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Chapter 14. Chemical Equilibrium. Chemical Equilibrium. In a chemical system, when the concentrations of both reactants and products reach a steady state, they are said to be in a dynamic equilibrium . This occurs when both the forward rate and the reverse rate are equal.
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Chapter 14 Chemical Equilibrium
Chemical Equilibrium • In a chemical system, when the concentrations of both reactants and products reach a steady state, they are said to be in a dynamic equilibrium. • This occurs when both the forward rate and the reverse rate are equal. • Reactants are still making products, but products are also making reactants. • Static equilibrium is when no change is occurring – like in a tug-o-war.
Chemical Equilibrium • Consider the equilibrium reaction of: N2O4(g) 2 NO2(g) • When equilibrium is established (no matter the starting amounts!), the forward and reverse rates are equal.
Chemical Equilibrium • This DOES NOT imply that the concentrations are equal! • Rather only the amounts remain constant. • Use of a double arrow () indicates an equilibrium.
Chemical Equilibrium • Analogy with traffic flow on a bridge. • More traffic coming in to the city than leaving. • But, if the rates coming in are constant and the rates going out are constant, then it is a dynamic equilibrium.
Equilibrium Constant, K • For the N2O4 reaction, the forward rate is: Ratef = kf [N2O4]1 • The reverse rate is: Rater = kr [NO2]2 • Setting these equal to each other: kf [N2O4]1 = kr [NO2]2
Equilibrium Constant, K • Re-arranging to: • Where the ratio of kr / kf = K, the equilibrium constant.
K • This concept of the equilibrium is also referred to as the law of mass action. • In general, for a reaction of the type: aA + bB cC + dD • The equilibrium constant will be:
K • Equilibrium amounts can be in moles per liter (Molarity) or as partial pressures, if gases are present. • The equilibrium constant, K, then is labeled as Kc or Kp. • As we do more types of equilibria, you will also see Ka, Kb, Ksp, etc. • K is always unitless, regardless of any reaction.
K • How do we prove that the law of mass action works? • Experiments, experiments, experiments.
K • Experiments 3 and 4 show that equilibrium can be achieved from either direction. • Experiment 3 starts with all products and Experiment 4 starts with all reactants.
K • Are Kc and Kp the same? • Recall that PV = nRT from Chapter 10. • Molarity = n / V, so… • P = M(RT) • Resulting in: • Kp = Kc(RT)Dn • Where Dn =
K • The magnitude of K can be used to predict whether the reactants or products are favored. • When K << 1, then the _____________ are predominant. • When K >> 1, then the _____________ are predominant.
K • What happens to K if a reaction is reversed? • N2O4 2 NO2 ; Kc = 0.212 • 2 NO2 N2O4 ; Kc = ??? • What happens to K if a reaction is multiplied by a factor? • 2 N2O4 4 NO2 ; Kc = ???
K • What happens when two reactions are added together like in Hess’ Law? • 2 NOBr 2 NO + Br2 ; Kc = 0.014 • Br2 + Cl2 2 BrCl ; Kc = 7.2 • ___________________________________ • 2 NOBr + Cl2 2 NO + 2 BrCl ; Kc = ???
K • When all substances in an equilibrium are of the same phase, then that equilibrium is homogeneous. • When two different phases are present, the equilibrium is said to be heterogeneous. • The amounts of solids and/or liquids in the presence of gases or aqueous compounds are in large excess. • Thus, their concentrations do not change.
K Consider the equilibrium for: CaCO3(s) CaO(s) + CO2(g)
Evaluating K • If given equilibrium amounts of all reactants and products, then a value for Kc can be found. • Or – can find an unknown concentration if Kcis known. • Can also use basic stoichiometric relationships in some problems. • These require an ICE table.
Evaluating K • ICE stands for Initial, Change, and Equilibrium. • Set-up as a table under the reaction. • Initial = starting amounts in Molarity. • Change = includes a sign plus a variable and reflects mole-to-mole amounts. • Equilibrium = combines the I and C lines. • Suppose a 4.0L flask has 0.20 moles of N2O4 initially added to it, the ICE table set-up would be:
Evaluating K • N2O4 2 NO2
Evaluating K • If the equilibrium amounts of the reactants and products are unknown, then an equation with a variable must be set-up and solved. • Must be given the value of K. • Two common problems are the perfect square problem and the quadratic problem.
Evaluating K • Problems with partial pressures are based on Dalton’s Law of Partial Pressures. • Ptotal = Pa + Pb + Pc + … • Thus, if the intital total pressure is known, then one gas has a pressure of “x” and the other is this pressure minus “x”.
Reaction Quotient • Given amounts of all reactants and products, which direction will the reaction move? • The Q expression is identical to K. • Can think of Q as a continuum. Q = 0, all reactants Q = , all products
Reaction Quotient • When Q < K, the system has too many reactants and not enough products.
Reaction Quotient • When Q > K, the system has too many products and not enough reactants.
Reaction Quotient • When Q = K, the system is in equilibrium.
LeChatelier’s Principle • If a system in equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position to counteract the disturbance. • Certain reactions, like the production of NH3 require the manipulation of an equilibrium to maximize the product.
LeChatelier’s Principle • Haber process – Fritz Haber looked at P, T effects on making NH3. • Reaction favors lower T and higher P. • Why?
LeChatelier’s Principle • Changes in reactant or product concentrations will force the equilibrium to re-balance just like a two pan balance. • Can add or remove components in the equilibrium system.
LeChatelier’s Principle • N2 + 3 H2 2 NH3 • Add some H2 will result in: • Removing NH3 as it is made will:
LeChatelier’s Principle • Pressure changes • As Boyle’s Law predicts, as P increases, V decreases. • For a system to reduce its volume, it must reduce the moles of gas present. • For reactions with different moles of products and moles of reactants, increasing the pressure will favor the side with fewer total moles of gases. • Of course, the opposite is true if the pressure is decreased. • Does this agree with Haber results?
LeChatelier’s Principle • Changing the pressure at constant temperature does NOT change the value of K. • As the total pressure is increased, the relative partial pressures increase, but not equally. • Thus, the K value remains the same.
LeChatelier’s Principle • Temperature changes • Depends on whether the reaction is endo- or exo-thermic. • Endothermic: Reactants + Heat Products • As T increases, reaction shifts to right and K will ___________ • Exothermic: Reactants Products + Heat • As T increases, reaction shifts to left and K will ____________
Is the reaction of: N2O4(g) 2 NO2(g), Exothermic or Endothermic?
LeChatelier’s Principle • If a catalyst is added to an equilibrium system, then it will speed up the reaction in both directions. • The equilibrium concentration, though, is unchanged.
Ammonia Production • N2(g) + 3H2(g) 2NH3(g); DH = -92 kJ • Favors high or low temperatures? • Favors high or low pressures? • Removal of NH3 as it is made • Catalyst – iron oxides • Over 100 million tons produced per year using 1-2% of the annual world’s energy supply. • Plants and some bacteria can perform the same reaction at much lower temperatures and pressures.