Value for K Large K 1 equilibrium will

Value for K • Large K (>1) = equilibrium will contain mostly products (equilibrium position is to the right) • Small K (<1) = equilibrium will contain mostly reactants (equilibrium position is to the left)

More… • K is NOT related to time (rate/activation energy) • When either the [reactants/products] = 0, it is easy to determine which way the reaction will shift • If not (some of each is present), Reaction Quotient (Q) must be determined

Reaction Quotient (Q) • Use the Law of Mass Action with initial concentrations instead of equilibrium concentrations. • N 2(g) + 3 H 2(g) <-> 2 NH 3(g) Q = [NH 3]02 [N 2]0[H 2]03 • Compare Q and K to determine which way the reaction will shift…

Q vs K • Q = K: System is at equilibrium - no shift • Q > K: System shifts to the left, consuming products and forming reactants • Q < K: System shifts to the right, consuming reactants and forming products

Example H 2(g) + I 2(g) <-> 2 HI(g), K = 7. 1 X 102 at 25°C • Predict the direction that the system will shift in order to reach equilibrium given each of the following initial conditions: – Q = 427 – Q = 1522 – [H 2]0 = 0. 81 M, [I 2]0 = 0. 44 M, [HI]0 = 0. 58 M – [H 2]0 = 0. 078 M, [I 2]0 = 0. 033 M, [HI]0 =1. 35 M – [H 2]0 = 0. 034 M, [I 2]0 = 0. 035 M, [HI]0 =1. 50 M

Calculating Equilibrium Pressures and [ ] N 2 O 4(g) <-> 2 NO 2(g) Kp = 0. 133, pressure of N 2 O 4 = 2. 71 atm, determine equilibrium pressure of NO 2(g)

ICE Tables • In order to determine concentrations at equilibrium, ICE tables can be used • I = INITIAL • C = CHANGE • E = EQUILIBRIUM

Example #1 • At a certain temperature a 1. 00 L flask initially contained 0. 298 mol PCl 3(g) and 8. 70 X 10 -3 mol PCl 5(g). After the system had reached equilibrium, 2. 00 X 10 -3 mol Cl 2(g) was found in the flask. Gaseous PCl 5 decomposes according to the reaction: PCl 5(g) <-> PCl 3(g) + Cl 2(g) • Calculate the equilibrium concentrations of all species and the value of K.

Example #2 • Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700 K the equilibrium constant is 5. 10. Calculate the equilibrium concentrations of all species if 1. 000 mol of each component is mixed in a 1. 000 L flask.

Example #3 • Assume that the reaction for the formation of gaseous hydrogen fluoride from hydrogen and fluorine has an equilibrium constant of 1. 15 X 102 at a certain temperature. In a particular experiment, 3. 000 mol of each component was added to a 1. 500 L flask. Calculate the equilibrium concentrations of all species.

Steps… 1. Write the balanced equation 2. Write the equilibrium expression using law of mass action 3. List initial concentrations 4. Calculate Q and determine the shift 5. Define change for equilibrium, apply the change to the initial concentrations 6. Substitute equilibrium concentrations into the equilibrium expression and solve for unknown 7. Check that calculated values make correct K

Problem-OH NO! • We are not always given easy numbers to work with and solve for • Sometimes the QUADRATIC EQUATION is necessary!! • When ax 2 + bx + c = 0 X = -b ± √(b 2 - 4 ac) 2 a You will get two values for x. Only ONE will work to give a positive value for concentration once subtracted from the original values

Example • For synthesis of HF from its elements, 3. 000 mol H 2 and 6. 000 mol F 2 are mixed in a 3. 000 L flask. Assume that the equilibrium constant for the synthesis reaction at this temperature is 1. 15 X 102. Calculate the equilibrium concentration of each component.

Example with Pressures • Assume that gaseous hydrogen iodide is synthesized from its elemental vapors at a temperature where the equilibrium constant is 1. 00 X 102. Suppose HI at 5. 000 X 10 -1 atm, H 2 at 1. 000 X 10 -2 atm, and I 2 at 5. 000 X 10 -3 atm are mixed in a 5. 000 L flask. Calculate the equilibrium pressures of all species.

Example with Small Equilibrium Constant • Usually these situations have you solving an x 3 equation • Gaseous NOCl decomposes to form gases NO and Cl 2. At 35°C the equilibrium constant is 1. 6 X 10 -5. In an experiment in which 1. 0 mol NOCl is placed in a 2. 0 L flask, what are the equilibrium concentrations? 2 NOCl(g) <-> 2 NO(g) + Cl 2(g) • For this assumption to work, the difference between the initial and equilibrium must be <5%