# Introduction To Dynamic Chemical Equilibria (Ch 19)

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Introduction To Dynamic Chemical Equilibria (Ch 19)
Equilibrium Suggested HW (Ch 19): 4, 10, 14, 18, 26, 36, 46, 52

Recap: Gibbs Free Energy and the Criteria for Spontaneity
In the last lecture, we were introduced to Gibbs free energy (ΔG) The sign of ΔG tells us if a reaction will occur spontaneously, and in which direction. A negative value of ΔG means that a reaction is favored and spontaneous in the forward direction. ∆G= ∆H−T∆S

Recap: Rate Laws Recall that for any reaction, the rate of the reaction is dependent on the reactant concentration. For a reaction A  B, the rate law is written as: As expressed by the equation above, as the concentration of A decreases, the rate (speed) of the reaction slows down. The reaction order, m, expresses exactly how the rate depends on [A] 𝑟𝑎𝑡𝑒=k[A ] m

Dynamic Equilibria To date, we have discussed reactions occurring in only one direction. However, in many cases, the forward and back reactions both occur at significant rates, and are important. For example: N2O4 g NO 2 (g) kF kB As the forward reaction proceeds, [N2O4] decreases, and [NO2] increases. The rate of the forward reaction slows down, while the rate of the back reaction increases. Once the rates are equal, the system is in equilibrium.

Dynamic Equilibria N2O4 g NO 2 (g) kF kB Equilibrium reaction are elementary, meaning that the reaction order follows the stoichiometry Chemical equilibrium is dynamic because the reactions never actually stop At equilibrium, NO2 is formed and consumed at the exact same rate, so there is no net change in the concentration of the products or reactants rate F = rate B 𝐤 𝐅 [ 𝐍 𝟐 𝐎 𝟒 ]= 𝐤 𝐁 [𝐍 𝐎 𝟐 ] 𝟐

Dynamic Equilibria Both NO2 and N2O4 are in the system at equilibrium.
In this instance, the vast majority of gas in the system is the product, NO2 (forward direction is the preferred direction). There is always some reactant remaining in any chemical reaction, no matter how small the amount. All reactions approach equilibrium.

Why Do Reactions Want to Approach Equilibrium?
Equilibrium is a counter-intuitive concept. Why wouldn’t a reaction just ‘go all the way’? If every reaction has a preferred direction, what is the driving force of the back reaction? The answer has to do with ENTROPY and GIBBS FREE ENERGY If ΔG is negative, the reaction will proceed forward (to the right). However, a point in every reaction is reached where it becomes entropically unfavorable (ΔS starts to become negative, too much order!) to form more product, which causes G to become more positive. Since the reaction can’t stop, the back reaction proceeds to prevent any further formation of product and to keep G at a minimum (maximize ΔG)

S G 100% 0% Reactants Products equilibrium
ΔG if reaction proceeds 100% S G ΔG at equilibrium Forward reaction A mix of reactants and products is entropically favored. No reaction goes 100% to completion. Back reaction

Equilibrium Constant, Kc
For any equilibrium reaction: 𝑘 𝐹 𝑎𝐴+𝑏𝐵 𝑐𝐶+𝑑𝐷 𝑘 𝐵 The rates of the forward and back reaction are equal when equilibrium is established. The reactions in an equilibrium process are elementary reactions. The rates can be described as: The equilibrium constant, K, is equal to the ratio of the rate constants of the forward and reverse reactions. It can be expressed in terms of either concentrations or pressures (subscripts are used to distinguish). 𝑟𝑎𝑡 𝑒 𝐹 = 𝑘 𝐹 [𝐴 ] 𝑎 [𝐵 ] 𝑏 𝑘 𝐹 [𝐴 ] 𝑎 [𝐵 ] 𝑏 = 𝑘 𝐵 [𝐶 ] 𝑐 [𝐷 ] 𝑑 𝑟𝑎𝑡 𝑒 𝐵 = 𝑘 𝐵 [𝐶 ] 𝑐 [𝐷 ] 𝑑 [𝐶 ] 𝑐 [𝐷 ] 𝑑 [𝐴 ] 𝑎 [𝐵 ] 𝑏 = 𝑘 𝐹 𝑘 𝐵 Kc

Equilibrium Constant, Kc
The value of Kc is constant at a given temperature. Reactants and products that are pure solids or pure liquids DO NOT appear in the equilibrium constant expression. Values of K larger than 1 indicate that the forward reaction is faster and more favored. Values of K less than 1 indicate that the reverse reaction is faster and more favored.

Examples Write equilibrium constant expressions of the following reactions K c = [NH 3 ] 2 N 2 [ H 2 ] 3 N 2 g +3 H 2 g NH 3 (g) 1 2 N 2 g H 2 g NH 3 (g) K c = [NH 3 ] [ N 2 ] 1/2 [ H 2 ] 3/2 PCl 3 L + Cl 2 g PCl 5 (s) K c = 1 [C l 2 ] K c = [C O 2 ] [C H 4 ] [ O 2 ] 2 C H 4 (g)+ 2O 2 (g) C O 2 g + 2H 2 O(L)

I.C.E. Tables: Calculating Equilibrium Concentrations
Suppose you are carrying out a reaction, and you wish to determine either the value of the equilibrium constant, or the equilibrium concentrations of your products and reactants (if K is given); To do this, you would plug the starting concentrations into an I.C.E (initial, change, equilibrium) table and tabulate the change in each concentration as the reaction proceeds

I.C.E. Table Calculations: Finding K
𝐇 𝟐 𝐠 + 𝐈 𝟐 𝐠 𝟐𝐇𝐈(𝐠) A closed system initially contains 1.0 x 10-3 M H2 and 2 x 10-3 M I2 at 448oC. The equilibrium concentration of HI is 1.87 x 10-3 M. Calculate Kc. Concentration (M) H2 I2 2 HI Initial .001 .002 Change Equilibrium -x -x +2x x x .00187 To determine the equilibrium concentrations of H2 and I2, we must determine the change in the initial concentration. We use the stoichiometry to do this We see that the change in HI is twice the change in H2 and I2 (1:2 stochiometry).

I.C.E. Table Calculations
𝑯 𝟐 𝒈 + 𝑰 𝟐 𝒈 𝟐𝑯𝑰(𝒈) x = M We can easily determine the equilibrium concentrations. Concentration (M) H2 I2 2 HI Initial .001 .002 Change -( ) +2( ) Equilibrium .00187 Now, we can calculate Kc 𝐾 𝑐 = [𝐻𝐼 ] 2 𝐻 2 [ 𝐼 2 ] = ( 𝑀) 𝑥 1 0 −5 𝑀 (1.065 𝑥 10 −3 𝑀) =50.5 The magnitude of Kc (much greater than 1) suggests that there is substantially more product than reactant at equilibrium. The reaction is favored to the right

I.C.E Tables: Calculating Equilibrium Concentrations from K
𝑁 2 𝑂 4 𝑔 𝑁 𝑂 2 𝑔 𝐾 𝑐 =0.20 Suppose that 2.00 moles of N2O4 are injected into a 1.00 L reaction vessel held at 100oC. Calculate the equilibrium concentrations. Set up the I.C.E. table. We don’t know the equilibrium concentrations, so we leave them in terms of x. Concentration (M) N2O4 2NO2 Initial 2.00 Change Equilibrium -x +2x 2.00-x 2x

We plug in our equilibrium concentrations and the given value of Kc
𝑁 2 𝑂 4 𝑔 𝑁 𝑂 2 𝑔 𝐾 𝑐 =0.20 Concentration (M) N2O4 2NO2 Initial 2.00 Change - x + 2x Equilibrium x 2x 𝐾 𝑐 = [𝑁 𝑂 2 ] 2 [ 𝑁 2 𝑂 4 ] We plug in our equilibrium concentrations and the given value of Kc Expand and cross multiply to get all terms on one side of the equation. We are now left with a QUADRATIC EQUATION (ax2+bx+c = 0). 0.20= (2𝑥 ) 2 (2−𝑥) 𝟒 𝒙 𝟐 +𝟎.𝟐 𝒙−𝟎.𝟒=𝟎

A Blast From the Past: Solving the Quadratic Equation
To solve a quadratic equation, you must use the quadratic formula Since all quadratic formulas are of the form ax2 + bx +c = 0, we identify these values and plug them into the formula. Quadratics yield two results. a = 4 b = 0.2 c = -0.4 𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 𝑥= −0.2 ± (0.2 ) 2 −4(4)(−0.4) 2(4) = −0.2± Check your values of x to determine the correct one. Only one value will work. 𝑥=0.29 𝑀 𝑥= −0.2±2.53 8 X 𝑥=−0.34 𝑀

I.C.E Tables: Calculating Equilibrium Concentrations from K
Plug the calculated value of x back into the I.C.E. table to determine the equilibrium concentrations (x = 0.29 M). DO NOT FORGET THIS STEP! Concentration (M) N2O4 2NO2 Initial 2.00 Change - (.29) + 2 (.29) Equilibrium 1.71 .58

Relating Concentrations to Partial Pressures
Recall from the gas laws that concentration is related to pressure: Therefore, equilibrium constants can be expressed in terms of partial pressures. Since solids and liquids do not contribute to the overall pressure, they are not included in the expression (as we learned previously). When using pressures, we write the equilibrium expression as Kp. The subscript p means that K is in terms of pressure. 𝑃 𝑅𝑇 = 𝑛 𝑉

Example A mixture of H2 and N2 is allowed to attain equilibrium at 472oC. The equilibrium mixture was found to contain 7.38 atm H2, 2.46 atm N2, and atm NH3. Find Kp? 𝑁 2 𝑔 +3 𝐻 2 𝑔 𝑁𝐻 3 (𝑔) 𝐾 𝑝 = ( 𝑝 𝑁 𝐻 3 ) 2 ( 𝑝 𝐻 2 ) 3 ( 𝑝 𝑁 2 ) = (.166 𝑎𝑡𝑚 ) 2 (7.38 𝑎𝑡𝑚 ) 3 (2.46 𝑎𝑡𝑚) =2.79 𝑥 1 0 −5 𝑎𝑡 𝑚 −2 The value of Kp is very small (less than 1), which tells us that the reaction is favored to the left.

LeChatlier’s Principle
LeChatlier’s Principle states that: If a chemical reaction at equilibrium is subjected to a change in conditions that displaces it from equilibrium, the reaction adjusts toward a new equilibrium state. The reaction will proceed in the direction that offsets the change.

LeChatlier’s Principle
For example, let’s say we have the following reaction at 25oC: Assume that our equilibrium concentration of CO is 6M, that of CO2 is 4M. As we learned earlier, we can express Kc as: Now, we add more CO2(g) to the system, until [CO2] = 8 M. WE ARE NO LONGER IN EQUILIBRIUM !!!. Thus, the state of the system is no longer described by the equilibrium constant Kc, but by the reaction quotient, Q. The subscript ‘o’ indicates a non-equilibrium concentration 𝐶 𝑠 +𝐶 𝑂 2 𝑔 𝐶𝑂 (𝑔) 𝐾 𝑐 =9 𝐾 𝑐 = [𝐶𝑂 ] 2 [𝐶 𝑂 2 ] = =9 𝑄= [𝐶𝑂 ] 𝑜 2 [𝐶 𝑂 2 ] 𝑜 = =4.5

Reaction Quotients, Q The direction of spontaneity is always toward equilibrium. The value of Q tells us the direction in which a system not at equilibrium will proceed to reach equilibrium.

Back to the Example At a given temperature, the equilibrium constant DOES NOT CHANGE. To reestablish equilibrium (Kc= 9), the reaction shifts right to consume some of the added CO2 This will increase [CO] and decrease [CO2] until Kc equals 9 again. When equilibrium is re-established, the new concentrations will be: 𝐾 𝑐 = [𝐶𝑂 ] 2 [𝐶 𝑂 2 ] = (6+𝟐𝒙) 2 8−𝒙 =9 x=.975M 𝐶𝑂 =7.95 𝐶 𝑂 2 =7.025

LeChatlier’s Principle Applied To Volume and Pressure
When you have gaseous reactants and products, changes to the volume and pressure of the system induce shifts in the equilibrium. The general rule is, for any equilibrium involving gases, decreasing the volume drives the equilibrium toward the side with the smaller number of moles of gas. From Boyle’s Law, we know that pressure and volume are inversely proportional, so increasing pressure has the same effect as lowering volume. N2O4 g NO 2 (g) kF kB 𝐾 𝑐 = [𝑁 𝑂 2 ] 2 [ 𝑁 2 𝑂 4 ] Equilibrium Decreasing V shifts reaction left Increasing V shifts reaction right

LeChatlier’s Principle Applied To Changing Temperature
Remember, all reactions must be either endothermic or exothermic In an endothermic process, heat is absorbed into the system in order for the reaction to proceed. Therefore, since heat is used in the reaction, heat may be considered a reactant Let’s use our previous example. We can rewrite this reaction as: Increasing the temperature of an endothermic process drives the equilibrium to the right because you are adding a reactant. ∆ 𝑯 𝒓𝒙𝒏 =𝟓𝟓 𝒌𝑱 𝒎𝒐𝒍 𝑁 2 𝑂 4 𝑔 𝑁 𝑂 2 𝑔 𝐾 𝑐 =0.20 𝑁 2 𝑂 4 𝑔 +𝒉𝒆𝒂𝒕 𝑁 𝑂 2 𝑔 𝐾 𝑐 =0.20

LeChatlier’s Principle Applied To Changing Temperature
The opposite must then be true for an exothermic process. Since heat leaves the system in an exothermic process, heat may be treated as a product For the following exothermic reaction: We can rewrite the reaction as: Increasing the temperature of an exothermic process drives the equilibrium to the left because you are adding a product. ∆ 𝑯 𝒓𝒙𝒏 =−𝟗 𝒌𝑱 𝒎𝒐𝒍 𝐻 2 𝑔 + 𝐼 2 𝑔 𝐻𝐼 𝑔 𝐻 2 𝑔 + 𝐼 2 𝑔 𝐻𝐼 𝑔 +𝒉𝒆𝒂𝒕

Calculations Based On LeChatlier’s Principle
Let’s take the reaction below. Assume that the reaction is at equilibrium. The equilibrium concentrations are given. We can calculate Kc from the equilibrium values Kc = .056 Now, additional CO2 is added to the system, raising [CO2] to 1.25 M. This will cause the equilibrium to shift right. But how far? If we can ‘freeze time’ right at the instant that the CO2 is added (before the shift), we would have the following: 𝐶 𝑂 2 𝑔 + 𝐻 2 𝑔 𝐶𝑂 𝑔 + 𝐻 2 𝑂(𝑔) 0.75 M 0.60 M 0.10 M 0.25 M Concentration(M) CO2 H2 CO H2O Initial 1.25 0.60 0.10 0.25 Change -x +x Equilibrium 1.25-x 0.60-x 0.10+x 0.25+x

Calculations Based On LeChatlier’s Principle
𝐶 𝑂 2 𝑔 + 𝐻 2 𝑔 𝐶𝑂 𝑔 + 𝐻 2 𝑂(𝑔) Concentration(M) CO2 H2 CO H2O Initial 1.25 0.60 0.10 0.25 Change -x +x Equilibrium 1.25-x 0.60-x 0.10+x 0.25+x Now we solve this as usual. Convert to quadratic form. .10+𝑥 (.25+𝑥) 1.25−𝑥 (.60−𝑥) =.056 0.944 𝑥 𝑥−.017=0 Use quadratic formula. Solve for x New equilibrium concentrations: [CO2]= M, [H2]= M [CO] = M, [H2O] = M x = M

Group Example 𝐴 𝑔 +𝐵 𝐿 𝐶(𝑔) The reaction above is enclosed in a vessel at 25oC. At equilibrium, you have atm of A and 4.89 atm of C. Then, the system is disturbed by a 35% decrease in the pressure of A. What are the new equilibrium pressures of A and C? When performing an equilibrium calculation involving LeChatlier’s principle, you must first answer the following What is the equilibrium expression? What is the value of Kp? What are the initial pressures (at the instant before the reaction proceeds)? In which direction does the disturbance shift the equilibrium?

No longer at equilibrium. Must shift LEFT to reestablish Kp
𝐾 𝑝 = ( 𝑃 𝐶 ) 2 𝑃 𝐴 = (4.89 𝑎𝑡𝑚 ) 2 (.171 𝑎𝑡𝑚) =139.8 𝑎𝑡𝑚 Pressure A 2C Initial (.111) 4.89 Change +x -2x Equilibrium .111+x 4.89-2x No longer at equilibrium. Must shift LEFT to reestablish Kp When we plug our new equilibrium pressures into Kp, the value of Kp MUST be 139.8 (4.89 −2𝑥 ) 2 (.111+𝑥) =139.8 4 𝑥 𝑥−8.38=0 x = atm PA = atm PC = atm