Step 3: Set up RICE TABLE Reaction: H2(g) + I2(g) 2HI(g) Initial 0.01 .005 .5 Change + x + x -2x___ Equilib. .01+x .005 + x .5 - 2x

Step 4: Plug into K expression Kp= 100= [PHI]2 = (.5 -2x)(.5-2x) [PH2][PI2] (.01+x)(.005 + x) 100= 4x2-2x+0.25 x2 + .0105x + 0.00005 100x2 +1.5x + 0.005= 4x2-2x+0.25 96x2 +3.5x - 0.245 =0 Use quadratic formula: x=0.0355 atm

Step 5: Plug #s in At equilibrium: PH2= 0.01 + x = 0.01 + 0.0355 = 0.0455 atm PI2= 0.005 + x = 0.0405 atm PHI= 0.5 – 2x= 0.5 – 2(.0355) = .429 atm

Step 6: CHECK Plug into original K expression to see if you get correct K value Original was 100 Kp= [.429]2 =99.99 [.0455][0.0405]

Systems With Small K’s See example on pages 635-636. When K is small, the change (x) is small compared to the initial concentration and a simplification can be made for the calculation. This approximation must be checked to see if it is valid. When K is two powers of ten or more smaller than the initial concentration, the approximation should be OK.

Le Châtelier’s Principle . . . if a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change.

Le Chatelier’s Principle If a reactant or product is added to a system at equilibrium, the system will shift away from the added component. If a reactant or product is removed, the system will shift toward the removed component.

As4O6(s) + 6C(s)  As4(g) + 6CO(g) Example: For the reaction: As4O6(s) + 6C(s)  As4(g) + 6CO(g) How will the following affect the direction of the equilibrium: Adding CO: Adding or Removing C and As4O6: Removing As4:

As4O6(s) + 6C(s)  As4(g) + 6CO(g) Example: For the reaction: As4O6(s) + 6C(s)  As4(g) + 6CO(g) How will the following affect the direction of the equilibrium: Adding CO: SHIFTS LEFT Adding or Removing C and As4O6: NO SHIFT, pure solids and liquids have no effect on equilibrium Removing As4: SHIFTS RIGHT

Effects of Changes on the System 1. Concentration: The system will shift away from the added component. 2. Temperature: K will change depending upon the temperature (treat heat as a reactant or product). If rxn is endothermic: adding heat will shift to right If Rxn is exothermic, adding heat will shift to left

Effects of Changes on the System (continued) 3. Pressure: a. Addition of inert gas does not affect the equilibrium position. b. Decreasing the volume shifts the equilibrium toward the side with fewer gaseous moles. c. INCREASING the volume shifts the equilibrium toward the side with more gaseous moles

The system of N2, H2, and NH3 are initially at equilibrium. When the volume is decreased, the system shifts to the right -- toward fewer molecules.

Example Predict the shift in equilibrium position that will occur for each of the following when the volume is reduced: P4(s) + 6Cl2(g)  4PCl3 (l) PCl3(g) + Cl2 (g)  PCl5 (g) PCl3(g) + 3NH3(g)  P(NH2)3 (g) + 3HCl (g)

Example Predict the shift in equilibrium position that will occur for each of the following when the volume is reduced: P4(s) + 6Cl2(g)  4PCl3 (l) Shifts right, toward less gaseous moles (6 vs 0) PCl3(g) + Cl2 (g)  PCl5 (g) Shifts right, less gaseous moles (2 vs 1) PCl3(g) + 3NH3(g)  P(NH2)3 (g) + 3HCl (g) No change, same # of gaseous moles (4 vs 4)

Equilibrium Constant, K For an exothermic reaction, if the temperature increases, K decreases. For an endothermic reaction, if the temperature increases, K increases.

Effect of Various Changes on Equilibrium Disturbance Net Direction of Rxn Effect of Value of K Concentration Increase (reactant) Towards formation of product None Decrease(reactant) Towards formation of reactant Increase (product) Decrease (product)

Effect of Pressure on Equilb. Increase P (decrease V) Towards formation of fewer moles of gas None Decrease P (Increase V) Towards formation of more moles of gas ( Add inert gas, no change in V) None, concentrations unchanged

Effect of Temperature on Equilb Increase T Towards absorption of heat Increases if endothermic Decreases if exothermic Decrease T Towards release of heat Increases if exothermic Decreases if endothermic Catalyst Added None, forward and reverse equilibrium attained sooner None