Equilibrium Equilibrium is the state where the RATE of the forward reaction is equal to the RATE of the reverse reaction

Equilibrium Alternate Definition of Equilibrium: The state where the concentration of all reactants and all products remain constant with time

Reactions are reversible A + B  C + D ( forward) A + B  C + D ( reverse) Preferred way of writing: Initially there is only A and B so only the forward reaction is possible in beginning As C and D build up, the reverse reaction speeds up while the forward reaction slows down. Eventually the rates are equal

Reaction Rate  Time Forward Reaction Reverse reaction Equilibrium Position Reaction Rate vs Time

Equilibrium Position Whether the reaction lies far to the left or right depends on 3 factors: 1.Initial concentrations – higher concentrations means more collisions – faster reaction 2.Relative energies of reactants and products – nature goes to minimum energy 3.Degree of organization of reactants and products – nature likes maximum disorder called entropy

What is equal at Equilibrium? Rates are equal Concentrations are NOT. Rates are determined by concentrations Dynamic: constant change, activity, or progress

Start with NO 2 Start with N 2 O 4 Start with NO 2 & N 2 O 4 equilibrium

Law of Mass Action For any reaction : This is called the EQUILIBRIUM EXPRESSION K is called the Equilibrium Constant. Sometimes written as K c or K eq MEMORIZE THIS EQUILIBRIUM EXPRESSION Carnegie Hall Time #17 p614

Playing with K If we write the reaction in reverse. Then the new equilibrium constant is K’ Notice Product always in numerator

Playing with K If we multiply the equation by a constant (n) Then the equilibrium constant is

K is CONSTANT No units Temperature affects rate. Each set of equilibrium concentrations is called an Equilibrium Position There are an unlimited number of Equilibrium Positions. Pure Solids do NOT appear in expressions Pure Liquids do NOT appear in expressions H 2 O (l) is pure so leave it out of expression, NOT H 2 O (g)

Calculate K N 2 + 3H 2  2NH 3 Initial At Equilibrium [N 2 ] 0 =1.000 M [N 2 ] = 0.921M [H 2 ] 0 =1.000 M [H 2 ] = 0.763M [NH 3 ] 0 =0 M [NH 3 ] = 0.157M Answer: 6.02 x 10 -2

Calculate K N 2 + 3H 2  2NH 3 Initial At Equilibrium [N 2 ] 0 = 0 M [N 2 ] = M [H 2 ] 0 = 0 M [H 2 ] = M [NH 3 ] 0 = M [NH 3 ] = 0.203M Answer: 6.02 x 10 -2

Calculate K N 2 + 3H 2  2NH 3 Initial At Equilibrium [N 2 ] 0 = 2.00 M [N 2 ] = 2.59 M [H 2 ] 0 = 1.00 M [H 2 ] = 2.77 M [NH 3 ] 0 = 3.00 M [NH 3 ] = 1.82 M Answer: 6.02 x Expression has same value regardless of amount of gases mixed together initially CH Time #19 p614

Equilibrium and Pressure Equilibria involving gases can be described using pressures. 1.PV = nRT (solve for P) 2.P = (n/V)RT mol / vol = M 3.P = MRT or P=CRT (solve for C) 4.C = P/RT C represents the molar concentration of the gas

Equilibrium and Pressure 2SO 2 (g) + O 2 (g)  2SO 3 (g) Write the Equilibrium Expression K p is equilibrium constant in terms of partial pressures of the gases. Use ( ) instead of [ ] in expression

Calculate K p The reaction for the formation of nitrosyl chloride 2NO (g) + Cl 2(g)  2NOCl (g) was studied at 25 o C. The pressures at equilibrium were found to be; P NOCl = 1.2 atm P NO =5.0 X atm P Cl 2 =3.0 X atm Write the Equilibrium Expression Calculate the value of K p for this reaction at 25 o C Answer: 1.9 X 10 3 CH Time #25 p615

Derivations of General Equations jA + kB  lC + mD  n= (l+m)-(j+k) = Change, in moles of gas, SOOOOooo

Heterogeneous Equilibria So far every example dealt with reactants and products where all were in the same phase called Homogeneous Equilibria Equilibria with more than 1 phase are called Heterogeneous Equilibria The position of a Heterogeneous Equilibria does not depend on amounts of pure solids or liquids present.

Heterogeneous Equilibria If the reaction involves pure solids or pure liquids the concentration of the solid or the liquid doesn’t change. As long as they are not used up we can leave them out of the equilibrium expression. Don’t put pure solid or liquid in Equilibrium Expression

For Example But the concentration of I 2 does not change, So this is the correct expression: What you think expression will look like:

Solving Equilibrium Problems Write the expressions for K c and K p for the following process: Solid phosphorus pentachloride decomposes to liquid phosphorus trichloride and chlorine gas. Answer: K c =[Cl 2 ] and K p =(P Cl 2 ) Deep blue solid copper(II) sulfate pentahydrate is heated to drive off water vapor to form white solid copper(II) sulfate Answer: K p = (P H 2 O ) 5 CHT #29 p615

Application of Equilibrium Constant Initial Condition CO=? H 2 O=? H 2 =? CO 2 =? New Condition CO=? H 2 O=? H 2 =? CO 2 =? K eq = 6.25 Is the New Condition at Equilibrium? Fill in the values for Initial and New Condition

Application of Equilibrium Constant Initial Condition CO = 7 H 2 O = 7 H 2 = 0 CO 2 = Equilibrium CO = 7-x H 2 O = 7-x H 2 = +x CO 2 = +x x CO Disappear x H 2 O Disappear x H 2 Form x CO 2 Form For the system to be at Equilibrium we must have the following ratio This is what Equilibrium looks like.

The Reaction Quotient Used to determine the direction system must shift to reach equilibrium Calculated the same as the equilibrium constant, but for a system NOT at equilibrium Compare Q to K

Compare Q : K 3 POSSIBLE CASES 1.If Q<K – Not enough products – Shift to right  more product is made 2.If Q>K – Too many products – Shift to left  more reactant is formed 3.If Q=K, system is at equilibrium

Example For the synthesis of ammonia at 500 o C, the equilibrium constant is 6.0 X Predict the direction in which the system will shift to reach equilibrium in the following case: [NH 3 ] 0 =1.0X10 -3 M; [N 2 ] 0 =1.0X10 -5 M; [H 2 ] 0 =2.00X10 -3 M Steps: 1.Write the balanced equation 2.Calculate Q 3.Compare Q to K: 1.3X10 7 > 6.0x Because Q > K the concentration of product must be decreased and the system will shift to the left 5.N 2 + 3H 2  2NH 3 CHT #33

Solving Equilibrium Problems Given the starting concentrations and one equilibrium concentration. Use stoichiometry to figure out other concentrations and K. Learn to create a table of initial and final conditions. ICE is nICE – coming soon to your paper!

Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700 K the equilibrium constant is Calculate the equilibrium concentration of all species if mol of each component is mixed in a 1.000L flask. Calculating Equilibrium Concentrations Calculate Initial Concentrations [CO] 0 =[H 2 O] 0 =[CO 2 ] 0 =[H 2 ] 0 = 1.000mol/1.000L = 1.000M

Calculate Q Compare Q to K: Because Q < K the system is not at equilibrium yet and needs to shift to the right. “But how much?” We define all the reactants and products in terms of x (change from initial conc.) Example This is nICE: Set up a table to enter information I nitial Concentration (mol/L) C hange (mol/L) E quilibrium Concentration (mol/L) [CO] 0 - x [CO] - x

I nitial Concentration C hange E quilibrium Concentration [CO] 0 = X x [H 2 O] 0 = X1.000 – x [CO 2 ] 0 = x x [H 2 ] 0 = x x Sign of X determined by the shift direction. Plug into the equilibrium expression. Solve for x = 0.387mol/L (It’s algebra folks) Plug the value for x back into the original equation and calculate the concentrations at equilibrium. [CO] = [H 2 O] = 0.613M and [CO 2 ] = [H 2 ] = 1.387M

ICE is n ICE Putting all the concentration values into a table is a convenient way of organizing the data. It’s called an ICE table. Be COOL - Use it on all problems! Balanced Equation A+B  2AB I nitial C hange-x +2x E quilibrium2.000-x x Carnegie Time # 46 p.616

Steps in Solving Equilibrium Problems 1.Write the balanced equation 2.Write the equilibrium expression 3.List the initial concentrations ( Ice Table) 4.Calculate Q & determine direction 5.Define change iCe ( ICe Table) 6.Sub. equilibrium conc. into equilibrium expression and solve icE ( IcE Table) 7.Check your answers Study Sample Exercise 13.11

What if you’re not given equilibrium concentration? The size of K will determine what approach to take. First let’s look at the case of a LARGE value of K ( >100). Allows us to make simplifying assumptions.

Example H 2 (g) + F 2 (g)  2HF (g) K = 1.15 x 10 2 at 25 º C Calculate the equilibrium concentrations if a 3.00 L container initially contains 6.06 g of H 2 and 228 g F 2 [ H 2 ] 0 = (6.06g/2.02)/3.00 L = 1.000M [I 2 ] 0 = (228g/38.)/3.00L = 2.000M [HI] 0 = 0

Q=0<K so more product will be formed because the reaction will shift to the right Assumption since K is large, reaction will go to completion. Stoichiometry tells us I 2 is LR, it will be smallest at equilibrium What is your mathematical proof? Set up ICE table.

Use x to represent the number of moles changed Change will be –x for the reactants because they are getting smaller by some amount called x Change in product is 2x because the product coefficient for HF is 2 and x represents the moles of HF produced H 2 (g)F 2 (g)2HF(g) Initial1.000M2.000 M0 M Change-x +2x Equilibrium

Concentration at Equilibrium is the sum of Initial and Change, plug it into Equilibrium Expression H 2 (g)F 2 (g)2HF(g) Initial1.000M2.000 M0 M Change-x +2x Equilibrium1.000-x2.000-x2x2x

We can NOT take the square root of both sides and arrive at a simple solution. Must use the quadratic equation

Now plug these values into the equilibrium expression and calculate K H 2 (g)F 2 (g)2HF(g) Initial1.000M2.000 M0 M Change-x +2x Equilibrium = =1.0322(.968)=1.936 The value is very close to the original K. So the calculations are correct.

Problems with small K AKA Approximation Process / 5% Rule K<.01

Approximation Process / 5% Rule Set up table of initial, change, and final concentrations. Choose X to represent a very small number. For a small K the product concentration is small This will greatly reduce the math needed

Example For the reaction: 2NOCl 2NO +Cl 2 K= 1.6 x (This is a small number) If 1.00 mol NOCl is placed in a 2.00 L container What are the equilibrium concentrations? Set up your ICE table

2NOCl (g)  2NO (g) +Cl 2(g) Initial Change-2x+2x+x Equilibrium0.50-2x2xx The equilibrium must satisfy the following equation: But when you cross multiply and collect all your terms on the same side of the equation you get an x 3 term which really complicates things.

1.K = 1.6 x is a small number 2.The system will NOT go very far to the right to reach equilibrium 3.The original reactant concentration will NOT get much smaller 4.SO the value for NOCl will not get smaller 5.Which means the value of x is relatively small 6.(0.50-x) can be approximated to (0.50) eliminating the x in the denominator The Logic Steps

Making the approximation allows us to simplify the expression Solve for x = 1.0 x How valid is the approximation? Plug the value of x into original (0.50-2(1.0 x ) = 0.48 The difference between 0.50 and 0.48 is 0.02 which is 4% of the original concentration of NOCl

Checking the assumption The rule of thumb is that if the value of X is less than 5% of all the other concentrations, our assumption was valid. If not we would have had to use the quadratic equation Our assumption was valid.

Practice Problem For the reaction 2ClO(g) Cl 2 (g) + O 2 (g) K = 6.4 x In an experiment mol ClO(g) 1.00 mol O x mol Cl 2 are mixed in a 4.00 L container. What are the equilibrium concentrations.

Mid-range K’s.01<K<10

No Simplification Choose X to be small. Can’t simplify so we will have to solve the quadratic (we hope) H 2 (g) + I 2 (g) HI(g) K=38.6 What is the equilibrium concentrations if mol H 2, mol I 2 and mol HI are mixed in a L container?

Problems Involving Pressure Solved exactly the same, with same rules for choosing X depending on K P For the reaction N 2 O 4 (g) 2NO 2 (g) K P =.131 atm. What are the equilibrium pressures if a flask initially contains atm N 2 O 4 ?

Le Chatelier’s Principle If a stress is applied to a system at equilibrium, the position of the equilibrium will shift to reduce the stress Can make a qualitative prediction of the effect of the change with LeChatelier’s Principle 3 Types of stress 1.Change Amount Of Species 2.Change Pressure 3.Change Temperature

Change Amounts of Species Adding product makes Q>K Removing reactant makes Q>K Adding reactant makes Q<K Removing product makes Q<K Determine the effect on Q, will tell you the direction of shift

Change Pressure Three Ways to change pressure: 1.Add or Remove gaseous reactant or product 2.Add inert gas 3.Change the Volume Add or Remove gaseous reactant or product is the same as Change amount of Species By adding an inert gas: – Partial pressures of reactants and product are not changed – No effect on equilibrium position

Change in Pressure By changing volume: – System will move in the direction that has the least moles of gas. – Because partial pressures (and concentrations) change a new equilibrium must be reached. – System tries to minimize the moles of gas. Pressure and Species Changes alter equilibrium position but they do not alter the equilibrium constant (K)

Change in Temperature Affects the rates of both the forward and reverse reactions. Temperature will effect the value of the equilibrium constant (K) Think as temperature as energy added to the reaction as either a reactant or product The direction of the shift depends on whether the reaction is exothermic or endothermic

Exothermic D H<0 Releases heat Think of heat as a product Rewrite the equation: A + B  C + D + heat Raising temperature pushes toward reactants. Shifts to left. Value of K decreases (Table 13.3)

Endothermic D H>0 Consumes ( Uses) heat Think of heat as a reactant Rewrite the equation: A + B + heat  C + D Raising temperature push toward products. Shifts to right. Value of K will increase

Practice For the reaction Cl 2 + O 2 2ClO(g) K = 156 In an experiment mol ClO, 1.00 mol O 2 and mol Cl 2 are mixed in a 4.00 L flask. If the reaction is not at equilibrium, which way will it shift? Calculate the equilibrium concentrations.