Chemical Equilibrium Chapter 14

Equilibrium: the extent of a reaction In stoichiometry we talk about theoretical yields, and the many reasons actual yields may be lower. Another critical reason actual yields may be lower is the reversibility of chemical reactions: some reactions may produce only 70% of the product you may calculate they ought to produce. Equilibrium looks at the extent of a chemical reaction.

Equilibrium is a state in which there are no observable changes as time goes by. Chemical equilibrium is achieved when: the rates of the forward and reverse reactions are equal and the concentrations of the reactants and products remain constant Physical equilibrium H 2 O (l) Chemical equilibrium N2O4 (g)N2O4 (g) 14.1 H 2 O (g) 2NO 2 (g)

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The Concept of Equilibrium Consider colorless frozen N 2 O 4. At room temperature, it decomposes to brown NO 2 : N 2 O 4 (g)  2NO 2 (g). At some time, the color stops changing and we have a mixture of N 2 O 4 and NO 2. Chemical equilibrium is the point at which the rate of the forward reaction is equal to the rate of the reverse reaction. At that point, the concentrations of all species are constant. Using the collision model: –as the amount of NO 2 builds up, there is a chance that two NO 2 molecules will collide to form N 2 O 4. –At the beginning of the reaction, there is no NO 2 so the reverse reaction (2NO 2 (g)  N 2 O 4 (g)) does not occur.

The Concept of Equilibrium As the substance warms it begins to decompose: N 2 O 4 (g)  2NO 2 (g) When enough NO 2 is formed, it can react to form N 2 O 4 : 2NO 2 (g)  N 2 O 4 (g). At equilibrium, as much N 2 O 4 reacts to form NO 2 as NO 2 reacts to re-form N 2 O 4 The double arrow implies the process is dynamic.

The Concept of Equilibrium As the reaction progresses –[A] decreases to a constant, –[B] increases from zero to a constant. –When [A] and [B] are constant, equilibrium is achieved.

N 2 O 4 (g) 2NO 2 (g) Start with NO 2 Start with N 2 O 4 Start with NO 2 & N 2 O 4 equilibrium 14.1

constant

The Equilibrium Constant No matter the starting composition of reactants and products, the same ratio of concentrations is achieved at equilibrium. For a general reaction the equilibrium constant expression is where K c is the equilibrium constant.

N 2 O 4 (g) 2NO 2 (g) = 4.63 x K = [NO 2 ] 2 [N 2 O 4 ] aA + bB cC + dD K = [C] c [D] d [A] a [B] b Law of Mass Action K >> 1 K << 1 Lie to the rightFavor products Lie to the leftFavor reactants Equilibrium Will 14.1

Homogenous equilibrium applies to reactions in which all reacting species are in the same phase. N 2 O 4 (g) 2NO 2 (g) K c = [NO 2 ] 2 [N 2 O 4 ] K p = NO 2 P2P2 N2O4N2O4 P In most cases K c  K p aA (g) + bB (g) cC (g) + dD (g) 14.2 K p = K c (RT) Δn Δn = moles of gaseous products – moles of gaseous reactants = (c + d) – (a + b)

Homogeneous Equilibrium CH 3 COOH (aq) + H 2 O (l) CH 3 COO - (aq) + H 3 O + (aq) K c = ‘ [CH 3 COO - ][H 3 O + ] [CH 3 COOH][H 2 O] [H 2 O] = constant K c = [CH 3 COO - ][H 3 O + ] [CH 3 COOH] =K c [H 2 O] ‘ General practice not to include units for the equilibrium constant. 14.2

The Equilibrium Expression Write the equilibrium expression for the following reaction:

The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl 2 (g) at 74 0 C are [CO] = M, [Cl 2 ] = M, and [COCl 2 ] = 0.14 M. Calculate the equilibrium constants K c and K p. CO (g) + Cl 2 (g) COCl 2 (g) Kc =Kc = [COCl 2 ] [CO][Cl 2 ] = x = 220 K p = K c (RT) Δn Δn = 1 – 2 = -1R = T = = 347 K K p = 220 x ( x 347) -1 =

The equilibrium constant K p for the reaction is 158 at 1000K. What is the equilibrium pressure of O 2 if the P NO = atm and P NO = atm? 2 2NO 2 (g) 2NO (g) + O 2 (g) 14.2 K p = 2 P NO P O 2 P NO 2 2 POPO 2 = K p P NO POPO 2 = 158 x (0.400) 2 /(0.270) 2 = 347 atm

Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. CaCO 3 (s) CaO (s) + CO 2 (g) K c = ‘ [CaO][CO 2 ] [CaCO 3 ] [CaCO 3 ] = constant [CaO] = constant K c = [CO 2 ] = K c x ‘ [CaCO 3 ] [CaO] K p = P CO 2 The concentration of solids and pure liquids are not included in the expression for the equilibrium constant. 14.2

P CO 2 = K p CaCO 3 (s) CaO (s) + CO 2 (g) P CO 2 does not depend on the amount of CaCO 3 or CaO 14.2

Writing Equilibrium Constant Expressions The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm. The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions. The equilibrium constant is a dimensionless quantity. In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature. If a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. 14.2

Calculating Equilibrium Concentrations 1.Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration. 2.Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x. 3.Having solved for x, calculate the equilibrium concentrations of all species. 14.4

At C the equilibrium constant (K c ) for the reaction Is 1.1 x If the initial concentrations are [Br 2 ] = M and [Br] = M, calculate the concentrations of these species at equilibrium. Br 2 (g) 2Br (g) Let x be the change in concentration of Br 2 Initial (M) Change (M) Equilibrium (M) x-x+2x x x [Br] 2 [Br 2 ] K c = ( x) x = 1.1 x Solve for x 14.4

K c = ( x) x = 1.1 x x x = – x 4x x = 0 ax 2 + bx + c =0 -b ± b 2 – 4ac  2a2a x = Br 2 (g) 2Br (g) Initial (M) Change (M) Equilibrium (M) x-x+2x x x x = x = At equilibrium, [Br] = x = Mor M At equilibrium, [Br 2 ] = – x = M 14.4

Example Problem: Calculate Concentration Note the moles into a L vessel stuff... calculate molarity. Starting concentration of HI: 2.5 mol/10.32 L = M 2 HI H 2 + I 2 Initial: Change: Equil: M00 -2x+x+x xxx What we are asked for here is the equilibrium concentration of H otherwise known as x. So, we need to solve this beast for x.

Example Problem: Calculate Concentration And yes, it’s a quadratic equation. Doing a bit of rearranging: x = or – Since we are using this to model a real, physical system, we reject the negative root. The [H 2 ] at equil. is M.

Example Problem: Calculate Keq This type of problem is typically tackled using the “three line” approach: 2 NO + O 2 2 NO 2 Initial: Change: Equilibrium:

Approximating If Keq is really small the reaction will not proceed to the right very far, meaning the equilibrium concentrations will be nearly the same as the initial concentrations of your reactants – x is just about 0.20 is x is really dinky. If the difference between Keq and initial concentrations is around 3 orders of magnitude or more, go for it. Otherwise, you have to use the quadratic.

Example Initial Concentration of I 2 : 0.50 mol/2.5L = 0.20 M I 2 2 I Initial change equil: x +2x 0.20-x 2x With an equilibrium constant that small, whatever x is, it’s near dink, and 0.20 minus dink is 0.20 (like a million dollars minus a nickel is still a million dollars) – x is the same as 0.20 x = 3.83 x M More than 3 orders of mag. between these numbers. The simplification will work here.

Example Initial Concentration of I 2 : 0.50 mol/2.5L = 0.20 M I 2 2 I Initial change equil: x +2x 0.20-x 2x These are too close to each other x will not be trivially close to 0.20 here. Looks like this one has to proceed through the quadratic...

A + B C + D C + D E + F A + B E + F K c = ‘ [C][D] [A][B] K c = ‘ ‘ [E][F] [C][D] [E][F] [A][B] K c = KcKc ‘ KcKc ‘‘ KcKc KcKc ‘‘ KcKc ‘ x If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. 14.2

N 2 O 4 (g) 2NO 2 (g) = 4.63 x K = [NO 2 ] 2 [N 2 O 4 ] 2NO 2 (g) N 2 O 4 (g) K = [N 2 O 4 ] [NO 2 ] 2 ‘ = 1 K = 216 When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant. 14.2

The reaction quotient (Q c ) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (K c ) expression. IF Q c > K c system proceeds from right to left to reach equilibrium Q c = K c the system is at equilibrium Q c < K c system proceeds from left to right to reach equilibrium 14.4

If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position. Le Châtelier’s Principle Changes in Concentration N 2 (g) + 3H 2 (g) 2NH 3 (g) Add NH 3 Equilibrium shifts left to offset stress 14.5

Le Châtelier’s Principle Changes in Concentration continued ChangeShifts the Equilibrium Increase concentration of product(s)left Decrease concentration of product(s)right Decrease concentration of reactant(s) Increase concentration of reactant(s)right left 14.5 aA + bB cC + dD Add Remove

Le Châtelier’s Principle Changes in Volume and Pressure A (g) + B (g) C (g) ChangeShifts the Equilibrium Increase pressureSide with fewest moles of gas Decrease pressureSide with most moles of gas Decrease volume Increase volumeSide with most moles of gas Side with fewest moles of gas 14.5

Le Châtelier’s Principle Changes in Temperature ChangeExothermic Rx Increase temperatureK decreases Decrease temperatureK increases Endothermic Rx K increases K decreases 14.5 colder hotter

uncatalyzedcatalyzed 14.5 Catalyst lowers E a for both forward and reverse reactions. Catalyst does not change equilibrium constant or shift equilibrium. Adding a Catalyst does not change K does not shift the position of an equilibrium system system will reach equilibrium sooner Le Châtelier’s Principle

Example

Chemistry In Action Life at High Altitudes and Hemoglobin Production K c = [HbO 2 ] [Hb][O 2 ] Hb (aq) + O 2 (aq) HbO 2 (aq)

Chemistry In Action: The Haber Process N 2 (g) + 3H 2 (g) 2NH 3 (g)  H 0 = kJ/mol

Le Châtelier’s Principle ChangeShift Equilibrium Change Equilibrium Constant Concentrationyesno Pressureyesno Volumeyesno Temperatureyes Catalystno 14.5