Equilibrium

Equilibrium

Notes

AP Chemistry

What is chemical equilibrium?

A dynamic state of a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction in a closed system, but no net effect is observable No change in macroscopic properties such as

pressure, color, or concentration at constant temperature

But at the atomic level, atoms, ions, or molecules are in constant reaction

Representing Equilibrium Reactions

N2O4(g) 2NO2(g)

The concept of equilibrium

Consider

Forward reaction: A B Rate = kf[A]

Reverse reaction: B A Rate = kr[B] At equilibrium kf[A] = kr[B]. Therefore, as the reaction progresses

[A] decreases to a constant, [B] increases from zero to a constant. When [A] and [B] are constant (but not

necessarily equal), equilibrium is achieved

Alternatively: kf[A] decreases to a constant,

kr[B] increases from zero to a constant.

When kf[A] = kr[B] equilibrium is achieved.

Equilibrium Constant Expressions

Mathematical expression of the equilibrium condition Results in a quantitative value for an equilibrium

constant under specified conditions Based on the law of mass action

The equilibrium constant (Keq) is computed as the ratio of the mathematical product of the concentrations of the products to the mathematical product of the concentrations of the reactants.

When molarities are used, the equilibrium constant is termed Kc

When partial gas pressures are used to represent concentrations, the constant is termed Kp

Equilibrium constant expressions using the law of mass action For the reaction

aA(aq) + bB(aq) cC(aq) + dD(aq)

Kc= [C]c [D]d

[A]a [B]b

If the reaction was

aA(g) + bB(g) cC(g) + dD(g)

Kp= PCc PD

d

PAa PB

b

Important things to remember about equilibrium constant expressions The reaction must be reversible The coefficients in the balanced equation are equal to the

exponents Products over reactants The concentrations must be the equilibrium concentrations Equilibrium constants

are unitless Are temperature dependent, since temperature

changes affect the forward and reverse reaction rates by different amounts, thus affecting the ratio of concentrations at equilibrium and thus the value of the equilibrium constant

Pure solids (s) and pure liquids (l) are excluded from the expression

Names and symbols for equilibrium constants Special names and symbols are used for different

types of chemical reactions Kc (most general form with molar concentrations) Kp (can be used with partial pressures when working

with a gas phase reaction) Ka (used for the dissociation of weak acids in water) Kb (used for the dissociation of weak bases in water) Kw (the equilibrium expression for the dissociation of

water into its ions) Ksp (used for the dissociation into ions of sparingly

soluble solids in water)

Kp vs. Kc

Mathematical relationship between Kp and Kc for gas phase reactions where either molar concentrations or partial pressures can be used

Kp= Kc(RT)Δn(gas)

R=the gas law constant (.0821 L atm/mol K)T=temperature in KΔn(gas)= moles of gaseous products - moles of gaseous

reactants (from coeffiecients)

**This is the “politically correct” (pc) equation- to remember the order of the K’s in the eqn

Size of Keq : The Equilibrium Position

If we say equilibrium lies far to the left or favors the reactants The reactant concentrations are much larger

than the product concentrations at equilibrium K<1

If we say equilibrium lies far to the right or favors the products The product concentrations are much larger

than the reactant concentrations at equilibrium K>1


Whether a reaction lies far to the right or to the left depends on three main factors Initial concentrations (more collisions- faster

reaction) Relative energies of reactants and products

(nature goes to minimum energy) Degree of organization of reactants and

products (nature goes to maximum disorder)


If Keq > 10 The reaction goes nearly to completion; products are favored;

equilibrium lies to the right The concentration of the products are much greater than the

concentration of the reactants, which simplifies calculations If Keq < 0.1

The reaction doesn’t go very far to completion; reactants are favored; equilibrium lies to the left

The concentrations of the reactants are much greater than the concentrations of the products, which simplifies calculations

If Keq =1 There are substantial amounts of both product and reactant present No approximations may be made

Other rules of interest

Coefficient rule If the coefficients of the reaction are increased or

decreased by a factor of n, then the K is raised to that power

Reciprocal rule If you reverse a reaction, you reciprocate (invert) the K

Multiple equilibria rule When two or more reaction equations are added

together, the equilibrium constant for the sum reaction is the product of the equilibrium constants of the added reactions

Coefficient rule: 1) 2A(aq) + B(aq) ↔ 4C(aq)

K1 = [C]4 .

[A]2[B]2) A(aq) + ½ B(aq) ↔ 2C(aq)

K2 = [C]2 .

[A][B]1/2 SO

K2 = (K1)1/2

Reciprocal Rule1) 2A(aq) + B(aq) ↔ 4C(aq)

K1 = [C]4 .

[A]2[B]2) 4C(aq) ↔ 2A(aq) + B(aq)

K2 = [A]2[B][C]4

SO

K2 = 1/K1

Multiple Equilibria Rule

2A(aq) + B(aq) ↔ 4C(aq) K1=[C]4/ [A]2[B]

4C(aq) + E(aq) ↔ 2F(aq) K2=[F]2/ [C]4[E]

2A(aq) + B(aq) + E(aq) ↔ 2F(aq) K3= K1 x K2

=[F]2/ [A]2[B][E]

Example Questions

Write the equilibrium constant expression for the following reaction (the Haber process)

N2(g) + 3H2(g) ↔ 2NH3(g)

Kc= [NH3]2

[N2][H2]3

Write the equilibrium constant expression for the following reaction

CaCO3(s) ↔ CaO(s) + CO2(g)

Kc = [CO2]

*remember pure solids are pure liquids are left out of the equilibrium constant expression

In the synthesis of ammonia from nitrogen and hydrogen, Kc= 9.60 at 300◦C.

N2(g) + 3H2(g) ↔ 2NH3(g)

Calculate Kp for this reaction at this temperature.

Kp= Kc(RT)Δn(gas) = (9.60)(0.0821x573)(2-4)

=4.34x10-3

The equilibrium constant for the reaction H2(g) + I2(g) ↔ 2HI(g) varies with temperature in the following way: Kc=794 at 298K Kc=54 at 700K

Is the formation of HI favored more at the higher or lower temperature?

At the lower temperature because of the larger Kc value (the larger Kc, the more products)

The value for the equilibrium constant for a reaction is 1.25x10-3 at a certain temperature. What is the value of the equilibrium constant for the reverse reaction at this temperature?

1/ 1.25x10-3=800 (K of the reverse reaction is reciprocal of the K of the forward reaction)

For the reaction N2O4(g) ↔ 2NO2(g) the equilibrium concentration of N2O4(g) is 0.00452M and that of NO2(g) is 0.0310M. Calculate the value of the equilibrium constant Kc.

Kc= [NO2]2 = (0.0310)2 = 0.213

[N2O4] (0.00452)

When a reaction is NOT at equilibrium- Predicting the direction of a reaction: the reaction quotient, Q For the general reaction

aA + bB cC + dD

Reaction quotient= Qc = [C]c[D]d

[A]a[B]b

Q has the appearance of K, but the concentrations aren’t necessarily equilibrium concentrations

The reaction quotient is found by substituting in the initial concentrations into the expression.

By finding the value of Q, we can predict the direction of a reaction If Q < K, the system is not at equilibrium.

To reach equilibrium, the reaction proceeds in the forward direction- “shifts right” -(the reactant concentrations will decrease, and the product concentrations will increase until Q=K)

If Q=K, then the reaction is at equilibrium If Q > K, the system is not at equilibrium

To reach equilibrium, the reaction proceeds in the reverse direction- “shifts left”- (the product concentrations will decrease, and the reactant concentrations will increase until Q=K)

Timesaver… If your initial concentration of any reactant is

zero, the reaction will proceed in the reverse direction (to the left)

If your initial concentration of any product is zero, the reaction will proceed in the forward direction (to the right)

Q vs K

example

For the synthesis of ammonia at 500◦C, the equilibrium constant is 6.0 x 10-2. Predict the direction in which the system will shift to reach equilibrium in each of the following cases:

a)[NH3] = 1.0 x 10-3M, [N2]=1.0 x 10-5M, [H2]= 2.0 x 10-3M

b) [NH3] = 2.00 x 10-4M, [N2]=1.5 x 10-5M, [H2]= 3.54 x 10-1M

c) [NH3] = 1.0 x 10-4M, [N2]=5.0M, [H2]= 1.0 x 10-2M

(Zumdahl 5th ed page 628)

Determining equilibrium concentrations using the RICE box

To determine equilibrium concentration given an initial set of conditions:

1. Set up the RICE box R=balanced reaction I=initial concentrations C=change in concentrations as reaction

proceeds E=Equilibrium concentrations

2. Underneath each formula, fill in the given information about concentrations

4. Calculate Q to determine which direction the reaction proceeds by substituting in initial concentrations in the equilibrium constant expression

5. Fill in the change line in the ICE box If Q<K, the reactant concentrations decrease by a

value of (coefficient)x and the initial concentration of each product is increased by a value of (coefficient)x

Or If Q>K, the reactant concentrations increase by a

value of (coefficient)x and the initial concentration of each product is decreased by a value of (coefficient)x

6. Fill in the missing information in the equilibrium line in terms of x (Sum of the “I” and “C” lines)

7. Substitute the equilibrium concentrations into the equilibrium expression

8. Solve for x May require use of quadratic formula, which will give

2 solutions, choose the logical one! Sometimes approximations can be made to make

solving for x easier If K <<1, x represents a relatively small number, and can

be treated as negligible when subtracting it out from an initial concentration; check the validity of the approximation after solving for x; a difference of less than 5% of the initial concentration is considered a valid assumption

9. Determine the equilibrium concentrations of the reactants and products by substituting in the value of x and solving

Examples

Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700K the equilibrium constant is 5.10. Calculate the equilibrium concentrations of all species if 1.000 mol of each component is mixed in a 1.000L flask. (Zumdahl 5th ed. P631)

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 1.500L flask. Calculate the equilibrium concentrations of all species. (Zumdahl 5th ed. Page633)

For a synthesis of hydrogen fluoride from hydrogen and fluorine, 3.000mol H2 and 6.000mol F2 are mixed in a 3.000L 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. (Zumdahl 5th ed page 635)

Determining K given initial concentrations and at least one equilibrium concentration To determine K given initial and one

equilibrium concentration:1. Write the balanced equation2. Set up the ICE box

I=initial concentrations C=change in concentrations as reaction

proceeds E=Equilibrium concentrations

3. Underneath each formula, fill in the given information about initial and equilibrium concentrations

4. Use the given equilibrium and initial concentration of one substance to determine the change in concentration that has occurred.

5. The change in concentration equals (coefficient)x, so solve for x.

6. Use the value of x along with the stoichiometry to determine the concentration changes and then the equilibrium concentrations of all other species in the reaction.

7. Substitute the equilibrium concentrations into the equilibrium constant expression and solve for K.

Example

A 1.00 L flask was filled with 2.00 mol gaseous SO2 and 2.00 mol gaseous NO2 and heated. After equilibrium was reached, it was found that 1.30 mol gaseous NO was present. Assume that the reaction

SO2(g) + NO2(g) SO3(g) + NO(g)

occurs under these conditions. Calculate the value of the equilibrium constant, K, for this reaction.

Le Chatelier’s Principle

If a closed system at equilibrium is subjected to a stress, a shift will occur to counteract the stress and establish a new equilibrium.

The most common “stresses” are changes in Temperature Concentration or partial pressure Volume

Note: Addition of a catalyst or an inert gas does not result in an equilibrium shift

A stress initially results in a nonequilibrium condition Because the rates of the forward and reverse

reactions are no longer equal A new equilibrium is then established over

time as the forward and reverse rates once again become equal.

Temperature Stresses

Increasing the temperature of an equilibrium system provides more available energy for the reaction increasing the rate of both the forward and

reverse reactions, but by different amounts Because the Ea of the forward reaction and

reverse reactions are not equal

Temperature increase

Endothermic (think of heat as a reactant) The forward rate will be increased more than the

reverse rate Resulting in a net shift of the reaction toward the

products Exothermic (think of heat as a product)

The reverse rate will be increased more than the forward rate

Resulting in a net shift of the reaction toward the reactants

NOTE: the reverse will happen with a temperature decrease

A temperature change is the ONLY change that will result in a new value for the equilibrium constant

Concentration or partial pressure stresses Note: there is no change in the value of

the equilibrium constant with changes in concentrations

The addition of a reactant in an equilibrium system will produce a shift toward the products (in the forward direction)

The addition of a product in an equilibrium system will produce a shift toward the reactants (in the reverse direction)

Increasing the concentration, or partial pressure, of a reactant in an equilibrium system will result in an increase in the rate of the forward reaction (assuming the reactant appears in the rate law)

Initially, the system is no longer at equilibrium Eventually, the forward and reverse reactions

will again be at the same rates, and equilibrium will be reestablished.

Similarly, the removal of a reactant in an equilibrium will cause a shift toward the reactants (in the reverse direction)

Removal of a product will cause a shift toward the products (in the forward direction)

3 common ways of removing species from equilibrium systems Neutralization

The addition of a strong acid to an equilibrium containing a base or the addition of a strong base to an equilibrium containing an acid

Precipitation The addition of an ion having low solubility with an ion

present in an equilibrium system Complex ion formation

The addition of a ligand that may bind some ion present in an equilibrium system

Volume change stresses

NOTE- volume changes do not change the value of the equilibrium constant

If the volume is decreased, the pressure will increase (Boyle’s law), if the volume is increased, the pressure will decrease.

An increase in pressure means the gas particles will be closer together, and an increase in collision frequency. Both the forward and reverse rates increase

Whether the forward or reverse reaction rate will be increased more than the other depends on the STOICHIOMETRY of the reaction In other words, the coefficients of the gaseous

reactants and products

If reactants and products have an equal number of moles, then changing the volume will have NO EFFECT on equilibrium

If the volume decreases, and the pressure increases, the system will attempt to decrease the pressure by shifting to the side with the fewest moles of gas

If the volume increases, and the pressure decreases, the system will shift to the side with more moles of gas.

Note- only the gas moles are considered, liquid or solid moles are ignored

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Equilibrium