Equilibrium

Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the forward reaction is possible 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

What is equal at Equilibrium? Rates are equal Concentrations are not. Rates are determined by concentrations and activation energy. The concentrations do not change at equilibrium. Or if the reaction is verrrry slooooow.

Law of Mass Action For any reaction jA + kB lC + mD jA + kB lC + mD K = [C] l [D] m PRODUCTS power [A] j [B] k REACTANTS power K = [C] l [D] m PRODUCTS power [A] j [B] k REACTANTS power K is called the equilibrium constant. is how we indicate a reversible reaction is how we indicate a reversible reaction Only gases and aqueous solutions are included in the law of mass action

Playing with K If we write the reaction in reverse. If we write the reaction in reverse. lC + mD jA + kB lC + mD jA + kB Then the new equilibrium constant is Then the new equilibrium constant is K ’ = [A] j [B] k = 1/K [C] l [D] m K ’ = [A] j [B] k = 1/K [C] l [D] m

Playing with K If we multiply the equation by a constant If we multiply the equation by a constant njA + nkB nlC + nmD njA + nkB nlC + nmD Then the equilibrium constant is Then the equilibrium constant is K’ = [A] nj [B] nk = ([A] j [B] k ) n = K n [C] nl [D] nm ( [C] l [D] m ) n

Manipulating K From a practical point of view, here is how K gets manipulated: If a reaction gets doubled, the value of K is squared. If it is tripled, K is raised to the third power. If the reaction gets cut in ½ then K is found by taking the square root of K.

The units for K This one is simple. This one is simple. K has no units. K has no units.

K is CONSTANT Temperature affects the equilibrium constant. Temperature affects the equilibrium constant. The equilibrium concentrations don’t have to be the same only K. The equilibrium concentrations don’t have to be the same only K. Equilibrium position is a set of concentrations at equilibrium. Equilibrium position is a set of concentrations at equilibrium. There are an unlimited number of K values so the temperature needs to be stated. There are an unlimited number of K values so the temperature needs to be stated.

Equilibrium Constant One for each Temperature

Calculate K N 2 + 3H 2 3NH 3 N 2 + 3H 2 3NH 3 Initial At Equilibrium Initial At Equilibrium [N 2 ] 0 =1.000 M [N 2 ] = 0.921M [N 2 ] 0 =1.000 M [N 2 ] = 0.921M [H 2 ] 0 =1.000 M [H 2 ] = 0.763M [H 2 ] 0 =1.000 M [H 2 ] = 0.763M [NH 3 ] 0 = 0 M [NH 3 ] = 0.157M [NH 3 ] 0 = 0 M [NH 3 ] = 0.157M

Calculate K N H 2 2 NH 3 (all gases) N H 2 2 NH 3 (all gases) 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 ] = 0.157M K is the same no matter what the amount of starting materials

Equilibrium and Pressure Don’t forget your gas laws, they are often needed to get some information for the problem. Especially PV = nRT Don’t forget your gas laws, they are often needed to get some information for the problem. Especially PV = nRT Some reactions are gaseous Some reactions are gaseous P = (n/V)RT P = (n/V)RT P = CRT P = CRT C is a concentration in moles/Liter C is a concentration in moles/Liter C = P/RT C = P/RT

Equilibrium and Pressure 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) Kp = (P SO3 ) 2 Kp = (P SO3 ) 2 (P SO2 ) 2 (P O2 ) (P SO2 ) 2 (P O2 ) K = [SO 3 ] 2 [SO 2 ] 2 [O 2 ] K = [SO 3 ] 2 [SO 2 ] 2 [O 2 ]

Equilibrium and Pressure: For those that are a glutton for punishment and want to see the full treatment. K = (P SO3 /RT) 2 (P SO2 /RT) 2 (P O2 /RT) K = (P SO3 /RT) 2 (P SO2 /RT) 2 (P O2 /RT) K = (P SO3 ) 2 (1/RT) 2 (P SO2 ) 2 (P O2 ) (1/RT) 3 K = (P SO3 ) 2 (1/RT) 2 (P SO2 ) 2 (P O2 ) (1/RT) 3 K = Kp (1/RT) 2 = Kp RT (1/RT) 3 K = Kp (1/RT) 2 = Kp RT (1/RT) 3

General Equation jA + kB lC + mD jA + kB lC + mD K p = (P C ) l (P D ) m = (C C x RT ) l (C D x RT ) m (P A ) j (P B ) k (C A x RT ) j (C B x RT ) k K p = (C C ) l (C D ) m x (RT) l+m (C A ) j (C B ) k x (RT) j+k K p = K (RT) (l+m)-(j+k) = K (RT)  n K p = K (RT) (l+m)-(j+k) = K (RT)  n  n= (l+m)-(j+k) = Change in moles of gas

Homogeneous Equilibria So far every example dealt with reactants and products where all were in the same phase. So far every example dealt with reactants and products where all were in the same phase. We can use K in terms of either concentration or pressure. We can use K in terms of either concentration or pressure. Units depend on reaction. Units depend on reaction.

Heterogeneous Equilibria If the reaction involves pure solids or pure liquids the concentration of the solid or the liquid doesn’t change. 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. As long as they are not used up we can leave them out of the equilibrium expression. For an example, check out the next slide. For an example, check out the next slide.

For Example H 2 (g) + I 2 (s) 2 HI(g) H 2 (g) + I 2 (s) 2 HI(g) K = [HI] 2 [ H 2 ][ I 2 ] But the concentration of I 2 does not change, so it factors out of the equilibrium expression and looks like this. But the concentration of I 2 does not change, so it factors out of the equilibrium expression and looks like this. K = [HI] 2 [ H 2 ] K = [HI] 2 [ H 2 ]

Solving Equilibrium Problems Type 1

The Reaction Quotient Tells you the direction the reaction will go to reach equilibrium Tells you the direction the reaction will go to reach equilibrium Calculated the same as the equilibrium constant, but for a system not at equilibrium Calculated the same as the equilibrium constant, but for a system not at equilibrium Q = [Products] coefficient [Reactants] coefficient Q = [Products] coefficient [Reactants] coefficient Compare value to equilibrium constant Compare value to equilibrium constant

What Q tells us If Q < K : Not enough products : Shift to right If Q > K : Too many products : Shift to left If Q = K system is at equilibrium. We like these ones – there is nothing to do !!

Example For the reaction: For the reaction: 2 NOCl(g) 2 NO(g) + Cl 2 (g) 2 NOCl(g) 2 NO(g) + Cl 2 (g) K = 1.55 x M at 35ºC K = 1.55 x M at 35ºC In an experiment 0.10 mol NOCl, mol NO(g) and mol Cl 2 are mixed in 2.0 L flask. In an experiment 0.10 mol NOCl, mol NO(g) and mol Cl 2 are mixed in 2.0 L flask. Which direction will the reaction proceed to reach equilibrium? Which direction will the reaction proceed to reach equilibrium?

Solving Equilibrium Problems Given the starting concentrations and one equilibrium concentration. Given the starting concentrations and one equilibrium concentration. Use stoichiometry to figure out other concentrations and K. Use stoichiometry to figure out other concentrations and K. Learn to create a table of initial and final conditions. Learn to create a table of initial and final conditions. Check out the next slide for an actual problem. Check out the next slide for an actual problem.

Consider the following reaction at 600ºC Consider the following reaction at 600ºC 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) In a certain experiment 2.00 mol of SO 2, 1.50 mol of O 2 and 3.00 mol of SO 3 were placed in a 1.00 L flask. At equilibrium 3.50 mol of SO 3 were found to be present. Calculate the equilibrium concentrations of O 2 and SO 2, K and K P

The ICE BOX The table we create is called the ICE Box. 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) Here is the information given in the problem. The blank spaces are what we have to determine. Some are determined and others are calculated. Initial Change Equilibrium 3.5

The ICE BOX The table we create is called the ICE Box. 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) Here is the information given in the problem. The blank spaces are what we have to determine. Some are determined and others are calculated. Initial Change - 2x - x + 2x Equilibrium 3.5

Where are we at ? The values in the CHANGE row are a combination of stoichiometry and equilibrium. Notice that the numbers are the same as the coefficients. We can’t forget stoichiometry. The [SO 3 ] increases so it has to be a + 2x. This tells us it shifts to the right and that the value of the reactants has to decrease. We combine the + or – sign along with the stoichiometry to set up the equation.

The ICE BOX The table we create is called the ICE Box. 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) Here is the information given in the problem. The blank spaces are what we have to determine. Some are determined and others are calculated. Initial Change - 2x - x + 2x Equilibrium 2 – 2x 1.5 – x 3.5

Where are we at ? Part 2 The values of the equilibrium are known algebraically. Here is our Law of Mass Action. We can solve that x =.25 K = [ SO 3 ] 2 = (3.5) 2 [ O 2 ] [ SO 2 ] 2 (1.25) (1.5) 2 K = 4.36 K = 4.36

Consider the same reaction at 600ºC 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) In a different experiment.500 mol SO 2 and.350 mol SO 3 were placed in a L container. When the system reaches equilibrium mol of O 2 are present. Calculate the final concentrations of SO 2 and SO 3 and K. A New Problem

Let’s set up the ICE Box 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) Initial Change + 2x + x - 2x Notice the signs. Since there is no oxygen present at the start, the reaction must shift to the left and that means products consumed and reactants created. Equilibrium.5 +2x – 2x This tells us that x =.045 and therefore the concentrations must be as follows. [SO 2 ] =.5 + 2x or =.59 [SO 3 ] =.35 – 2x or =.26 K = (.26) 2 / (.59) 2 (.045) = 4.32

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). We can make simplifying assumptions.

5 % rule Now is as good a time as any to introduce the 5 % rule. Equilibrium concentrations are often a bit of an approximation. Therefore it’s allowable to simplify the math to make it more workable. The 5 % rule states that if x is small in relation to the stated amount, and the K, then it can be ignored in the calculation. Generally works if the K eq is less than or greater than 10 3

Example  H 2 (g) + I 2 (g)  2 HI (g) K = 7.1 x 10 2 at 25ºC Calculate the equilibrium concentrations if a 5.00 L container initially contains 15.9 g of H 2 and 294 g I 2. [ H 2 ] 0 = (15.7g/2.02)/5.00 L = 1.56 M [ I 2 ] 0 = (294g/253.8)/5.00L = M [ HI ] 0 = 0

Q > K so more product will be formed. Since K is large, rxn. will go to completion. Another great clue is that when there is a component with a concentration of zero, the reaction must shift in that direction. Stoichiometry tells us I 2 is LR, it will be smallest at equilibrium let it be x Set up table of initial, final and change in concentrations. Let’s set up the Ice Box !!

Choose x so it is small. Since we know the I 2 is going to lose most of its concentration, x is actually going to be a very high percentage of the amount of I 2 There is a way around this dilemma. H 2 (g) + I 2 (g)  2 HI(g) initial 1.56 M M 0 M change - x - x + 2x equilibrium x x 2x This equation needs manipulating

Here we go. H 2 (g) + I 2 (g)  2 HI(g) initial 1.56 M.232 M 0 M change equilibrium In the work below, we are going to assume that the reaction goes to completion. Let’s combine the following facts and ideas and solve the problem.

Piecing it together 1. We didn’t change the conditions of the rxn. (ie. Temp. ) so the K eq remains unchanged. 2. The K eq is large so we know there will be more product than reactant at equilibrium. 3. Since we assumed the reaction went to completion we now have no Iodine and the reaction must move to the left. 4. X must be small because there needs to be more product to satisfy Keq. X will be negligible compared to the amounts given. 5. However, x can’t be ignored for the I 2 because it is at zero and x is 100% of the total amount of I 2

Here is our new Law of Mass Action after we remove the ‘x’ according to the 5% rule. Remember we can’t remove the x in the I 2 because that is all that there is of it. 710 = (.464) 2 / (1.328) ( x ) = 2.28 x H 2 (g) + I 2 (g)  2 HI(g) initial change + x+ x -2x equilibrium x x.464 – 2x

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 More on this later. Our assumption was valid.

Practice For the reaction Cl 2 + O 2  2 ClO(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? Q = ? Calculate the equilibrium concentrations of all components.

Problems with small K K<.01

Process is the same Set up table of initial, change, and final concentrations. Your ICE Box Choose X to be small. Sounds familiar so far, I hope !! For this case it will be a product. For a small K the product concentration is small.

For the reaction 2 NOCl  2 NO + Cl 2 K= 1.6 x Make note the small K favors more reactant If 1.20 mol NOCl, 0.45 mol of NO and 0.87 mol Cl 2 are mixed in a 1 L container What are the equilibrium concentrations ? Q = [NO] 2 [Cl 2 ] = (0.45) 2 (0.87) = 0.15 [NOCl] 2 (1.20) 2 Q > K so it shifts to the left. Let’s start the set up.

We can now use the Equilibrium concentrations to rework the problem. Remember that the K does not change. K favors the reactants. Having [ NO ] = 0 means that the reaction will shift to the right. Combining the two means that the change will be small so x will be small and we can use the 5 % simplification rule. This is a Happy Day ! 2 NOCl  2 NO + Cl 2 Initial Change Equilibrium Here is the ICE Box set up Once again we can manipulate the components to make the math easier and we do this by assuming the reaction goes to completion according to stoichiometry.

Where are we at ? We can now use the Equilibrium concentrations to rework the problem. Remember that the K does not change. K favors the reactants. Having [ NO ] = 0 means that the reaction will shift to the right. Combining the two means that the change will be small so x will be small and we can use the 5 % simplification rule. This is truly a Happy Day !

Solving: 1.6 x = (.645) (2x) 2 (1.65) 2 The algebra is trickier but doable x = NOCl  2 NO + Cl 2 Initial Change- 2x +2x+x Equilibrium1.65 – 2x 2x x Equilibrium (5%) x.645 Here is the new ICE Box

l If x =.0042, we compare that to the concentration /.87 =.48 % l We are good !! Always verify the 5 % rule

Practice Problem For the reaction 2 ClO (g)  Cl 2 (g) + O 2 (g) K = 6.4 x In an experiment mol ClO(g), 1.00 mol O 2 and 1.00 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 formula H 2 (g) + I 2 (g)  2 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?

Here is the set - up Q = 2.34 so Q < K & it shifts to the right. H 2 (g) + I 2 (g)  2 HI (g) Initial Change-x -x + 2x Equilibrium.9 – x.8 – x x K = 38.6 = ( x) 2 / (.9 – x) (.8 – x) = ?? 1. 1.Fortunately, we don’t see them this type often You are encouraged to discover the many uses of your TI – 84 or equivalent.

Problems Involving Pressure Solved exactly the same, with same rules for choosing X but use pressures due to 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 ? N 2 O 4 (g)  2 NO 2 (g) Initial 1 0 Change- x+ 2x Equilibrium 1-x1+ 2x

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, and establish a new equilibrium. 3 Types of stress Temperature changes Pressure changes Concentration changes

The Basic Steps Here is the best way to always get LeChatelier’s problems correct. Sample Problem A + B  C + D How is the system going to change if more B is added ? 1. Identify the stress: The addition of B 2. What is the opposite: The removal of B 3. Which way must the reaction shift to achieve step 2: It must shift to the right

Change amounts of reactants and/or products 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: remember that it is important to know what Q is equal to.

Change in Pressure # 1 This is usually done by changing volume. Decreasing the volume increases the pressure System shifts in the direction that has the less moles of gas: less gas = less pressure Because partial pressures (and concentrations) change a new equilibrium must be reached and established. System tries to minimize the moles of gas.

Change in Pressure # 2 This is usually done by changing volume. Increasing the volume decreases the pressure System shifts in the direction that has more moles of gas: more gas = more pressure Because partial pressures (and concentrations) change a new equilibrium must be reached and established. System tries to maximize the moles of gas.

Change in Pressure # 3 By the addition of an inert gas. Partial pressures of reactants and product are not changed. No effect on equilibrium position. This is a common question on the AP exam so be aware of it. It should be easy points. It might phrase it as an inert gas or it might use one of the Noble Gases: He Ne Ar Kr Xe Rn

Change in Temperature Affects the rates of both the forward and reverse reactions. Doesn’t just change the equilibrium position, changes the equilibrium constant. The direction & magnitude of the shift depends on whether the reaction is exothermic or endothermic. On many questions, it states a temperature that K is measured in. This is more to make the question technically correct. Look back on our solved problems. Temperature was not calculated in.

Change in Temperature :Exothermic DH < 0 It is a negative number. DH < 0 It is a negative number. The reaction releases heat. The reaction releases heat. Think of heat as a product since you know an equation can be written with a heat term: Think of heat as a product since you know an equation can be written with a heat term: A + B C + D + Heat A + B  C + D + Heat Raising temperature (adding heat) effectively increases the amount of heat (product) Shifts to left. Remember the steps to solving LeChatelier’s problems.

Change in Temperature :Endothermic DH > 0 It is a positive number. The reaction absorbs heat. Think of heat as a reactant since you know an equation can be written with a heat term. A + B + Heat C + D A + B + Heat  C + D Lowering temperature push toward reactants. Lowering temperature push toward reactants. Shifts to left. Remember the steps to solving LeChatelier’s problems.

Changes in Temperature: Summary 1.Place the heat term in the reaction. 2.Determine if heat is being added or taken away from the reaction. If heat is being added to an exothermic reaction, the reaction shifts to the left. If heat is being added to an exothermic reaction, the reaction shifts to the left. If heat is being added to an endothermic reaction, the reaction shifts to the right. If heat is being added to an endothermic reaction, the reaction shifts to the right. If heat is being taken away from an exothermic reaction, the reaction shifts to the right If heat is being taken away from an exothermic reaction, the reaction shifts to the right If heat is being taken away from an endothermic reaction, the reaction shifts to the left. If heat is being taken away from an endothermic reaction, the reaction shifts to the left.