Chapter 14 : Chemical Equilibrium Essentials of General Chemistry Ebbing • Gammon • Ragsdale 2nd Edition
Dr. Azra Ghumman Memorial University of Newfoundland Chem 1011 Winter 2008
Chapter 14 Chemical Equilibrium Dr. Ghumman
Chapter 14 : Chemical Equilibrium 14.1
Chemical Equilibrium
14.2 14.3
The Equilibrium Constant Heterogeneous Equilibria: Solvents in Homogeneous Equilibria Qualitatively Interpreting the Equilibrium Constant Predicting the Direction of Reaction Calculating Equilibrium Concentration Removing Products or Adding Reactants Changing the Pressure or Adding Reactants Changing the Pressure and Temperature Effect of a Catalyst
14.4 14.5 14.6 14.7 14.8 14.9 14.10
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Chemical Equilibrium Some reactions go to completion and represented by a single arrow e.g. Catalytic methanation CO(g) + 3H2(g) Æ CH4(g) + H2O (g) Steam reforming CH4 (g) + H2O (g) Æ CO(g) + 3H2(g) Reversible reaction Some reactions do not go to completion and as soon as some products form, they react to yield back reactants. So the reaction mixture contain both reactants and products. CO(g) + 3H2(g) CH4 + H2O (g) • Chemical equilibrium: is the state reached by a reaction mixture when the rates of the forward and reverse reactions have become equal. Chem 1011 Winter 2008
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Dynamic equilibrium • Dynamic equilibrium. – A dynamic equilibrium consists of a forward reaction, in which substances react to give products, and a reverse reaction, in which products react to give back the original reactants. • Characteristics of dynamic equilibrium – The concentrations of reactants and products remain constant over time (at constant T and P). – There are no visible changes in the system. – At molecular level, reactant molecules continue to form products while product molecules yield reactant molecules. Chem 1011 Winter 2008
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Dynamic equilibrium of water Dynamic equilibrium of water in a closed container. H2O(l) H2O (g) (physical equilibrium) • At equilibrium rate of condensation = rate of evaporation – It is the characteristic of the dynamic equilibrium. • Once the equilibrium is established the number of molecules per unit volume in vapor does not change with time.
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Catalytic Methanation Reaction Equilibrium At Equilibrium, CO(g) + 3H2(g) CH4 + H2O (g) • rate of forward reaction = rate of reverse reaction
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A Dynamic Equilibrium Catalytic methanation reaction CO(g) + 3H2(g) CH4 + H2O (g) • Initially the rate of forward reaction is large and rate of reverse reaction is zero, with time reverse happens and a state is reached when both forward and reverse rate becomes equal – It establishes an equilibrium state where all four species are present.
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Applying Stoichiometry to an Equilibrium Mixture • Suppose we place 1.000 mol N2 and 3.000 mol H2 in a reaction vessel at 450oC and 10.0 atmospheres of pressure. The reaction is:
N 2 ( g ) + 3H 2 ( g )
2NH 3 (g)
– What is the composition of the equilibrium mixture if it contains 0.080 mol NH3?
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The Equilibrium Constant • Equilibrium concentrations (or partial pressures) for a reaction at a given temperature are related by a quantity called the equilibrium constant • Equilibrium Constant Kc is the numerical value obtained for the equilibrium constant expression when equilibrium concentrations (partial pressures)are substituted.
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The Equilibrium Constant • Equilibrium Constant Expression Kc for a reaction is an expression obtained by multiplying the concentrations of the products, dividing by the concentrations of the reactants, and raising each concentration term to a power equal to the coefficient in the chemical equation. – Product conc. appear in the numerator and reactant conc. in the denominator. – Coefficients in the balanced equation are the exponents of reactants and products concentrations in balanced equation. Chem 1011 Winter 2008
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Writing the Equilibrium Constant Expression Kc • For the following general reaction
aA + bB
cC + dD
•The equilibrium constant expression is
[C]c [ D]d Kc = [ A ]a [B ]b The molar concentration of a substance is denoted by writing its formula in square brackets.The exponents are the coefficients in the balanced equation •In Kc, the subscript ‘c’ means that it is defined in terms of concentrations Chem 1011 Winter 2008
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The Law of Mass Action The Law of Mass Action is a relation that states that the value of the equilibrium-constant expression Kc is constant for a particular reaction at a given temperature, whatever equilibrium concentrations are substituted. – A large Kc indicates large concentrations of products at equilibrium. – A small Kc indicates large concentrations of unreacted reactants at equilibrium. • An early term used for the concentration was ‘Active mass’ hence the term mass action is used. Chem 1011 Winter 2008
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Calculating Kc • Equilibrium concentrations for a reaction must be obtained experimentally and then substituted into the equilibriumconstant expression in order to calculate Kc • Consider the catalytic methanation reaction below ( see Fig 14.4)
CO ( g ) + 3 H 2 (g)
CH 4 (g) + H 2 O(g)
• Suppose we started with initial concentrations of CO and H2 of 0.100 M and 0.300 M, in a 10.0 L flask respectively.
– When the system finally settled into equilibrium we determined the equilibrium concentrations to be as follows: Chem 1011 Winter 2008
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Figure 14.4: Some Equilibrium Compositions for the Methanation Reaction at 1200K
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Calculating Kc •
The equilibrium-constant expression for methanation reaction
Kc = •
By substituting equilibrium concentrations from Expt # 1,(Fig. 14.4) we can obtain
Kc =
•
[CH 4 ][ H 2O ] [CO ][ H 2 ]3
( 0 . 0387 M )( 0 . 0387 M ) = 3 . 93 ( 0 . 0613 M )( 0 . 1839 M ) 3
There are no units with Kc values Why?
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Calculating Kc • Regardless of the initial concentrations (whether they be reactants or products), the law of mass action dictates that the reaction will always settle into an equilibrium where the equilibrium-constant expression equals Kc. – Whether we start with reactants or products, the system establishes the same ratio.
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Relationship between chemical equilibrium and kinetics • Kc is the ratio of the rate constants for forward and reverse reaction • If the forward and reverse reaction rates in a system at equilibrium are equal, then it follows that their rate laws would be equal. • Consider the decomposition of N2O4, dinitrogen tetroxide.
N 2O 4 ( g )
2NO 2 (g)
• We can write the equation with rate constant for the forward and reverse reaction kf
N 2O 4 ( g )
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kr
2NO 2 (g)
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Equilibrium: A Kinetics Argument • We can call the decomposition of N2O4 the forward reaction and the formation of N2O4 the reverse reaction. • These are elementary reactions, and you can write the rate law for each elementary reaction as
Rate (forward) = k f [ N 2O 4 ] Rate (reverse) = k r [ NO 2 ]2
kf and kr are rate constants for forward and reverse rates
• At equilibrium, Rate (forward) = Rate (reverse) Chem 1011 Winter 2008
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Equilibrium: A Kinetics Argument •
Therefore we can write
k f [ N 2 O 4 ] = k r [ NO 2 ]2 •
Combing the constants we can identify the equilibrium constant, Kc, as the ratio of the forward and reverse rate constants.
Kc
k = f kr
[ NO 2 ]2 = [ N 2O 4 ]
• If value of kf > kr then forward reaction (products) is favoured and Kc>1 • If value of kf< kr then reverse reaction(reactants) is favoured and K< 1 • At equilibrium kf = kr and Kc = 1 Chem 1011 Winter 2008
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Gas Phase Equilibria • In discussing gas-phase equilibria, it is often more convenient to express concentrations in terms of partial pressures rather than molarities.( units in atm). • We can see from the ideal gas equation that the partial pressure of a gas is proportional to its molarity at constant temperature. PV = nRT and rearranging for P
P = Chem 1011 Winter 2008
( n ) RT V
= MRT
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The Equilibrium Constant, Kp • Equilibrium constant KpIt is written in terms of partial pressures of gases instead of molar concentration (we use atm instead of mol L-1). For the methanation reaction
N 2 (g ) + 3H 2 (g )
2NH
3 (g)
2
Kp =
(PNH3) (PN2)(PH2 )3
• As long as the temperature and volume of the container remains unchanged, the partial pressure of the gases (equilibrium) remains unchanged with time Chem 1011 Winter 2008
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Relationship between Kc and Kp From the relationship n/V=P/RT, we can show that: Kp = Kc(RT)∆n where ∆n = sum of the moles of products - the sum of the moles of reactants. R = gas constant (0.0821 L.atm mol-1 K-1) K = temperature in kelvin (K) In general, In general Kp ≠ Kc Kp = Kc(0.0821xT)∆n but when ∆n = 0 then Kp = Kc Chem 1011 Winter 2008
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Example • Consider the reaction
2SO 2 (g ) + O 2 (g )
2 SO 3 (g)
– Kc for the reaction is 2.8 x 102 at 1000 K. Calculate Kp for the reaction at this temperature.
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Calculating the value of Kc • From equilibrium concentrations (simple case) – Substitute values in Kc expression and calculate. • From initial concentration and one equilibrium concentration. – Or from initial concentrations only.
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Changing the equilibrium equation and Kc • The expression of Kc depends upon the form of the chemical equation written to describe the equilibrium system. • K is meaningless unless accompanied by a chemical equation. there are two general rules • Coefficient rule: If the coefficient of the balanced equation is multiplied by a factor n, the equilibrium constant is raised to the nth power • Consider the following reaction N2O4 (g) 2NO2 (g) – many other equations can be written for this reaction e.g.(next slide) Chem 1011 Winter 2008
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Changing the equilibrium equation and Kc 1. N2O4 (g)
2NO2 (g )
2. 2N2O4 (g)
4NO2 (g)
3. 1/2N2O4 (g) NO2 (g)
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K =
K′ =
K ′′ =
( PNO 2 ) 2 ( PN 2 O 4 )
( PNO2 ) 4 ( PN 2O4 ) 2 PNO2 ( PN 2 O4 )1/ 2
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=Κ
= Κ2 = K 1/2
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Reciprocal rule • Reciprocal rule:The equilibrium constant for the forward and reverse reactions are the reciprocal of each other. N2O4(g) 2NO2(g) K = 11 • By reversing the equation we can write 2NO2(g) N2O4(g) and K'= 1/11 • K for reverse equation is the reciprocal of K of original equation, the numerator and denominator have been inverted K' = 1/ K and if K = 11 • then K' = 1/ K = 1/11 = 0.091 For gaseous equilibria Kp is usually written as K Chem 1011 Winter 2008
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Equilibrium Constant for the Sum of Reaction • Rule of multiple equilibria states, if a reaction can be expressed as the sum of two or more reactions (as in Hess’s Law), K for the overall reaction is the product of the equilibrium constants of the individual reactions. • It means if the reaction 3 = reaction 1 + reaction 2 – The K3 = K1 x K2 •
Consider the following reaction at 1200K
1. CO (g) + 3H2 (g) CH4 (g) + H2O (g) 2. CH4(g) + 2H2S (g) CS2 (g) +4H2 (g)
K1 =3.92 K2 = 3.3 x 104
3. CO(g) + 2H2S (g) CS2 + H2O(g)+ H2 (g) K3 = 1.3 x 105 Chem 1011 Winter 2008
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Multiple Equilibria According to rule K3 = K1 x K2 this is easy to verify
K 1K 2 =
K3 =
( PCH 4 )( PH 2 O ) ( PCO )( PH 2 ) 3
K1K 2 =
x
( PH 2 ) 4 ( PCS 2 ) ( PCH 4 )( PH 2 S ) 2
( PH 2 O )( PH 2 )( PC S 2 ) ( PC O )( PH 2 S ) 2
K3 = 3.92 x (3.3 x 104) Chem 1011 Winter 2008
•
= 1.3 x 105
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Homogeneous Equilibria • A Homogeneous equilibria is an equilibrium that involves reactants and products in a single phase. • For example synthesis of NH3 in Haber process
N 2 (g ) + 3H 2 (g )
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2NH 3 (g)
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Heterogeneous Equilibria; Solvents in Homogeneous Equilibria • A Heterogeneous equilibrium is an equilibrium that involves reactants and products in more than one phase. – The equilibrium of a heterogeneous system is unaffected by the amounts of pure solids or liquids present, as long as some of each is present. – The concentrations of pure solids and liquids are always considered to be “1” and therefore, do not appear in the equilibrium expression. • In thermodynamic equilibrium constant, the activity of pure liquid and pure solid is 1 Chem 1011 Winter 2008
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Writing Kc for reactions with pure solids or Liquids • Consider the following reaction
C(s) + H 2O(g )
CO(g) + H 2 (g)
–The equilibrium-constant expression contains terms for only those species in the homogeneous gas phase (H2O, CO, and H2).
Kc =
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[ CO ][ H 2 ] [ H 2O ]
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Rules to express K for any system • In the expression for K • Gases enter as their partial pressures in atmospheres • Pure liquids or solids do not appear in the eqilibrium expression. • Species in water (ions or molecules) in water solution enter as their molar concentrations
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Applications of Equilibrium Constant (Kc) 1. Qualitative interpretation of equilibrium constant –
Reaction favours products or reactants?
2. Predicting the direction of the reaction – in which direction reaction will proceed 3. Calculating equilibrium concentrations for any set of starting concentrations using value of Kc.
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Qualitatively interpreting Kc
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Qualitatively interpreting Kc • Kc is the ratio of product concentrations to reactant concentrations at equilibrium, hence – if the value of Kc is large, the products are favoured at equilibrium. – if the value of Kc is small, the reactants are favoured at equilibrium • e.g. Kc for the following reaction at 25°C = 4.1 x 108
N 2 (g ) + 3H 2 (g )
2NH 3 (g)
It means the product conc. is 4.1 x 108 times higher than the reactant concentration. • The value of Kc is temperature dependent. Chem 1011 Winter 2008
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Predicting the Direction of Reaction • The reaction quotient, Qc is an expression that has the same form as the equilibrium-constant expression but whose concentrations are not necessarily those at equilibrium. • For a general reaction
aA + bB • The Qc expression is
cC + dD
[ C ]ci [ D ]di Qc = [ A ]ai [ B ]bi
• The subscript ‘i’ indicates the conc. at a particular instant Chem 1011 Winter 2008
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Predicting the direction of the reaction; Reaction Quotient •
To predict the direction of reaction, compare the value of Qc to equilibrium constant Kc then
•
If Qc > Kc the reaction will go to the left (R)
•
If Qc< Kc the reaction will go to the right (P)
•
If Qc = Kc the reaction mixture is at equilibrium
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Example • Consider the following equilibrium:
N 2 ( g ) + 3H 2 ( g )
2NH 3 (g)
• A 50.0 L vessel contains 1.00 mol N2, 3.00 mol H2, and 0.500 mol NH3. In which direction (toward reactants or toward products) will the system shift to re-establish equilibrium at 400°C? – Kc for the reaction at 400 °C is 0.500. Chem 1011 Winter 2008
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Calculating Equilibrium Concentrations • The Kc for a reaction can be used to calculate the concentrations of substances in an equilibrium mixture. 1. The simplest case, when one equilibrium concentration (or partial pressure) is calculated knowing all the others. 2. Kc is used to calculate equilibrium concentrations (or partial pressures) of reactants and products, when original concentrations (partial pressures) of reactants and products are known.
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Steps involved in equilibrium problem solving
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Calculating Equilibrium Concentrations • Calculating Equilibrium Concentrations of one species when the equlibrium conc. of all others and the value of Kc is known. • For example, consider the following equilibrium:
CO(g ) + 3 H 2 (g)
CH 4 (g) + H 2O(g)
– Suppose a gaseous mixture contained 0.30 mol L-1 CO, 0.10 mol L-1 H2, 0.020 mol L-1 H2O, and an unknown amount of CH4 L-1 . – What is the concentration of CH4 in this mixture? The equilibrium constant Kc equals 3.92. Chem 1011 Winter 2008
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Calculating Equilibrium Concentrations •
First, calculate concentrations from moles of substances.
CO ( g ) + 3 H 2 (g) 0.30 mol
CH 4 (g) + H 2O(g)
0.10 mol
??
0.020 mol
Write the equilibrium constant expression for the reaction
K
c
=
[ CH 4 ][ H 2 O ] [ CO ][ H 2 ] 3
–Substituting the known concentrations and the value of Kc gives: Chem 1011 Winter 2008
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Example 3.92 =
[CH 4 ](0.020M ) ( 0.30M )(0.10M )3
•You can now solve for [CH4].
[CH4 ] =
(3.92)(0.30M)(0.10M)3 = 0.059 (0.020M)
–The concentration of CH4 in the mixture is 0.059 mol L-1. Chem 1011 Winter 2008
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Calculating Equilibrium Concentrations when the Initial concentrations are given • Suppose we begin a reaction with known amounts of starting materials and want to calculate the quantities at equilibrium. • Consider the following equilibrium.
CO ( g ) + H 2O ( g )
CO 2 (g) + H 2 (g)
– Suppose you start with 1.000 mol each of carbon monoxide and water in a 50.0 L container. Calculate the molarity of each substance in the equilibrium mixture at 1000 °C. – Kc for the reaction is 0.58 at 1000 °C Chem 1011 Winter 2008
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Solving an equilibrium problem • First, calculate the initial molarities of CO and H2O. • Let x be the moles per liter of product formed.
CO(g) + H 2O(g)
CO2 (g) + H 2 (g)
Initial Change
0.0200 -x
0.0200 -x
0 +x
0 +x
Equilibrium
0.0200-x
0.0200-x
x
x
•The equilibrium constant for the reaction is
Kc = Chem 1011 Winter 2008
[ CO 2 ][ H 2 ] [ CO ][ H 2 O ]
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Solving an equilibrium problem – Substituting the values for equilibrium concentrations, we get:
0.58 =
( x )( x ) (0.0200 − x )(0.0200 − x )
x2 0 . 58 = ( 0 . 0200 − x ) 2 –Taking the square root of both sides we get: 0 . 76 =
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x ( 0 . 0200 − x )
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Solving an equilibrium problem – Rearranging to solve for x gives:
x =
0 . 0200 × 0 . 76 = 0 . 0086 1 . 76
– If you substitute for x in the last line of the table you obtain the following equilibrium concentrations:
0.0114 mol L-1 CO 0.0114 mol L-1H2O
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0.0086 mol L-1 CO2 0.0086 mol L-1 H2
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Solving an equilibrium Problem involving Quadratic equation • In some cases it is necessary to solve a quadratic equation to obtain equilibrium concentrations. • The next example illustrates how to solve such an equation ( will be solved in class). • Consider the following equilibrium. H 2 (g ) + I 2 (g ) 2 HI(g) – Suppose 1.00 mol H2 and 2.00 mol I2 are placed in a 1.00-L vessel. How many moles per liter of each substance are in the gaseous mixture when it comes to equilibrium at 458°C? – Kc at this temperature is 49.7. Chem 1011 Winter 2008
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Changing the Reaction Conditions • Obtaining the maximum amount of product from a reaction depends on the proper set of reaction conditions. • Le Châtelier’s principle states that when a system in a chemical equilibrium is disturbed by a change of temperature, pressure, or concentration, the equilibrium will shift in a way that tends to counteract this change.
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Changing the Reaction Conditions • •
By changing these conditions, you can increase or decrease the yield of product You can alter the equilibrium composition of a gaseous reaction mixture in three ways and possibly increase the yield of the product 1. By removing product or adding reactant 2. Changing the partial pressure of gaseous reactant and products by changing the volume. 3. Changing the temperature
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Removing Products or Adding Reactants • By removing products (H2O vapour) from the reaction vessel, the reaction would shift to the right (products side) until the equilibrium was re-established.
– e.g. CO(g) + 3H2(g) Æ CH4(g) + H2O(g) – At equilibrium Qc = Kc – By removing H2O, Qc < Kc and reaction will proceed in forward direction to restore the equilibrium Chem 1011 Winter 2008
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Effect on Chemical Equilibrium of Changing the Concentration of a Substance
Figure 14.6: HgI2(s) + 2I-(aq) HgI42-(aq) if more reactant (I-)is added the reaction would again shift to the right (colour of HgI2 disappears)until equilibrium is re-established. Chem 1011 Winter 2008
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Effect of Pressure Change • A pressure change obtained by changing the volume of the reaction vessel can affect the yield of products in a gaseous reaction only if the reaction involves a change in the total moles of gas present • If the products in a gaseous reaction contain fewer moles of gas than the reactants, it is logical that they would require less space. – So, reducing the volume of the reaction vessel would, therefore, favor the products. • If the reactants require less volume (that is, fewer moles of gaseous reactant), then decreasing the volume of the reaction vessel would shift the equilibrium to the left (toward reactants). Chem 1011 Winter 2008
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Effect on Chemical Equilibrium of Changing the Pressure Figure 14.7 •Methanation reaction •By increasing the pressure,volume is decreased and products are favoured as moles of products are less compared to moles of reactants. Chem 1011 Winter 2008
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Effect of pressure change • Literally “squeezing” the reaction will cause a shift in the equilibrium toward the fewer moles of gas. • Immediate effect of compression is to increase the the concentration of the molecules. To counteract this reaction goes to the side with less number of moles (products). • In the event that the number of moles of gaseous product equals the number of moles of gaseous reactant, vessel volume will have no effect on the position of the equilibrium.
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Effect of Temperature Change • Reaction rates generally increase with an increase in temperature. Consequently, equilibrium is established sooner. – The numerical value of the equilibrium constant Kc varies with temperature. • For endothermic reaction , increase in temperature favours the products. • For an exothermic reaction, increase in temperature favours the reactants.
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Effect on Chemical Equilibrium of Changing the Temperature •Co(H2O)62+ (aq) + 4Cl-(aq) Æ CoCl42-(aq) +6H2O(l) • Figure 14.8 It is an endothermic reaction. At room temperature the equilibrium mixture is blue.
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Effect of Temperature Change •For an endothermic reaction, the amount of products are increased at equilibrium by an increase in temperature. •For an endothermic reaction ∆H° is positive and Kc is larger at higher T • For an exothermic reaction ∆H° is negative and Kc is larger at lower T
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Effect of Temperature Change • If we consider heat as a reactant in endothermic reactions and a product in exothermic reactions; – the increase in the temperature is analogous to adding more product (in the case of exothermic reactions) – the increase in the temperature is analogous to adding more reactant (in the case of endothermic reactions). • This ultimately has the same effect as if heat were a physical entity.
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Effect of Temperature Change • For example, consider the following generic exothermic reaction:
Reactants
Products + “heat” (∆ ∆H is negative)
• Increasing temperature would be analogous to adding more product, causing the equilibrium to shift left. – Since “heat” does not appear in the equilibriumconstant expression, this change would result in a smaller numerical value for Kc. – For an endothermic reaction
Reactants + “heat” Chem 1011 Winter 2008
Products (∆ ∆H is negative)
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Effect of a Catalyst • A catalyst has no effect on the equilibrium composition of a reaction mixture – A catalyst is a substance that increases the rate of a reaction but is not consumed by it. – A catalyst merely speeds up the attainment of equilibrium (increasing both forward and reverse rate. Even though catalyst doesn’t change the position of an equilibrium, a catalyst can significantly influence the choice of optimum conditions for a reaction. Chem 1011 Winter 2008
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Operational Skills • Applying stoichiometry to an equilibrium mixture • Writing equilibrium-constant expressions • Obtaining the equilibrium constant from reaction composition • Using the reaction quotient • Obtaining one equilibrium concentration given the others • Solving equilibrium problems • Applying Le Châtelier’s principle
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