Gaseous Equilibrium Gaseous equilibrium is the state at which the concentrations of reactants and products remain constant with time. This is a dynamic equilibrium because the rate of the forward reaction equals the rate of the reverse reaction.  Reactant molecules are converted to product molecules at the same rate as product molecules are converted to reactant molecules.

In the following slide, A is the reactant and [A] is its concentration.  Initially, only reactant molecules are present, so only reactant molecules can collide.  As the product concentration, [B], starts to increase, collisions will take place leading to the reverse reaction.  [A] decreases with time while [B] increases with time.

 At ≈ 650 s, [A] and [B] remain constant. At equilibrium, they need not be equal but must remain constant. What must be equal at equilibrium is the forward rate and the reverse rate of reaction.  (molL -1 s -1 ) f = (molL -1 s -1 ) r

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Two last important points:  The reactants and products are present at all times.  The equilibrium can be approached from either the forward or reverse reaction.

Equilibrium Constant/Expression For the chemical reaction at equilibrium and at constant temperature shown below, aA(g) + bB(g) cC(g) + dD(g) the following expression can be written: K eq =KpKp = P C c × P D d P A a × P B b or

K eq =KcKc = [C] c × [D] d [A] a × [B] b K p and K c are related by the expression: KpKp = KcKc (RT) Δ ng where Δ n g = moles of product gas – moles of reactant gas.

There are several noteworthy points:  The ratio of the product pressures to the reactant pressures is called the equilibrium expression.  The expression is also referred to as the law of mass action.  At constant temperature, the ratio remains constant and is symbolized by K eq.

 Because all the reactant and products are in one phase, it is called homogeneous equilibria.  When there is more than one phase present, it is called heterogeneous equilibria. The position of the equilibrium is independent of the amount of solid or liquid as long as some is present. Gases are entered as partial pressures.

Pure liquids, solids, and the solvent for a dilute solution do not appear in the equilibrium expression. Ions and molecules in aqueous solutions appear as their molarities. Write the equilibrium expression for the following reaction: 2PbS(s) + 3O 2 (g) 2PbO(s) + 2SO 2 (g) K p = P(SO 2 ) 2 /P(O 2 ) 3

 K eq is not affected by changes in concentration, a change in pressure, or by the addition of a catalyst.  K eq is only affected by a change of temperature. An increase in temperature favors the endothermic reaction.  A catalyst does not change the position of the equilibrium, but only speeds up the rate of both the forward and reverse reactions.

 If only a small amount of product is formed, the equilibrium lies far to the left and K << 1.  If only a small amount of reactant remains, the equilibrium lies far to the right and K >> 1.  When working with K values, there are no units. Each pressure or concentration is divided by 1 atm or 1 M. This prevents dealing with units of varying exponents.

Typical Equilibrium Problems Ammonium chloride is heated until it decomposes as shown in the following reaction, NH 4 Cl(s) NH 3 (g) + HCl(g) When equilibrium is established, the temperature is 202°C and the total pressure is 4.40 atm. (a) What is K p ?

T = 202°C P T = 4.40 atm P T = P(NH 3 ) + P(HCl) 4.40 = 2 × P(NH 3 ) P(NH 3 ) + P(HCl) = 2.20 atm NH 3 and HCl are in a 1:1 mole ratio, making their partial pressures equal.

K p = P(NH 3 ) × P(HCl) K p = = 4.84 (b) Ammonia is added to the system with the temperature held constant. When equilibrium is reestablished, the partial pressure of NH 3 is three times that of HCl. Determine the partial pressure of NH 3 and HCl.

KpKp = P(NH3) × P(HCl) 4.84 = 3x x = 3x 2 × P(NH 3 ) = 3.81 atm, P(HCl) = 1.27 atm (c) mole of NH 3 and mole of HCl are injected into a 1.00 L container. If the temperature remains at 202°C, how many moles of NH 4 Cl are produced? n(NH 3 ) = n(HCl) = mol V = 1.00 L, T = 202°C

NH 4 Cl(s) NH 3 (g) + HCl(g) [ ] i [ ] c -x -x [ ] e x x K p = K c (RT) Δ ng 4.84 = K c ( × 475) 2 K c =

K c = [NH 3 ][HCl] = (0.130-x) 2 x 2 – 0.260x = 0 x = n(NH 4 Cl) = mol (d) If the temperature of the system is raised to 525°C, describe the change, if any, on the K eq. An increase in temperature favors the endothermic reaction, so by the equilibrium shifting to the right, K eq increases.

There are many key points concerning this problem.  Solids and liquids are never included in the equilibrium expression.  An equilibrium table using the i (initial), c (change), and e (equilibrium) convention is preferred.  The equilibrium constant expression only depends on stoichiometry.

 If moles of gas are given rather than partial pressures, the Ideal Gas Law (PV = nRT) is used to calculate the partial pressures before proceeding.  There are programs available for a TI calculator to solve quadratic equations. The TI has a built in program called Solver which works equally as well.  Part (c) of the problem was intended to show that equilibrium can be achieved from either the forward or reverse reaction.

Nitrogen monoxide with a partial pressure of atm and gaseous bromine with a partial pressure of atm are injected into a 1.00-L flask at 67°C. The following reaction occurs: 2NO(g) + Br 2 (g) 2NOBr(g) If the equilibrium partial pressure of nitrosyl bromide is atm at 67°C, what is K eq ? Eq Problems (Continued)

P i (NO) = atm, P i (Br 2 ) = atm, P eq (NOBr) = atm 2NO(g) + Br 2 (g) 2NOBr(g) P i Δ P c P e K p = P(NOBr) 2 /(P(NO) 2 × P(Br 2 )) K p = /( × 0.115) = 3.73

Eq Problems (Continued) At 460°C, the following equilibrium exists: SO 2 (g) + NO 2 (g) NO(g) + SO 3 (g) If the partial pressure of SO 2 and NO 2 is atm and the K p = 85.0, what are the gas pressures at equilibrium?

SO 2 (g) + NO 2 (g) NO(g) + SO 3 (g) P i Δ P c -x -x +x +x P e x x x x KpKp = P(NO) × P(SO 3 ) P(SO 2 ) × P(NO 2 ) 85.0 = (0.075 – x) x × x ×

Here’s the time to use a legitimate shortcut! Taking the square root of both sides will make the computation easier. x = P e (SO 2 ) = P e (NO 2 ) = atm P e (NO) = P e (SO 3 ) = atm

Eq Problems (Continued) The following reaction occurs at 1200°C. H 2 (g) + C 2 N 2 (g) 2HCN(g)K p = 1.50 If hydrogen with a partial pressure of atm, cyanogen with a partial pressure of atm, and hydrogen cyanide with a partial pressure of atm are injected into a 2.00 L flask, calculate the equilibrium pressures.

H 2 (g) + C 2 N 2 (g) 2HCN(g) P i P c -x -x +2x P e x x x KpKp = P(HCN) 2 P(H 2 ) × P(C 2 N 2 ) 1.50 = ( x) 2 (0.200-x) × (0.150-x)

x = P(H 2 ) = atm, P(C 2 N 2 ) = atm, P(HCN) = atm

Reaction Quotient A reaction that is not at equilibrium proceeds to a state of equilibrium. For a given set of pressures, phosphorus(V) chloride decomposes according to the following reaction: PCl 5 (g) PCl 3 (g) + Cl 2 (g) How do you determine if the reaction is at equilibrium?

Using the law of mass action and Q the reaction quotient gives Q = P(PCl 3 ) × P(Cl 2 ) P(PCl 5 ) The pressures in the expression for Q will not be equilibrium pressures unless the system is at equilibrium. The value of Q will be compared to K eq to determine the direction of the reaction.

 If Q = K eq, the system is at equilibrium and no shift occurs.  If Q < K eq, the reaction shifts to the right, forming more product and consuming some of the reactant.  If Q > K eq, the reaction shifts to the left, forming more reactant and consuming some of the product.  The shift continues until Q = K eq.

Reaction Quotient Problem A gaseous mixture made of SO 2, Cl 2, and SOCl 2 is enclosed in a 2.0 L container. The pressures are 0.30, 0.16, and 0.50 atm respectively. K p = and the reaction is: SO 2 Cl 2 (g) SO 2 (g) + Cl 2 (g) (a) Is the system at equilibrium? Explain. Q = P(SO 2 ) × P(Cl 2 ) P(SO 2 Cl 2 )

Q = P(SO 2 ) × P(Cl 2 ) P(SO 2 Cl 2 ) Q = 0.30 × = No, because Q > K p (b) If the system is not at equilibrium, in which direction will the system move to reach equilibrium? Q > Kp, reaction shifts to the left.

LeChatelier’s Principle When a system is in a dynamic equilibrium and a stress is applied to the system, the system reacts to oppose the effect of the change. The stress can be caused by a change in concentration, a change in volume, or a change in pressure of one of the reactants or products, or a change in temperature of the system.

A change in either reactant or product concentration: 4KO 2 (s) + 2H 2 O(g) 4KOH(s) + 3O 2 (g) What happens when the [KO 2 ] is increased? Because solids and liquids do not appear in the equilibrium expression, increasing or decreasing the concentration of KO 2 has no effect on the equilibrium.

KO 2 (s) + 2H 2 O(g) 4KOH(s) + 3O 2 (g) What happens when the [O 2 ] is increased? In response to increasing [O 2 ], the system will shift to the left to consume the additional product. What happens when 1.0 mol of argon is added to the system? The argon will increase the total pressure but it will not have any effect on the concentration or partial pressure of the reactants or products.

A change in the volume of the container: N 2 (g) + 3H 2 (g) 2NH 3 (g) What happens if the volume of the container is increased? The equilibrium shifts to the left to increase the total number of moles or molecules present. When the volume of the container is increased, the system responds by increasing its own volume.

N 2 (g) + 3H 2 (g) 2NH 3 (g) What happens if the volume of the container is decreased? The equilibrium shifts to the right to decrease the total number of moles or molecules present. When the volume of the container is decreased, the system responds by decreasing its own volume.

N 2 (g) + O 2 (g) 2NO(g) What happens if the volume of the container is decreased? Nothing! If Δ n g = 0, volume changes do not effect the equilibrium.

A change in the pressure on the system:  For an increase in pressure by decreasing the volume or a decrease in pressure by increasing the volume, see A change in the volume of the container.

A change in temperature of the system: N 2 O 4 (g) 2NO 2 (g) Δ H° = 57.2 kJ What effect does changing the temperature have on the reaction? For an endothermic reaction, think of the enthalpy change as a reactant. N 2 O 4 (g) kJ 2NO 2 (g)

N 2 O 4 (g) kJ 2NO 2 (g) When the temperature is increased, the equilibrium shifts to the right, increasing the value of K eq. Think of increasing temperature as adding energy and remember the system will respond to oppose the change. Keep in mind that an increase in temperature always favors the endothermic reaction.

N 2 O 4 (g) kJ 2NO 2 (g) When the temperature is decreased, the equilibrium shifts to the left, decreasing the value of K eq. Think of decreasing temperature as removing energy and remember the system will respond to oppose the change. Keep in mind that an decrease in temperature always favors the exothermic reaction.

N 2 (g) + 3H 2 (g) 2NH 3 (g) Δ H° = kJ What effect does raising the temperature have on the reaction? For an exothermic reaction, think of the enthalpy change as a product. N 2 (g) + 3H 2 (g) 2NH 3 (g) kJ

N 2 (g) + 3H 2 (g) 2NH 3 (g) kJ When the temperature is increased, the equilibrium shifts to the left, decreasing the value of K eq. Think of increasing temperature as adding energy and remember the system will respond to oppose the change. Keep in mind that an increase in temperature always favors the endothermic reaction.

N 2 (g) + 3H 2 (g) 2NH 3 (g) kJ When the temperature is decreased, the equilibrium shifts to the right, increasing the value of K eq. Think of decreasing temperature as removing energy and remember the system will respond to oppose the change. Keep in mind that an decrease in temperature always favors the exothermic reaction.

Van’t Hoff Equation The value of K eq for any reaction depends only on temperature. To determine the new value of K eq, the van’t Hoff equation is used. = - ( ) ln K2K2 K1K1 Δ H° R - 1 T2T2 1 T1T1

Given the following reaction, PCl 3 (g) + Cl 2 (g) PCl 5 (g) Δ H° = kJmol -1 If K p is 5.38 at 200.°C, what is K p at 250.°C? = - ( ) ln K2K2 K1K1 Δ H° R - 1 T2T2 1 T1T1 = - ln K2K ( ) K2K2 = 0.561

Multiple Equilibria What happens to K eq when we reverse a chemical equation? Given the following equation, CO(g) + Cl 2 (g) COCl 2 (g) K p = 1.5 × 10 8 write the equilibrium expression. KpKp = P(COCl 2 ) P(CO) x P(Cl 2 ) = 1.5 × 10 8

Reversing the reaction gives, COCl 2 (g) CO(g) + Cl 2 (g)K p = ? You will notice the equilibrium expression of the reverse reaction is the reciprocal of the forward reaction. This results in taking the reciprocal of K p. KpKp = P(COCl 2 ) = (1.5 × 10 8 ) -1 P(CO) x P(Cl 2 ) K p = 6.7 x 10 -9

What happens to K eq when we multiply the coefficients of a balanced equation by a number? Given the following equation, ½CO(g) + ½Cl 2 (g) ½COCl 2 (g) K p = ? write the equilibrium expression. KpKp = P(COCl 2 ) ½ P(CO) ½ x P(Cl 2 ) ½ = (1.5 × 10 8 ) ½ KpKp = 1.2 × 10 4

What happens to K eq when we add two or more reactions? N 2 (g) + O 2 (g) 2NO(g)(1) 2NO(g) + Br 2 (g) 2NOBr(g)(2) We will add equations (1) and (2) and determine the K p for the resulting reaction. N 2 (g) + O 2 (g) 2NO(g)(1) 2NO(g) + Br 2 (g) 2NOBr(g)(2) N 2 (g) + O 2 (g) + Br 2 (g) 2NOBr(g)

KpKp = P(NOBr) 2 P(N 2 ) × P(O 2 ) × P(Br 2 ) We will write the equilibrium expression for equations (1) and (2) and multiply them. KpKp = P(NO) 2 P(N 2 ) × P(O 2 ) (1) KpKp = P(NOBr) 2 P(NO) 2 × P(Br 2 ) (2)

K p = K p1 × K p2 KpKp = P(NO) 2 P(N 2 ) × P(O 2 ) × P(NOBr) 2 P(NO) 2 × P(Br 2 ) KpKp = P(N 2 ) × P(O 2 ) × P(NOBr) 2 P(Br 2 ) This is the same equilibrium expression as obtained from adding equations (1) and (2).

We have therefore shown that the equilibrium constant for a reaction that consists of two or more steps is the product of the equilibrium constants for the individual steps.