Chapter 15 - Chemical Equilibrium Homework: 9, 10, 13, 14, 15, 16, 20, 21, 27, 28, 30, 31, 35, 36, 38, 39, 40, 43, 47, 48, 51, 52, 53, 54, 57, 63, 67, 69

15. 1 - The Concept of Equilibrium Imagine the following reaction N2O4(g)  2 NO2(g) N2O4 is colorless NO2 is brown We can analyze this reaction using kinetics (last chapter) Call decomposition of N2O4 the forward reaction Call decomposition of NO2 the reverse reaction In both cases, these reactions are elementary

N2O4(g)  2 NO2(g) Forward Reverse Rate =kf[N2O4] f for forward Reverse 2NO2(g)  N2O4(g) Rate = kr[NO2]2 r for reverse At equilibrium, the rate at which products are formed from reactants = the rate at which reactants are formed from the products Dynamic equilibrium

Therefore... kf[N2O4] = kr[NO2]2 Rearrange this So the ratio of the concentrations equals a constant This constant will be called the equilibrium constant, which we will deal with next section Does not matter if we start with N2O4 or with NO2, or even some mixture of the two. At equilibrium, the ratio will equal a specific value

Once equilibrium has been established Concentrations of substances no longer changes Have created a dynamic equilibrium Still reacting, just at the same rate

3 Things to Remember From this The fact that a mixture of reactants and products is formed in which concentrations no longer change with time indicates the reaction has reached a state of equilibrium. For equilibrium to occur, neither reactants nor products can escape from a system At equilibrium, the particular ratio of concentration terms equals a constant

15.2 - The Equilibrium Constant Opposing reactions naturally lead to an equilibrium, regardless of how complicated the reaction might be and regardless of the kinetics of the reaction We can approach equilibrium either by starting with the reactants or by starting with the products

Law of Mass Action For any reaction, the relationship between the concentration of the reactants and the products can be expressed by a constant. Related to the ratio of the product and reactants in equilibrium

Equilibrium-Constant Expression Given the following general equilibrium equation aA + bB  dD + eE The equilibrium condition is expressed by This is called the equilibrium-constant expression Kc is the equilibrium constant c means that the concentrations are in molarity

In general The numerator (top number) is the product of the concentrations of the products raised to their coefficient. Denominator (bottom number) is the product of the concentrations of the reactants raised to their coefficient. So once we know the balanced equation, we can write the equilibrium-constant expression Depends only on stoichiometry, not as the mechanism

The value of the equilibrium constant AT A GIVEN TEMPERATURE does NOT depend on the initial amount of reactants and products Does not matter if other substances are present, as long as they do not react with our reactants or products. Equilibrium constant only depends on the particular reaction and the temperature

Example Write the equilibrium expression for the following reactions 2 O3(g)  3 O2(g) 2 NO(g) + Cl2(g)  2 NOCl(g) Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq)

Equilibrium Constants in Terms of Pressure, Kp When reactants and products are gases, we can set up the equilibrium-constant expression in terms of partial pressures Instead of the normal molar concentrations When we use partial pressures, we call the equilibrium constant Kp (p for pressure)

Using aA + bB  dD + eE, but with Pressure... Px = partial pressure of X Generally partial pressure here is in atmospheres But units do not matter, as long as you are consistent

Kp  Kc We can perform algebra magic to convert between them magicmagicmagicmagicmagic Kp = Kc(RT)Δn Δn is the change in the number of moles of gas in the equation for the reaction Is equal to the sum of the coefficients of the gaseous products – the sum of the coefficients of the gaseous reactants Δn = (moles of gaseous product) – (moles of gaseous reactant)

Example For the following reaction N2(g) + 3 H2(g)  2 NH3(g) Kc= 9.60 at 300ºC. Find Kp for this reaction at this temperature

N2(g) + 3 H2(g)  2 NH3(g) Kp = Kc(RT)Δn Given Kc, need to find the change in moles/coefficient 2 moles of gaseous product (2 NH3) and four moles of gaseous reactant (1 N2 and 3 H2) Therefore, Δn = 2-4 = -2 Delta always means product - reactant

Kp = Kc(RT)Δn We just plug everything in! Kp = 9.60(8.314)(573)-2 Kp= 4.34x10-3

Equilibrium Constants and Units Generally No units

15.3 - Interpreting and Working with Equilibrium Constants We are going to look at how to analyze a reaction based upon the value of its equilibrium constant

The Magnitude of Equilibrium Constants Equilibrium constants can vary from very large to very small To analyze what the constant means, we look at where it comes from Given aA + bB  cC + dD If we have a very large equilibrium constant, the numerator must be much larger than the denominator When this happens, we say the equilibrium lies to the right, or, favors the product

If K >> 1: Equilibrium lies to the right; mostly products If, however, the equilibrium constant is very small, then we have a large denominator compared to the numerator. Reaction is said to lie to the left, or favor the reactants If K >> 1: Equilibrium lies to the right; mostly products If K << 1: Equilibrium lies to the left; mostly reactants

Example The reaction of N2 with O2 to form NO is considered a means of fixing nitrogen N2(g) + O2(g)  2 NO(g) Kc at 25ºC is 1x10-30 Is it feasible to fix nitrogen by forming NO at 25ºC? No Since constant is so small, reaction favors the reactants, and very little NO will be produced at equilibrium

The Direction of the Chemical Equation and K Because we can approach equilibrium from either direction in a reaction (from reactants or from products), the way we write the chemical equation at equilibrium is arbitrary

N2O4(g)  2 NO2(g) Could also write 2 NO2(g) N2O4(g)

We can see from the previous example that the equilibrium-constant expression for a reaction written in one direction is the reciprocal of the one for the reaction written in the reverse direction. Therefore, if we know the equilibrium constant for one reaction, we can take its reciprocal to find the equilibrium constant for its reverse reaction Note: Assuming temperature remains constant

Example Given 2 NO(g)  N2(g) + O2(g) Kc for the reverse reaction is 1x10-30 Write the equilibrium-constant expression for Kc for this reaction Find the equilibrium-constant for this reaction

2 NO(g)  N2(g) + O2(g) Kc for the reverse reaction is 1x10-30

Relating Chemical Equations and Equilibrium Constants Just like the equilibrium constants of forward and reverse reactions are reciprocals of each other, the equilibrium constants of related reactions are also related.

Example Original equation was New equation is N2O4(g)  2 NO2(g) Equilibrium-constant expression would be Simply the square of the equilibrium-constant expression of the original Because the new expression equals the original squared, the new constant equals the original squared.

Hess Law type problem In problems that have multiple steps (like Hess Law), we must use equations made up of two or more steps in the whole process We find the net equation by adding the individual equations and canceling identical terms

Example Consider 2 NOBr(g)  2 NO(g) + Br(g) Br2(g) + Cl2(g)  2 BrCl(g) Overall Equation 2 NOBr(g) + Cl2(g)  2 NO(g) + 2 BrCl(g)

The total equilibrium-constant expression is the product of the expressions for the individual steps Similarly, the equilibrium-constant for the total is the product of the two individual equilibrium constants

Summary The equilibrium constant of a reaction in the reverse direction is the inverse of the equilibrium constant of the reaction in the forward direction The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power equal to that number The equilibrium constant for a net reaction made up of two or more steps is the product of the equilibrium constants for the individual steps

15.4 - Heterogeneous Equilibria If, during equilibria, all of the substances are in the same phase, it is called a homogeneous equilibria. Likewise, if the substances are in different phases, we get a heterogeneous equilibria. Question: What is the concentration of pure water?

Why is this a problem? Consider the following reaction PbCl2(s)  Pb2+(aq) + 2 Cl-(aq) When we try writing out the equilibrium-constant expression, we run into a problem We cannot express the molarity of a solid! Because there is no volume of solution or gas! How do we solve this problem? Ignore the solid!

PbCl2(s)  Pb2+(aq) + 2 Cl-(aq) So the equilibrium-constant expression for the above reaction would just be Kc = [Pb2+][Cl-]2 Even though PbCl2 is not shown in the expression, it must still be present for there to be equilibrium. Likewise, we can also ignore pure liquids Such as water

Why can we ignore these? Concentration of a solid or pure liquid has a constant value If we double the mass of a solid, we also double its volume So its concentration (relates mass to volume) stays the same Since equilibrium-constant expressions only have terms for things whose concentrations can change, these are left out

Example Write the equilibrium-constant expressions for Kc for each of the following CO2(g) + H2(g)  CO(g) + H2O(l) SnO2(s) + 2 CO(g)  Sn(s) + 2 CO2(g)

Gas-Style Reaction CaCO3(s)  CaO(s) + CO2(g) Kc = [CO2] and Kp = PCO2 So at a given temperature, an equilibrium between CaCO3, CaO and CO2 will ALWAYS lead to the same partial pressure of CO2 As long as all three components are present

Example Given this reaction CaCO3(s)  CaO(s) + CO2(g) If each of the following mixtures was placed in a closed container and allowed to stand, which could attain the above equilibrium? pure CaCO3 CaO and CO2 pressure greater than the value of Kp Some CaCO3 and CO2 pressure greater than the value of Kp CaCO3 and CaO

15.5 - Calculating Equilibrium Constants The simplest way to calculate the equilibrium constant is when the concentrations of all substances is known

Example 7.38 atm H2, 2.46 atm N2, 0.166 atm NH3 N2(g) + 3 H2(g)  2 NH3(g) Create the equilibrium-constant expression, solve for Kp And that is it!

Unfortunately We oftentimes dont know the equilibrium concentrations of everything If we know the equilibrium concentration of at least ONE thing, we can generally use stoichiometry of the reaction to find the concentrations of the others Then once we have found the equilibrium concentrations of the others, we can solve for the equilibrium constant.

Steps to Solve for Unknown Equilibrium Concentrations Tabulate all the known initial and equilibrium concentrations For those that we have both the initial and equilibrium concentrations, find the change in concentration Use stoichiometry to find the changes for the rest

Example A closed system initially contains 1.000x10-3 M H2, and 2.000x10-3M I2 at 448ºC. It is allowed to reach equilibrium. We find that at equilibrium, the concentration of HI is 1.87x10-3M. Calculate Kc at this temperature for this reaction H2(g) + I2(g)  2 HI(g)

H2(g) + I2(g)  2 HI(g) ICE = Initial, Change, Equilibrium H2(g) I2(g) All of these are concentration values H2(g) I2(g) HI(g) Initial 1.000x10-3M 2.000x10-3 M 0 M Change Equilibrium 1.87x10-3 M

If we know the initial and final of ANY of the substances, we can find the change in concentration Change in [HI] 1.87x10-3 M - 0 = 1.87x10-3 M Once we know the change, we use stoichiometry to find the change in the others

H2(g) + I2(g)  2 HI(g) Change in HI = 1.87x10-3 M So in this case, both reactants had a change of 0.935x10-3 M

H2(g) I2(g) HI(g) Initial 1.000x10-3M 2.000x10-3 M 0 M Change Equilibrium 0.065x10-3 M 1.065x10-3 M 1.87x10-3 M

So we now have our equilibrium concentrations Can now find the equilibrium constant Does this reaction favor the products or reactants?

Example 2 Sulfur trioxide decomposes at high temperature in a sealed container 2 SO3(g)  2 SO2(g) + O2(g) Initially, the container is charged at 1000 K with SO3(g) at a partial pressure of 0.500 atm. At equilibrium, the SO3 partial pressure is 0.200 atm. Calculate the value of Kp at 1000 K.

Workspace SO3(g) SO2(g) O2(g) Initial .500 atm 0 atm Change Equilibrium 0.200 atm

Workspace - Find change

SO3(g) SO2(g) O2(g) Initial .500 atm 0 atm Change Equilibrium 0.200 atm

Workspace - Stoichiometry

Workspace - Calculation

15.6 - Applications of Equilibrium Constants We know that the magnitude of K shows how a reaction will proceed If K is large, reaction tends to proceed toward the product If K is small, reaction tends to proceed towards the reactants We can also use equilibrium constant to determine the direction a reaction will go (based on a reaction mixture) find the concentrations of reactants and products once equilibrium has been achieved.

Predicting the Direction of Reaction We know which direction of a reaction is favored (based upon value of K) But what if we have a mixture of products and reactants, and we want to see which way the reaction goes to achieve equilibrium? Meaning do we produce more reactants or more products?

Reaction Quotient The reaction quotient, Q, lets us find which direction a reaction goes To find Q, we substitute initial reactant and product concentrations (or partial pressures) into our normal equilibrium-constant expression So we find Qc (for concentration) or Qp (for pressure)

Determining Reaction Direction We compare the value of Q with the K value for the reaction If Q = K: We are already at equilibrium. If Q > K: Concentration of products too large. Reaction moves from right to left (products to reactants) If Q < K: Concentration of products too small. Reaction moves by forming products (left to right)

Example At 448ºC the equilibrium constant Kc for the following reaction is 50.5. H2(g) + I2(g)  2 HI(g) Predict in which direction the reaction will proceed to reach equilibrium at this temperature if we start with 2.0x10-2 mol of HI, 1.0x10-2 mol of H2, and 3.0x10-2 mol of I2 in a 2.00 L container

Step 1: Find actual concentrations [HI] [H2] [I2]

Step 2: Find Q and compare

Calculating Equilibrium Concentrations Oftentimes we must find the amount of reactants and products at equilibrium. We do this similar to how we found the equilibrium constant We use ICE method or just the equilibrium values, depending on information given

Example Kp for the following process is 1.45x10-5 at 500ºC N2(g) + 3 H2(g)  2 NH3(g) In an equilibrium mixture of the three gases at this temperature, the partial pressure of H2 is 0.928 atm, and that of N2 is 0.432 atm. What is the partial pressure of NH3 in this equilibrium mixture?

Start with equilibrium-constant expression Rearrange to solve for x x2 = (1.45x10-5)(0.432)(0.928)3 x2 = 5.01x10-6 x = 2.24x10-3 atm = PNH3

We then solve for the equilibrium amounts. Usually we know the value of the equilibrium constant and the initial amount of everything. We then solve for the equilibrium amounts. We use the change in concentration as a variable So NOW we use ICE

Example A 1.000-L flask is filled with 1.000 mol of H2and 2.000 mol of I2 at 448ºC. Kc for this reaction is 50.5 H2(g) + I2(g)  2 HI(g) What are the equilibrium concentrations of H2, I2and HI in molarity?

Start the ICE tabulation H2(g) I2(g) HI(g) Initial 1.000 M 2.000 M 0 M Change Equilibrium

Use stoichiometry We use the stoichiometric relationships to determine the change in concentration Since H2 and I2 have a 1:1 mole ratio, they will both lose the same concentration H2(g) I2(g) HI(g) Initial 1.000 M 2.000 M 0 M Change -x +2x Equilibrium

Use change to find equilibrium Use the same data to find out the equilibrium amount H2(g) I2(g) HI(g) Initial 1.000 M 2.000 M 0 M Change -x +2x Equilibrium (1.000-x)M (2.000 - x)M 2x M

Substitute equilibrium amounts into equation Now substitute the amounts of the equilibrium into the equilibrium-constant expression Using algebra, solve for x 4x2 = 50.5(x2 - 3.000x + 2.000) 46.5x2 - 151.5x + 101.0 = 0 x = 2.323 or 0.935 (quadratic equation gives two answers)

x = 2.323 or 0.935? Substitute each value into the equilibrium amounts, and see if it makes sense. Start with 2.323 = x [H2] = 1.000 -x = 1.000 -2.323 = a negative concentration We cannot have a negative concentration, therefore the correct value of x is hopefully 0.935 [H2] = 1.000-0.935 = 0.065 M [I2] = 2.000-0.935 = 1.065 M [HI] = 2x = 1.870 M Whenever we do this, we will always have to choose between two answers. Only one of them will be chemically meaningful

15.7 - Le Châteliers Principle Le Châteliers principle is stated as: If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance. This means if we mess with a reaction at equilibrium, the system will change to accommodate our change

Change in Reactant or Product This is the easiest shift If a reaction is in equilibrium, and we add a substance (either reactant or product), the reaction will shift to reestablish equilibrium by consuming part of the added substance. Similarly if we remove a substance, the reaction moves in the direction that forms more of that substance

What This Means Consider this familiar equation What would happen if I N2(g) + 3 H2(g)  2 NH3(g) What would happen if I Added more nitrogen? Reaction shifts to the right, forming more ammonia (but using more hydrogen) Added more hydrogen? Same as above. Shifts to the right, forms more ammonia, using more nitrogen Added more ammonia? Shifts to the left, forming more nitrogen and hydrogen Removed ammonia? Shifts to the right, forming more ammonia, using up nitrogen and hydrogen

Effect of Volume and Pressure Change If volume of system is decreased (increasing pressure), system will shift equilibrium to reduce the pressure We reduce pressure by reducing the total number of gas molecules More gas molecules = more pressure So lowering the volume of a gaseous equilibrium mixture causes the system to shift in the direction that reduces the number of moles of gas

Example: N2O4(g) 2 NO2(g) If I decrease the volume of the container this reaction occurred in Reaction shifts to the side of fewer moles So the reaction would shift towards the reactants So increasing pressure (decreasing volume) causes reaction to produce more N2O4(g)

How about N2(g) + 3 H2(g)  2 NH3(g) There are 4 reactant molecules to every 2 NH3 molecules That means that in increase in pressure (decrease in volume) shifts reaction towards side with fewer molecules So reaction moves to the right What if we decreased the pressure in the reaction? Reaction would move to the left

Remember Pressure-volume changes DO NOT change the K value as long as T remains constant All we do is change the partial pressures of the substances

Example N2(g) + 3 H2(g)  2 NH3(g) At 472ºC, Kp = 2.79x10-5 7.38 atm H2, 2.46 atm N2 and 0.166 atm NH3 at equilibrium Consider what will happen when we suddenly reduce the volume by ½. If the equilibrium did not shift, then the partial pressure of each would double Giving us 14.76 atm H2, 4.92 atm N2 and 0.332 atm for NH3

These new pressures would mean the reaction quotient would no longer equal the equilibrium constant Qp  Kp Because Qp < Kp Reaction no longer at equilibrium Equilibrium reestablished by increasing PNH3 and decreasing the reactants Therefore, reaction shifts to the right

Pressure Change w/o Volume Change? We can change pressure of system without changing the volume If additional amounts of any of the reacting components is added/removed Adding a gas not involved in reaction Since this would not change any of the partial pressures, no shift in equilibrium

Effect of Temperature Change Almost every equilibrium constant changes with temperature changes Look at temperature dependence through Le Châteliers principle. Treat heat as if it were a chemical reagent In an endothermic reaction, treat heat as a reactant In an exothermic reaction, treat heat is a product

When the temperature of a system at equilibrium is increased, it is like we added a reactant to an endothermic reaction or a product to an exothermic reaction The equilibrium will shift in the direction that consumes the excess reactant (or product), meaning heat

What This Means Endothermic: Exothermic: Increasing T results in an increase in K Exothermic: Increasing T results in a decrease in K Cooling is just the opposite of heating

Example 1 Consider the equilibrium N2O4(g)  2 NO2(g) ΔHº = 58.0 kJ In which direction will the equilibrium shift when N2O4is added? To the right NO2 is removed

N2O4(g)  2 NO2(g) ΔHº = 58.0 kJ The total pressure is increased by addition of N2? No shift at all N2 does not change the partial pressures of either of these The volume is increased? Shifts towards the right The temperature is decreased? To the right Endothermic reaction, heat is treated as a reactant

The Effect of Catalysts Remember, when we add a catalyst, we lower the activation energy between reactants and products Also lowers the activation energy for the reverse reaction Catalysts DO NOT change the composition of an equilibrium It just changes the rate at which equilibrium is achieved