chemical equilibrium

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Chapter 17

Equilibrium: The Extent of Chemical Reactions

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Figure 17.1

Reaching equilibrium on the macroscopic and molecular levels.

N2O4(g) 2NO2(g)

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Equilibrium - the condition in which the concentrations of all the reactants and products in a closed system cease to change with time

At equilibrium: rateforward = ratereverse

no further net change is observed because changes in one direction are balanced by changes in the other but it doesn’t mean that the reaction had stopped.

The amount of reactants and products are constant but they are not necessarily equal.

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If rateforward = ratereverse then

kforward[reactants]m = kreverse[products]n

= = K the equilibrium constantkforward

kreverse

[products]n

[reactants]m

The values of m and n are those of the coefficients in the balanced chemical equation. The rates of the forward and reverse reactions are equal, NOT the concentrations of reactants and products.

K is dependent only of the temperature.

Note: The terms for pure solids or pure liquids do not appear in the equilibrium constant expression

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22 eqfwd

rev 2 4 eq

[NO ]k =

k [N O ]

Equilibrium constant expression for

N2O4(g) 2NO2(g)

K =

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Practice Problem

04/08/23 6

3 42 2

c 1 53 8 2

[CO ] [H O]K =

[C H ] [O ]

The subscript “c” in Kc indicates the equilibrium constant is based on reactant and product concentrations

The value of “K” is usually shown as a unitless number,

BUT IT ACTUALLY DOES HAVE A UNIT EXPRESSION

Write the Equilibrium Constant for the combustion of Propane gas C3H8(g) + O2(g) CO2(g) + H2O(g)

1. Balance the EquationC3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

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Equilibrium Constants in Terms of Partial Pressures• For reactions involving gases, concentrations are generally

reported as partial pressures, and the equilibrium expression is often written:

• where the partial pressure of each reactant and product are given as Px in units of atmospheres (atm), and Kp is the equilibrium constant when concentration is given in partial pressures.

qp

P Q

p a b

A B

P PK

P P

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Figure 17.2 The range of equilibrium constants

small K

large K

intermediate K

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Thus, a mixture of N2 and O2 will react to a very small extent to produce NO at equilibrium.

302( ) 2( ) ( )2 1.0 10g g g cN O NO K x

Thus, a mixture of N2 and H2 will almost completely be converted to NH3 at equilibrium.

82( ) 3( )2( )

3 2 5.0 10g g cgN H NH K x

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Relationsip between Kc and Kp

Kp = Kc (RT)n(gas)

Δn = (no. of moles of gaseous products) – (no. of moles of gaseous reactants)

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Sample Problem 17.4 Converting Between Kc and Kp

PROBLEM: A chemical engineer injects limestone (CaCO3) into the hot flue gas of a coal-burning power plant for form lime (CaO), which scrubs SO2 from the gas and forms gypsum. Find Kc for the following reaction, if CO2 pressure is in atmospheres.

CaCO3(s) CaO(s) + CO2(g) Kp = 2.1x10-4 (at 1000K)

PLAN: We know Kp and can calculate Kc after finding ngas. R = 0.0821 L*atm/mol*K.

SOLUTION: ngas = 1 - 0 since there is only a gaseous product and no gaseous reactants.

Kp = Kc(RT)n Kc = Kp/(RT)n = (2.1x10-4) / (0.0821 x 1000)1 = 2.6x10-6

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Q - The Reaction Quotient

The reaction quotient, Q, is defined in the same way as the equilibrium constant Kc except that the concentrations in the equilibrium constant expression are not necessarily equilibrium

We use the molar concentrations of the substances in the reaction. This is symbolized by using square brackets - [ ].

For a general reaction aA + bB cC + dD where a, b, c, and d are the numerical coefficients in the balanced equation, Q (and K) can be calculated as

Q = [C]c[D]d

[A]a[B]b

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Figure 17.3 The change in Q during the N2O4-NO2 reaction.

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Figure 17.5 Reaction direction and the relative sizes of Q and K.

Equilibrium: no net change

reactants products

forward

Reaction Progress

reactants products

backward

Reaction Progress

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Sample Problem 17.1

PROBLEM:

SOLUTION:

Writing the Reaction Quotient from the Balanced Equation

Write the reaction quotient expression, Qc, for each of the following reactions:

(a) The decomposition of dinitrogen pentoxide, N2O5(g) NO2(g) + O2(g)

(b) The combustion of propane gas, C3H8(g) + O2(g) CO2(g) + H2O(g)

PLAN: Be sure to balance the equations before writing the Qc expression.

42(a) N2O5(g) NO2(g) + O2(g) Qc = [NO2]4[O2]

[N2O5]2

3 45(b) C3H8(g) + O2(g) CO2(g) + H2O(g) Qc = [CO2]3[H2O]4

[C3H8][O2]5

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Sample Problem 17.5

SOLUTION:

Comparing Q and K to Determine Reaction Direction

PROBLEM: For the reaction N2O4(g) 2NO2(g), Kc = 0.21 at 1000C. At a point during the reaction, [N2O4] = 0.12M and [NO2] = 0.55M. Is the reaction at equilibrium. If not, in which direction is it progressing?

PLAN: Write an expression for Qc, substitute with the values given, and compare the Qc with the given Kc.

Qc =[NO2]2

[N2O4]=

(0.55)2

(0.12)= 2.5

Qc is > Kc, therefore the reaction is not at equilibrium and will proceed from right to left, from products to reactants, until Qc = Kc.

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Sample Problem 17.10

SOLUTION:

Predicting Reaction Direction and Calculating Equilibrium Concentrations

PROBLEM: Given the reaction:

CH4(g) + 2H2S(g) CS2(g) + 4H2(g)

In one experiment, 4.0 M of CH4, 4.0 M of CS2, 8.0 M of H2S, and 8.0Mof H2 are initially mixed in a vessel at 9600C. At this temperature, Kc = 0.036 In which direction will the reaction proceed to reach equilibrium?

[CH4]initial = 4.0M

[H2S]initial = 8.0M [H2]initial = 8.0M

[CS2]initial = 4.0M

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Sample Problem 17.10 Predicting Reaction Direction and Calculating Equilibrium Concentrations

continued

Qc = [CS2][H2]4

[CH4][H2S]2=

[4.0][8.0]4

[4.0][8.0]2= 64

A Qc of 64 is >> than Kc = 0.036

The reaction will progress to the left.

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LeChatelier’s PrincipleLeChatelier’s Principle

When a chemical system at equilibrium

is subjected to a stress,

the system will return to equilibrium

by shifting to reduce the stress.

If the concentration increases, the system reacts to consume some of it.

If the concentration decreases, the system reacts to produce some of it.

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Factors Affecting Equilibrium

• Concentration Changes

• Pressure Changes

• Temperature Changes

• Addition of a Catalyst

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If the conc. of a substance is increased, the equilibrium will shift in a way that will decrease the conc. of the substance that was added.

I. Concentration Changes

2( ) 2( ) ( )2g g gH I HI

H2

I2

HI

Decrease HI - forward

Decrease I2 - backward

Increase H2 -forward

Increase HI - backward

Where will the reaction shift?

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Note:

adding/removing a solid/liquid to an equilibrium system will not cause any shift in the position of equilibrium.

addition of an inert gas such as He, Ar, Kr, etc. at constant volume, pressure and temperature does not affect the equilibrium.

3( ) ( ) 2( )s s gCaCO CaO CO

Increase CO2 :

Increase CaO

Decrease CO2

Adding Kr

- backward

- No effect

- forward

- No effect

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Practice Exercise: Predicting the Effect of a Change in Concentration on the Equilibrium Position

PROBLEM: To improve air quality and obtain a useful product, chemists often remove sulfur from coal and natural gas by treating the fuel contaminant hydrogen sulfide with O2;

2H2S(g) + O2(g) 2S(s) + 2H2O(g)

In what direction will the rection shift if

(a) O2 is added? (b) O2 is added?

(c) H2S is removed? (d) sulfur is added?

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+

lower P(higher V)

more moles of gas

higher P(lower V)

fewer moles of gas

I. Pressure Changes

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An increase in pressure (decrease in volume) shifts the position of the equilibrium in such a way as to decrease the number of moles of gaseous component.

When the volume is increased (pressure decreased), a net reaction occurs in the direction that produces more moles of gaseous component

The effect of pressure (volume) on an equilibrium system.

2( ) 2( ) 3( )3 2g g gN H NH

Decrease in pressure :

Decrease in volume:

backward

forward

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SOLUTION:

Predicting the Effect of a Change in Volume (Pressure) on the Equilibrium Position

PROBLEM: For the following reactions, predict the direction of the reaction if the pressure is increased:

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

(b) S(s) + 3F2(g) SF6(g)

(c) Cl2(g) + I2(g) 2ICl(g)

(a) CO2 is the only gas present. The equilibrium will shift to the direction with less moles of gas. Answer: backward

(b) There are more moles of gaseous reactants than products.

Answer: forward

(c) There are an equal number of moles of gases on both sides of the reaction. Answer: no effect

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The Effect of a Change in Temperature on an Equilibrium

Consider heat as a product or a reactant.

In an exothermic reaction, heat is a product, H0rxn = negative

In an endothermic reaction, heat is a reactant, H0rxn = positive

3( ) ( ) 2( ) 92.4s s gCaCO CaO CO H kJ

3( ) ( ) 2( ) 92.4s s gCaCO CaO CO kJ

CO2 92.4 kJ

Increase temperature: backward

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SOLUTION:

Predicting the Effect of a Change in Temperature on the Equilibrium Position

PROBLEM: In what direction will the reaction shift if there is a decrease in temperature

(a) CaO(s) + H2O(l) Ca(OH)2(aq) H0 = -82kJ

(b) CaCO3(s) CaO(s) + CO2(g) H0 = 178kJ

(c) SO2(g) S(s) + O2(g) H0 = 297kJ

(a) CaO(s) + H2O(l) Ca(OH)2(aq) heat

A decrease in temperature will shift the reaction to the right

(b) CaCO3(s) + heat CaO(s) + CO2(g)

The reaction will shift to the left

(c) SO2(g) + heat S(s) + O2(g)

The reaction will shift to the left

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Addition of a Catalyst

The presence of a catalyst has no effect on the position of the chemical equilibrium, since a catalyst affects the rates of the forward and reverse reactions equally.

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chemical equilibrium

Technology

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equilibrium expression

equilibrium constantexpression

concentrations of reactants

kcrtn kc

kc rtngasn

equilibriumconstant

reaction direction