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Equilibrium Notes Ch 15:

Homework:

Read Chapter 15 Work out sample/practice exercises in the sections,

Chapter 15: 23, 27, 29, 31, 39, 41, 45, 51, 57, 63, 77, 83, 97

Check for the MasteringChemistry.com assignment and complete before due date

For fun:

How is chemical equilibrium similar to a successful relationship? Consider

pressures, temperature and the environment.

Equilibrium State:

1. Chemical Equilibrium occurs for reversible reactions when the forward and

reverse reactions rates are equal and there is no net conversion of reactants to

products.

2. These equilibrium reactions do not go to completion, all species from reactants

and products exist at equilibrium.

3. A double arrow is used as reactions occur in both directions at the same rate, .

4. It does not matter if you start with only reactants, or only products, or some

mixture of all. Most reactions end in a dynamic equilibrium condition.

5. Irreversible reactions are those that proceed nearly to completion so the end

result contains almost all products and almost none of the limiting reactants.


Chemical Equilibrium:

Chemical Equilibrium answers the question…To what extent will a reaction go?

Most chemical reactions do not go to completion. Closed systems after a time

generally reach a point of dynamic equilibrium in which the forward rate and the

reverse rate are equal.

For the generic reaction… aA + bB cC + dD

The forward rate at equilibrium is equal to the reverse rate, making the following

rate law expressions equal to each other…

Rate = kforward[A]a[B]b = kreverse[C]c[D]d

Rearranging the equilibrium rate law expressions creates a new constant with a

capitalized K that we call the equilibrium constant, or the Law of Mass Action.

K = kforward/kreverse = [C]c[D]d/[A]a[B]b

Heterogeneous Equilibria:

When reactions have solid, pure liquids, or a solvent as either a reactant or

product, these species have an activity = 1 and are not part of the equilibrium

constant, providing some of each species must be present to establish equilibrium.

The Equilibrium Constant:

Equilibrium constant K = product concentrations to the power of their

coefficients divided by reactant concentrations to the power of their

coefficients… K = kforward/kreverse = [C]c[D]d/[A]a[B]b

Unlike the rate law expression from kinetics in which the slowest step affects

the orders, a reaction at equilibrium will have all steps ( forward, reverse, and

elementary steps ) going at the same rate while remaining in a dynamic

equilibrium. For this reason the coefficients in the overall reaction will be in

the equilibrium constant, capital K.

All the K constants use activities and are therefore unitless. The activity

value for pure solids, pure liquids and solvents are equal to 1. The activities

for concentrations in M divide by 1M, for pressure it is divided by 1

atmosphere. Therefore, no units are involved in a capital K, equilibrium

constant.


The important values to determine K come from gases and solutions.

Kc: concentrations bases on Molarity

Kp : concentrations based on partial pressures, atmospheres. All species in

the equilibrium constant must be gas (No aqueous Molarities, but solid

whose activity is 1 is OK)

Keq: Molarity for aqueous solutions and partial pressure for gases.

K is similar to the rate constant k, and will vary with temperature, but it is

independent of time and initial concentrations. An important difference is

that it is unaffected by a catalyst.

Equilibrium can be reached from either direction. K will be reached whether

you start with all reactants, all products, or a little of each substance in the

reaction.

Understanding K will help in determining how a reaction may be

manipulated in order to increase a desired substance.

PV = nRT from the gas laws… so P = MRT

We can calculate and find… Kp = Kc(RT)n or Kc = Kp(RT)-n

Where R is the gas constant 0.0821 (L atm)/(mol K) and n is the change in

number of moles of gases only.

Magnitude of K: The equilibrium constant is always a positive number

between 0 to infinite. Large constant, K >> 1, indicates mostly products will

exist at equilibrium. Small equilibrium constant , K << 1, indicates mostly

reactants

If the reaction doubles, K is squared… K2

If the reaction reverses K is inversed…K-1

In a multistep reaction, multiply each K to get the overall… K1 x K2 x K3


K can be determined experimentally if all equilibrium concentrations are

known, or calculated from thermochemical data (Keq = e-G°/RT) discussed

later when we study thermochemistry.

RICE represents…

Reaction: write the balanced chemical reaction

Initial: initial concentrations/amounts

Change: change taking place to reach equilibrium

Equilibrium: equilibrium concentrations/amounts

Applying Equilibrium:

Example 1:

For the reaction… H2 (g) + I2 (g) 2 HI (g) at 229 °C in a 1.00 L vessel

a) Initially, 0.500 mole of H2 gas and 0.500 mole of I2 gas is injected into the 1.00L

vessel. Once equilibrium has been established, 0.780 mol of HI is found in the

vessel. Apply RICE to determine the equilibrium Molarities of all substances in

this reaction and the equilibrium constant, Kc.

b) This same reaction… H2 (g) + I2 (g) 2 HI (g) at 229 °C… is now initiated

by injecting 1.00 mol of HI in a 1.00 L evacuated vessel. Using the calculated

value of Kc, apply RICE to determine the equilibrium Molarities of all

substances.


For the same reaction… H2 (g) + I2 (g) 2 HI (g) at 229 °C in a 1.00 L vessel

c) Graph concentration of all substances verses time for both part a and b.

d) Calculate the value of Kp for this reaction at 229°C using the equation…

Kp = Kc(RT)n (this R is 0.0821 L atm/mol K)

e) Using the Kc that was calculated in part (a), determine the value of Kc for the

reaction written below…

HI (g) ½ H2 (g) + ½ I2 (g) at 229°C


Example 2:

Write all possible (Kc, Kp, Keq) equilibrium constant expressions for the following.

(Not all reactions have a Kp)

a) PCl5 (g) PCl3 (g) + Cl2 (g)

b) 4 NH3 (g) + 5 O2 (g) 4 NO (g) + 6 H2O (g)

c) N2 (g) + 3 H2 (g) 2 NH3 (g)

d) CaCO3 (s) CaO (s) + CO2 (g)

e) Mg (s) + 2 HCl (aq) MgCl2 (aq) + H2 (g)

f) Cu(NO3)2 (aq) + 2KOH (aq) Cu(OH)2 (s) + 2KNO3 (aq)


Example 3:

Prove that Koverall = K1 x K2 for the two step reaction below. (assume all are Kc)

Step 1: NO2 (g) + O3 (g) NO3 (g) + O2 (g) K1

Step 2: NO3 (g) + NO2 (g) N2O5 (g) K2

Overall: Koverall

Example 4:

For the reaction… 2 SO2 (g) + O2 (g) 2 SO3 (g) ; Kp = 2.5 x 109

a) Do you expect mostly reactants or products at equilibrium? Explain.

b) What is the numerical value of Kp for the reverse reaction?

c) What is the numerical value of Kp if the original reaction is doubled?

d) What is the numerical value of Kc for the reaction at 25°C?


Example 5:

At 27°C, 0.80 mol N2 and 0.90 mol H2 were injected in an empty 1.00 L vessel.

Once equilibrium was established, 0.20 mol NH3 (g) was present. Calculate Kc and

Kp for the reaction… N2 (g) + 3 H2 (g) 2 NH3 (g)

The Reaction Quotient, Q, and Predicting Direction of the Reaction: The Reaction Quotient, Q, is the same formula as the equilibrium constant, K,

except Q is obtained using any starting concentrations that have generally not

reached the equilibrium point.

Q < K, indicates that reactant concentrations are higher than those

needed for equilibrium.

Which direction will the reaction shift to reach equilibrium?

Q = K, indicates that all concentrations have reached equilibrium.

Q > K, indicates that product concentrations are higher than those

needed for equilibrium.

Which direction will the reaction shift to reach equilibrium?


Example 6:

For the reaction… H2 (g) + I2 (g) 2 HI (g) at 450°C, Kc = 49

0.22 mole of I2, 0.22 mole H2 and 0.66 mole HI are injected in an evacuated

4.00 L vessel.

a) Solve for the reaction Quotient, Qc

b) What must happen to establish equilibrium?

c) Solve for the equilibrium concentrations.


Le Chatelier’s Principle and How to Manipulate the Reaction:

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 as to counteract the effect of the disturbance.

Changes to Consider:

Concentration of any substance:

Adding a reactant or removing a product …

will shift the reaction _____________ (which direction?)

Removing a reactant or adding a product

will shift the reaction _____________(which direction?)

Volume / Partial Pressures of any gas substance:

Increasing the volume decreases all the partial pressures and the

reaction will shift to create more pressure if possible…

will shift in the direction that has (Choose one: more/less)

moles of gas.

No effect occurs when an inert gas is added that does not participate in

the reaction or when the number of moles of gas are equal on either

side of the chemical reaction.

Temperature:

Increasing the temperature will favor the (Choose one:

endothermic/exothermic) reaction to reduce the extra energy added by

increasing the temperature.

Decreasing the temperature will favor the (Choose one:

endothermic/exothermic) reaction.

Catalyst/Inhibitor:

How will this affect a system at equilibrium? Explain.


For the Reaction below,

N2O4 (g) 2 NO2 (g) H =

When N2O4 gas is added to a reaction vessel at equilibrium, more NO2 gas will

form.


Heat also will effect this reaction:

Cause and Effect for a System at Equilibrium that undergoes a Change…

Example 7:

Consider the synthesis of ammonia from the elements nitrogen and hydrogen, the

Haber Process… N2 (g) + 3 H2 (g) 2 NH3 (g); H = -92 kJ/mol

What conditions will increase the production of ammonia on an industrial scale?

Explain.

a) Add or Remove NH3

b) Add or Remove N2 and H2

c) Changing the pressure of the system by…

(i) Addition of an inert gas

(ii) Altering the volume that contain the gases (Increase or decrease the

overall pressure?)

(d) and (e) are tricky: Consider kinetics as well as equilibrium. Which may be more

important in the industrial production of ammonia?

d) Increase or decrease the temperature?

e) Addition of a catalyst.


The Haber Process: … N2 (g) + 3 H2 (g) 2 NH3 (g); H = -92 kJ/mol

Effect of Temperature and Pressure on the Yield of NH3

°C Kc Mol% NH3 at

10 atm Ptotal

Mol% NH3 at

100 atm Ptotal

Mol% NH3 at

1000 atm Ptotal

209 650 51% 82% 98%

467 0.5 4% 25% 80%

758 0.014 0.5% 5% 13%

The production of ammonia is generally run at conditions around 450°C and

between 200-1000 atm pressure. The reaction is much too slow at lower

temperatures, and if the temperature is raised too high the mole % of ammonia

decreases dramatically.

The Quadratic Equation may be necessary for some calculations:

Example 8:

For the reaction… A (g) B (g) + C (g)

a) An equilibrium mixture from the reaction above is found to contain 0.20 M A,

0.30 M B, and 0.30M C. Determine the equilibrium constant, Kc.

R: A (g) B (g) + C (g)

E: 0.20M

Kc =


For the same reaction… A (g) B (g) + C (g)

b) If the above equilibrium reaction at the same temperature is expanded into a

vessel double the original size… Determine the reaction quotient, Q. Will the

reaction shift toward products, reactants, or stay the same? Solve for the new

equilibrium concentrations.

R: A (g) B (g) + C (g)

E: 0.20M

C: double the vessel volume has affect on M

I: 0.10M

C:

E:

c) If the first equilibrium reaction at the same temperature is compressed into a

vessel half the original size… Determine the reaction quotient, Q. Will the

reaction shift toward products, reactants, or stay the same? Solve for the new

equilibrium concentrations.


Example 9:

For the reaction… CO (g) + Cl2 (g) COCl2 (g)

A 2.00 L vessel in equilibrium contains 1.20 mol COCl2, 0.60 mol CO, and

0.20 mol Cl2

a) Determine the equilibrium constant, Kc.

b) If 0.80 mol Cl2 is added to the above equilibrium reaction at the same

temperature, will the reaction shift toward products, reactants, or stay the

same? Determine the new equilibrium concentrations.

Example 10:

NOBr is 34.0% dissociated at 25°C in a vessel whose total pressure is 0.25 atm.

What is Kp?

2 NOBr (g) 2 NO (g) + Br2 (g)


Example 11:

BrCl may replace Cl2 as water disinfectant someday. 1.00 mol Br2 and 0.75 mol

Cl2 are added to a 5.00 L container and allowed to reach equilibrium.

Cl2 (g) + Br2 (g) 2 BrCl (g) Kc = 4.7 x 10-2

a) Solve for the equilibrium concentrations.

b) What % of the original Cl2 has reacted?

Example 12:

For the reaction… 2 SO3 (g) O2 (g) + 2 SO2 (g) ; H = + 198 kJ

If this reaction is already at equilibrium, what will the following do to it?

a) Increase Temperature

b) Increase pressure

c) Remove O2

d) Add a catalyst

e) Increase the volume


Example Answers:

1a,b) 0.110 M, 0.110M, 0.780M, 50.3 1c) graph, 1d) 50.3, 1e) 0.141, 4a) products,

4b) 4.0 x 10-10, 4c) 6.3 x 1018, 4d) 6.1 x 1010, 5) Kc = 0.26, Kp = 4.4 x 10-4, 6a) Qc=

9, 6b)shift to product, 6c) 0.031 M H2, 0.031M I2, 0.214M HI, 7a) remove, 7b) add,

7ci) no affect, 7cii) increase, 7d,e) needs thoughtful explanation 8a) 0.45 8b)

forward to more products, Q = 0.225, A = 0.07M, B = 0.18M, C = 0.18M 8c)

reverse toward more reactants, Q = 0.90, A = 0.52M, B = 0.48M, C = 0.48M 9a)

20 9b) CO = 0.12 M, Cl2 = 0.32 M, COCl2 = 0.78 M, 10) 0.25 atm =[ PNOBr + PNO

+ PBr2 ] = x[0.66+0.34+0.17], x = 0.21 atm, Kp = 9.3 x 10-3 11a) 0.13 M Cl2,

0.18 M Br2, 0.0334 M BrCl 11b) 11%, 12 a,c,e) toward product, 12b) toward

reactant, 12d) no affect

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