chemical equilibrium

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Chemical Equilibrium. The Equilibrium Condition. For all of your chemistry career, especially when performing stoichiometry calculations, we assumed that all reactions proceed to completion . In other words, until one of the reactants ran out. - PowerPoint PPT Presentation

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Page 1: Chemical Equilibrium
Page 2: Chemical Equilibrium

•For all of your chemistry career, especially when performing stoichiometry calculations, we assumed that all reactions proceed to completion.•In other words, until one of the reactants ran out.

Page 3: Chemical Equilibrium

•Many reactions, however, do not run to completion.•Consider the dimerization of nitrogen dioxide:

NO2(g) + NO2(g) → N2O4(g)•The reactant, NO2, is a dark brown gas, and the product, N2O4, is a colorless gas.•Notice when NO2 is placed in a sealed container, the initial brown color decreases in intensity as it is converted to colorless N2O4. •However, the contents of the

container do not become colorless. •Instead the intensity of the brown color becomes constant, which mean the concentration of NO2 is no longer changing.

Page 4: Chemical Equilibrium

•This reaction is illustrated below.•The observation that the brown color does not completely fade is an indication the reaction has not reached completion (all NO2 is gone).•The system has come to equilibrium.•Chemical equilibrium: the state where the concentrations of all reactants and products remain constant with time.

Page 5: Chemical Equilibrium

•Consider the flow of cars across a bridge connecting two cities with the traffic flow in both directions even.•There is motion since the cars are moving, but the number of cars in each city is not changing because equal numbers of cars are entering and leaving.

•The result is no net change in the car population.

Page 6: Chemical Equilibrium

•The same concept applies to a reaction at equilibrium.•It may appear that everything has stopped since no changes occur in the concentrations of reactants or products.•However, on the molecular level, there is frantic activity because both the forward and reverse reactions are occurring at an equal rate.

Page 7: Chemical Equilibrium

•Consider the reaction between H2O and CO: H2O(g) + CO(g) ⇌ H2(g) + CO2(g)

⇌ (double arrows) indicate reaction can occur in either direction

In (a) H2O and CO are mixed in equal numbersand begin to react (b) to form CO2 and H2.After time has passed, equilibrium is reached (c)and the numbers of reactant and product molecules then remain constant over time (d).

Page 8: Chemical Equilibrium

H2O(g) + CO(g) ⇌ H2(g) + CO2(g)•Once equilibrium occurs, the concentration of the reactants and products remain unchanged.•Unless the system is disturbed, no further changes in concentrations will occur.•Remember that the reaction is still occurring but the forward and reverse rates are equal.

Page 9: Chemical Equilibrium

•In 1864 Guldberg and Waage proposed the law of mass action as a general description of the equilibrium condition.•For a reaction of the type

jA + kB ⇌ lC = mDwhere A, B, C, and D represent chemical species and j, k, l, and m are their coefficients, the law of mass action is represented by the following equilibrium expression:

•The square brackets indicate the concentration of the chemical species at equilibrium and K is a constant called the equilibrium constant.

kj

ml

K]B[]A[]D[]C[

Page 10: Chemical Equilibrium

Summary of Conclusions about the Equilibrium Expression:•The equilibrium expression for a reaction is the reciprocal of that for the reaction written in reverse.

•When the balanced equation is multiplied by a factor n, the equilibrium expression for the new reaction is the original expression raised to the nth power. Thus Knew = (Koriginal)n.

•K values are customarily written without units.

Page 11: Chemical Equilibrium

•Important!! For a given reaction at a certain temperature, the value of K is the same regardless of the amounts of the reactants that are mixed together individually.•The value of K depends on the ratio of the concentrations which remains the same (within experimental error).•The subscript 0 indicate initial concentrations.•Each set of equilibrium concentrations is called an equilibrium position.•Only one equilibrium constant for a particular reaction at a particular temperature but an infinite number of equilibrium positions.

Page 12: Chemical Equilibrium

Pressure point massage

•Equilibria involving gases can also be described in terms of pressure.•Remember the ideal gas law?

•PV = nRTor

CRTRTVn

P

where C equals n/V, or the number of moles n of gas per unit volume V. Thus C represents the molar concentration of the gas.

Page 13: Chemical Equilibrium

N2(g) + 3H2(g) ⇌ 2NH3(g)

or in terms of the equilibrium partial pressures of the gas, that is,

Both K and Kc are used commonly for equilibrium constants in terms of concentration (your book always uses K). Kp represents an equilibrium constant in terms of partial pressures.

c3

HN

2

NH

3

22

2

3

))((]H][N[]NH[

22

3 KCC

CK

))(( 3

HN

2

NH

p

22

3

PP

PK

Page 14: Chemical Equilibrium

•For a general reactionjA + kB ⇌ lC + mD

the relationship between K and Kp isKp = K(RT)Δn

where Δn = (l + m) – (j + k), the difference in the sums of the coefficients for the gaseous products and reactants.

Page 15: Chemical Equilibrium

•Homogeneous equilibria: where all reactants and products are in the gaseous phase.•Heterogeneous equilibria: involves more than one phase.•Consider the reaction below:

CaCO3(s) ⇌ CaO(s) + CO2(g)

•Experimental results show that the position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present.

Page 16: Chemical Equilibrium

•Fundamental reason: concentrations of pure solids or liquids present cannot change.•For any pure liquid or solid, the ratio of the amount of substance to volume of substance is a constant.•Therefore the equilibrium expression for

CaCO3(s) ⇌ CaO(s) + CO2(g)is written as

K = [CO2]