Chemistry SEAS Q, K eq, and Equilibrium. K ; where G o = -RTlnK eq eq.

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1 Chemistry SEAS Q, K eq, and Equilibrium At a given temperature and set of conditions (pressures or concentrations), we can tell if a reaction is already at equilibrium, or which way it will approach equilibrium, by comparing the reaction quotient (Q) to the equilibrium constant (K eq ). Note: when comparing Q to K eq, you must compare the same version of Q and K eq. That is, if you determine Q in terms of pressures (Q P ), then you must compare to K P ; if you determine Q in terms of concentrations, (Q C ), then you must compare to K C. Be careful! The first table (Table 1) below relates Q, its corresponding value of G, K eq and the subsequent approach to equilibrium. Remember the important relationships from which this table derives: G = G o + RTln(Q) = RTln Q K ; where G o = -RTlnK eq eq. Remember, the approach to equilibrium is such that Q will either increase or decrease (depending on the situation) until Q = K eq. In other words, the equilibrium constant (K eq ) is the parameter to which Q must adjust in order for the reaction to establish equilibrium. The second table (Table 2) below relates the magnitude of K eq "products" "reactants". Table 1: Q versus K eq and the direction of approach to equilibrium Relationship of Q to K eq Sign of G Comments Q < K eq - Rxn will proceed to products to establish equilibrium. Q > K eq + Rxn will proceed to reactants to establish equilibrium. Q = K eq 0 Rxn is already at equilibrium at the specified conditions. Table 2: K eq and the Composition of the equilibrium reaction mixture Magnitude of K eq Sign of G o Composition of equilibrium rxn mixture K eq > 1 - Greater proportion of products. K eq < 1 + Greater proportion of reactants. K eq = 1 0 Rxn is at equilibrium at standard conditions. Each reactant and product gas is at 1.0 atm Each reactant and product (aq) species is at 1.0 M. Thus, Tables 1 & 2 allow us to look at Q and K eq, instead of G and G o, in order to investigate the approach to and composition of the equilibrium system. Remember however, Q and K eq give us the same information as G and G o - just in a different form. 1

2 Chemistry SEAS Van't Hoff equation - K eq as a function of Temperature Equations: G o = -RTlnK eq and G o = H o - T S o. Assumptions: H o and S o are independent of temperature for chemical reactions (as discussed before). Derivation: Setting the two equations above equal to each other, since they each equal G o, we have: -RTlnK eq = H o - T S o. Consider two temperatures, T 1 and T 2 (which must be expressed in kelvin) and their corresponding equilibrium constants K eq,1 and K eq,2, respectively. [Note: for a gas phase reaction, K eq must be K P ; and for an aqueous reaction, K eq must be K C.] Thus: -RT 1 lnk eq,1 = H o - T 1 S o and -RT 2 lnk eq,2 = H o - T 2 S o Dividing each equation through by R x temperature, we have: -lnk eq,1 = Ho RT 1 - S o R and -lnk eq,2 = H o RT 2 - S o R Subtracting equation 2 (at temperature T 2 ) from equation 1 (at equation T 1 ), the S o R terms cancel. we have: -lnk eq,1 + lnk eq,2 = H o RT 1 - H o RT 2 - S o R + So R Rearranging: = H o RT 1 - H o RT 2 ln K eq,2 K = H o eq,1 R 1 1 T - 1 T = H o 2 R T 2 - T 1 T 2 T 1 - van't Hoff eqn. An example calculation follows. 2

3 Consider the equilibrium between NO 2 (g) and N 2 O 4 (g): 2 NO 2 (g) N 2 O 4 (g) ; G o 298 = _6 kj and Ho 298 = _58 kj. Determine K P and K C at 25 o C (298 K) and at 227 o C (500 K). Solution: Let's find K P at each temperature, first. [Remember, for a gas phase reaction (or a heterogeneous reaction where there are only gases, pure solids and pure liquids), G o gives K P and the van't Hoff equation will use K P. For aqueous reactions (homogeneous or heterogeneous) G o gives K C and the van't Hoff equation will use K C. Be careful! ] At 298 K: ln(k P,298 ) = - G o 298 RT = -(-6 x 10 3 J) J = 2.42 K (298 K) Thus, K P,298 = antiln(2.42) or e 2.42 = Using van't Hoff, we now find K P,500. Let T 2 = 500 K and T 1 = 298 K. Thus, ln K P,500 K = H o P,298 R 1 1 T - 1 T = H o 2 R T 2 - T 1 T 2 T. Plugging in: 1 ln K P, = -58 x 10 3 J J K 500 K K (500 K)(298 K) = K P, = antiln(-9.46) or e-9.46 = 7.79 x Finally, K P,500 = (11.42)( 7.79 x 10-5 ) = 8.90 x Now, to get K C at each temperature, we use: K C = K P (RT) _ n g. Here, n g = [1-2] = -1. R must be the version, not the version. First, we find K C,298. K C,298 = K P,298 (RT) - n g = (11.42)[(0.0821)(298)] -(-1) = Now we'll find K C,500. K C,500 = K P,500 (RT) - n g = (8.90 x 10 _4 )[(0.0821)(500)] -(-1) = 3.65 x Note: K P (or K C ) decreases as temperature increases. This is expected for an exothermic ( H o < 0) reaction. For an endothermic ( H o > 0) reaction, K P (or K C ) will increase as temperature increases - as expected. Check it out for yourself! 3

4 Chemistry SEAS Solving a Typical Equilibrium Problem - An Example Equilibrium problems, that is, determining information about and/or from the composition of an equilibrated chemical reaction, are very important in chemistry. The typical equilibrium problem makes use of the reaction quotient (Q) and its relationship to the equilibrium constant (K eq ) of a chemical reaction. As discussed previously, Q and K eq are really just convenient parameters which help us relate G (and hence the Second Law of Thermodynamics) to the solution of equilibrium problems - in terms of measurable quantities such as pressure or concentration. We will solve such a typical sample problem below, keeping careful attention to the details. Example: Consider a 1.00 L bulb filled only with 10 torr of SO 2 (g) and 10 torr of Cl 2 (g), at 25 o C (298 K). 10 torr SO 2 10 torr Cl 2 V = 1.00 L T = 298 K Maintaining constant volume and constant temperature, reaction is initiated (via a catalyst), and the following equilibrium is established - in which SO 2 Cl 2 (g) (Sulfuryl chloride) is produced: SO 2 (g) + Cl 2 (g) SO 2 Cl 2 (g). At 298 K, K P = 2840 for the above equilibrium. Determine the equilibrium partial pressure of each gas, when equilibrium is established. Solution: The 10 steps are listed below. (1) Write down the equilibrium reaction and begin the construction of a table beneath the reaction. (2) In the first row of the table, write down the initial amount of each species - in the appropriate units. If you are working with K P, all entries (throughout the table) must be in atmospheres. If you are working with K C, all entries (throughout the table) must be in Molarity (M) units. If we use the appropriate units for all entries, we will not only ensure that the answers come out right, but we will ensure that K P (K C ) is unitless. Initial (atm): SO 2 (g) + Cl 2 (g) SO 2 Cl 2 (g)

5 (3) Calculate Q, the reaction quotient, based on your listed initial amounts in the first row of the table. Compare Q with the appropriate K eq value. If Q is in terms of concentrations (Q C ) - compare with K C. If Q is in terms of atmospheres (Q P ) - compare with K P. If Q < K eq - reaction will proceed toward products to establish equilibrium If Q > K eq - reaction will proceed toward reactants to establish equilibrium If Q = K eq - reaction is already at equilibrium Note: If Q = 0 (one or more products not present initially, but all reactants are present initially), Q has to be less than K eq and thus, reaction will proceed toward products to establish equilibrium. This is the case in this problem. P SO2 Cl 2,o Q = rxn quotient (in terms of initial pressures, P o ) = P SO2,o P. Cl 2,o 0 Q = = 0 Q < K P - Proceeds to products to establish equilibrium. (4) Define a parameter - x - to reflect the changes that occur to establish equilibrium. Define x in such a way so that it reflects positive change in the direction that the reaction wants to proceed to establish equilibrium, i.e., the direction indicated by Q (see (3)). Let x = increase in atmospheres of SO 2 Cl 2 (g), in order to reach equilibrium. (5) Then, the changes are not independent! They are related by the stoichiometry of the reaction. [This is true because, at constant volume and constant temperature (for an ideal gas), P n and concentration n.] Hence, the changes can all be expressed in terms of the single parameter - x. Write down these changes, as a function of x, under each species, in the second row of the table. Note: make sure x reflects correctly whether the species will increase in amount ( +x ) or decrease in amount ( -x ) as equilibrium is established (see (3)). SO 2 (g) + Cl 2 (g) SO 2 Cl 2 (g) Initial (atm): Change (atm) -x -x +x (6) The third row is the algebraic sum of the initial and change (first and second rows of the table) of each species. Thus, the third row clearly expresses the equilibrium amounts. SO 2 (g) + Cl 2 (g) SO 2 Cl 2 (g) Initial (atm): Change (atm) -x -x +x 10 Equil. (atm) x x +x 5

6 (7) These equilibrium amounts are then plugged into the algebraic expression for K eq. ( P eq denotes equilibrium pressure) K P = 2840 = P SO2 Cl 2,eq P SO2,eq P Cl 2,eq = x x 2 (8) This then provides you with one equation in one unknown ( x ). This equation must be solved for x, using suitable algebraic techniques. In principle, although sometimes mathematically difficult, the equation can be solved for x. Here, we have a quadratic equation in x. By suitable re-arrangement we can put the above expression in the form: ax 2 + bx + c = 0 ; and then apply the quadratic formula: -b ± b x ± = 2-4ac 2a. [Note: you should become fluent with use of the quadratic formula, since many equilibrium expressions will give rise to quadratic equations!] Putting the equilibrium expression in x in suitable form, we have: 2840 x x = 0. (Check this for yourself!) (a x 2 b x c ) Thus, by correspondence, we have: a = ; b = ; c = Substituting into the quadratic formula, we have: x ± = -(-75.74) ± (-75.74) 2 - (4)(2840)(0.4917) 2(2840) = ( ± ) atm We have two (2) roots (as usual) when we solve a quadratic equation; thus: x + = ( ) atm = atm = ( atm) 760 torr 1 atm. = torr; x - = ( ) atm = atm = ( atm) 760 torr 1 atm. = 8.49 torr. Since x was defined as a positive quantity (See step (4)), it should be easy to choose the meaningful solution. But wait! Both solutions are positive!! This situation is not uncommon. Which solution is correct? Are they both correct? There is only one (1) meaningful solution. In this case, we must go back to the table. Remembering that P α n here, we realize that if the reaction went to completion, the maximum pressure of SO 2 Cl 2 (g) that could be produced is only 10 torr (Why?). Also, according to the table, x represents the (total) atmospheres of SO 2 Cl 2 (g) produced. Thus, torr is meaningless, since it indicates a greater pressure of SO 2 Cl 2 (g) than is stoichiometrically possible! See how important it is to set up the table clearly!! x - = atm. = 8.49 torr is the proper solution. 6

7 (9) The parameter x is then used to determine the equilibrium amounts (atm or M) of each species in the table. This can be achieved by utilizing the third row of the table (see (6)). According to the third row of the table: P SO2 Cl 2,eq = x = atm = 8.49 torr P SO2,eq 10 = x = atm = 1.51 torr P Cl2,eq = x = atm = 1.51 torr. (10) This step is optional, but useful. Plug the determined equilibrium amounts into the K eq expression and evaluate it. If the result equals K eq (it should!) you can be reasonably assured that you probably have not made a mistake. If the result does not equal K eq, ( approximately ) you have made a mistake!! Go back and check! P SO2 Cl 2,eq K P = P SO2,eq P = Cl 2,eq ( ) 2 = 2821 (??) But K P = 2840!! Have we made a mistake? No! Due to round-off errors during the course of the calculation, our equilibrium pressures do not give K P exactly. But it is close enough! The above steps can be followed for any equilibrium problem (the specifics will, of course, be different). Skipping steps or taking careless shortcuts will invariably lead to disaster! Note: According to Dalton's Law, the total equilibrium pressure (see third row of the table): P total,eq = P SO2 Cl 2,eq + P SO 2,eq + P Cl 2,eq = x x x 20 P total,eq = x = atm = torr. And, the total initial pressure (see first row of the table): P total,o = P SO2 Cl 2,o + P SO 2,o + P Cl 2,o = P total,o = atm = 20 torr. Even though we have a closed system at constant volume and constant temperature, P total,o P total,eq. Why? Addendum: Consider what happens if the above equilibrated system is connected to a previously empty bulb of equal volume (1.00 L) and the equilibrated gaseous mixture is allowed to fill both bulbs (with a total volume = 2.00 L). After some time, equilibrium is re-established at 25 o C (298 K). See diagrams. 7

8 Before valve is opened torr Cl torr SO torr SO Cl 2 2 Evacuated Equilibrium 298 K V = 1.00 L V = 1.00 L After valve is opened. 298 K Cl 2 SO K Cl 2 SO 2 SO Cl 2 2 SO Cl 2 2 V total = 2V = 2.00 L Questions: (a)what is the partial pressure of each gas, after the gases fill both bulbs, but before equilibrium is established? (b)is the system at equilibrium after the expansion? Calculate Q to verify your answer. (c) What is the equilibrium partial pressure of each gas, after the expansion? Solution: (a) In order to find out if the system is at equilibrium (or not) after the expansion occurs, we must calculate the pressures after the gases expand. These pressures will serve as the initial pressures (P o ) - which will allow us to determine if the system is now at equilibrium. Since each gas is expanding at constant temperature (assumed) and constant amount (no gas leaks out); we can use Boyle's Law. Thus, since each gas expands from V to 2V, the new pressure of each gas will be half of its equilibrated pressure before expansion. 8

9 P SO2 P SO2 Cl 2,o = Cl 2,eq 2 = P SO2,o = P SO2,eq 2 = P Cl2,o = P Cl2,eq 2 = 8.49 torr 2 = torr = x 10 _3 atm 1.51 torr 2 = torr = x 10 _4 atm 1.51 torr 2 = torr = x 10 _4 atm (b) Following the suggestion, we calculate Q and compare to K P (= 2840). P SO2 Cl 2,o Q = rxn quotient (in terms of initial pressures, P o ) = P SO2,o P Cl 2,o. Q = x 10-3 (9.934 x 10-4 )(9.934 x 10-4 ) = Q > K P The system is not at equilibrium and will proceed to reactants to establish equilibrium. (c) We will follow the same procedure as before, but this time we'll just give the skeleton version. You should check through it yourself to make sure you understand! Let x = increase in atmospheres of SO 2 (g) (or Cl 2 (g)), in order to reach equilibrium. Remember, the reaction proceeds to the left. Constructing the table, we have: _ SO 2 (g) + Cl 2 (g) SO 2 Cl 2 (g) Initial (atm): x x x 10-3 Change (atm) +x +x -x Equil. (atm) x x x x x x K P = 2840 = P SO2 Cl 2,eq P SO2,eq P Cl 2,eq = x x (9.934 x x) 2. Again, we have a quadratic equation in x. By suitable re-arrangement we can put the above expression in the form: ax 2 + bx + c = 0 ; and then apply the quadratic formula: x ± = -b ± b 2-4ac 2a. Putting the equilibrium expression in x in suitable form, we have: 2840 x x x 10-3 = 0 (Check this for yourself!) Thus, by correspondence, we have: a = ; b = ; c = x Substituting into the quadratic formula, we have: x ± = -( ) ± ( ) 2 - (4)(2840)( x 10-3 ) 2(2840) 9

10 x ± = ( x 10-3 ± x 10-3 ) atm We have two (2) roots, thus: x + = ( x x 10-3 ) atm = x 10-4 atm x - = ( x x 10-3 ) atm = x 10-3 atm Since x was defined as a positive quantity it is easy, this time, to choose the meaningful solution. Only x + is positive. Thus, x = x 10-4 atm is the correct choice. According to the third row of the table: P SO2 Cl 2,eq = x x = x 10-3 atm = 3.97 torr P SO2,eq P Cl2,eq = x x = x 10-3 atm = 1.03 torr = x x = x 10-3 atm = 1.03 torr. Now, plugging into the K P expression to check (Remember, it should equal 2840): K P = P SO2 Cl 2,eq P SO2,eq P Cl 2,eq = x 10-3 (1.356 x 10-3 ) 2 = 2840 (Perfect!). Note: When we expanded the equilibrated system from V to 2V (at constant temperature), we initially decreased the pressure (via Boyle's Law) by increasing the volume. The system re-equilibrated by going backwards, i.e., toward reactants. In other words, it shifted to the side with the greater moles of gas (2 moles of gas on the reactant side and 1 mole of gas on the product side). We will see that this is exactly what is predicted qualitatively by a principle due to LeChâtelier, i.e., LeChâtelier's Principle! More about that later! Also: Comparing the equilibrium pressures after the volume change with the equilibrium pressures before the volume change, we see that the original equilibrium pressures were not attained! The equilibrium pressures after the volume change are all lower than the equilibrium pressures before the volume change. In other words, compensation is not complete! Why? 10

11 Chemistry SEAS Approximate Solution of Equilibrium Problems - Neglecting x Sometimes the algebraic equilibrium expression, i.e., the equation which relates the equilibrium constant (K P or K C ) to the equilibrium amounts (partial pressures or concentrations) - in terms of x, is intractable. In other words, the exact algebraic solution of the problem is either not possible or else (most times) the algebraic process of solution is complicated and/or tedious. Thus, we wish (or must) seek approximate methods to find x. There are several approaches - we will consider only one here; the technique of neglecting x. We will actually choose a problem that can be solved exactly, in very easy fashion, via the quadratic formula. The reason for this is so we can better assess our approximation(s). The best way to demonstrate this method is by example. Consider the following reaction, which can equilibrate at a certain temperature, T: N 2 F 4 (g) 2 NF 2 (g) ; K C = 8.10 at this temperature. Determine the equilibrium concentration (in M) of NF 2 (g) and of N 2 F 4 (g) if mole of N 2 F 4 (g) and mole of NF 2 (g), at temperature T, are initially placed in a previously empty ml sturdy container. We set up our table, in moles/l (M), remembering to calculate Q. N 2 F 4 (g) 2 NF 2 (g) moles [initial] L = 0.10 M moles L = 1.00 M Q = [NF 2 ] 2 o [N 2 F 4 ] o = (1.00)2 (0.10) = 10 Q > K C - reaction proceeds toward reactants (to the left ) to reach equilibrium. Let x = increase in concentration (in M) of N 2 F 4 (g) in order to establish equilibrium. {Thus, decrease in concentration (in M) of NF 2 (g) in order to establish equilibrium is _2x.} Setting up the complete table, we have: N 2 F 4 (g) 2 NF 2 (g) [initial] 0.10 M 1.00 M [change] +x -2x [equil] x x K C = 8.10 = [NF 2 ] 2 eq [N 2 F 4 ] eq = (1.00-2x) 2 ( x). This is a quadratic equation in x, which, of course, we can solve exactly. However, for purposes of demonstration, we will (first) solve this expression by neglecting x. The algebraic equation is: 8.10 = (1.00-2x) 2 ( x). 11

12 The method of neglecting x consists of assuming (initially) that x is small when compared to the initial amount from which it is added/subtracted. In other words, we can see from the above algebraic expression, that we have factors of the form: (A o ± ax), where A o is the initial amount and a is the factor that multiplies x (in the above algebraic equation, a = 2 in the numerator and a = 1 in the denominator. Thus, for a term of the form (A o ± ax), if we want to neglect x, we are approximating this term as: (A o ± ax) A o. [Obviously, we can't neglect x in every term of our algebraic equilibrium expression - if we did this, then we would eliminate x from the problem entirely! ] Now, since we assume that x was small, that is, that we could neglect the term ax with respect to A o ; we must check to make sure that the term ax is small enough. Criteria vary - the one we will use is the following. If we approximate (A o ± ax) as A o, then the approximate value of x, x approx that we subsequently find must pass the following test - termed the 5 % rule: If ax approx A o 0.05 ; (A o ± ax) A o is valid. In other words: ax approx (not just x approx ) must be less than or equal to 5 % of A o, in order to warrant the neglecting of ax with respect to A o. If this criterion is not met, our neglecting of x is not warranted, and we must go back to the original form of the algebraic equilibrium equation. Let's apply this method to the sample problem above. Looking at the algebraic expression once again, we have: (1.00-2x) 8.10 = 2 ( x). We can neglect x in either of two places (not both); Neglect 2x with respect to 1.00 in the numerator, or neglect x with respect to 0.10 in the denominator. Clearly, since 1.00 is ten times larger than 0.10, it should be a better approximation to neglect 2x with respect to 1.00 than to neglect x with respect to So, let's neglect 2x with respect to 1.00 and solve = x (1.00-2x) 2 ( x) (1.00) 2 ( x) ; where our x is now x approx. Solving, we have: (0.10)(8.10) 8.10 = M. Now we apply the 5 % rule, to check. We must compare 2x approx to 1.00, i.e., 2x approx 1.00 M = (2)( M) 1.00 M = which is Thus, the approximation is warranted. The equilibrium concentrations (approximately) are: [N 2 F 4 ] eq = x = ( ) M = M and [NF 2 ] eq = x = ( (2)(0.0235)) M = M. [Just to see, let's, instead, neglect x with respect to Thus, the algebraic equilibrium expression becomes: 8.10 = (1.00-2x) 2 ( x) _ (1.00-2x) 2 (0.10) ; where our x is now x approx. Solving: 12

13 x = (0.10) = M. Now we apply the 5 % rule, to check. We must compare x approx to 0.10, i.e., x approx 0.10 M = M 0.10 M = 3.58 which is not Thus, the approximation is not warranted.] [If the problem were solved exactly, you should verify the following. The algebraic equilibrium expression, in quadratic form (i.e., ax 2 + bx + c = 0) becomes: 4 x x = 0. Also, by use of the quadratic formula, you should verify that: x ± = -b ± b 2-4ac 2a = -(-12.10) ± (-12.10) 2 - (4)(4)(0.19) 2(4) x ± = ( ± ) M. Verify that x + = M and x - = M. Finally, prove that x - = M is the only meaningful solution. Thus, the exact equilibrium concentrations are: [N 2 F 4 ] eq = x = ( ) M = M and [NF 2 ] eq = x = ( (2)(0.0158)) M = M.] As you can see, the approximate and exact solutions are close, thanks to the 5 % rule! 13

14 Chemistry SEAS Heterogeneous Equilibria - Examples Although (pure) solids and (pure) liquids do not enter (i.e., enter as 1 ) into the K eq expressions, they do play a role in chemical equilibrium. Remember, all species in a chemical reaction are connected by the stoichiometric chemical equation. Thus, pure solids and pure liquids are affected by the species that are contained in the K eq expression (aqueous and gaseous species) - albeit indirectly. An example or two may help. Example #1: Consider the following equilibrium, between liquid methanol (CH 3 OH(l)), H 2 (g) and CO(g): CH 3 OH(l) 2 H 2 (g) + CO(g) K C = at 500 K. When 2.00 moles of CH 3 OH(l), 6.00 moles of H 2 (g), and 3.00 moles of CO(g) are placed in a 5.00 L container at 500 K, how many moles of H 2 (g), CO(g), and CH 3 OH(l) are present in the container at equilibrium? [Assume that the liquid methanol occupies a negligible volume of the container.] Solution: We will work the problem in terms of K C (since it is given) and then convert the (gaseous) concentrations to moles via: moles = concentration x volume. [We could equally well have stated the problem in terms of K P and solved the entire problem with partial pressures. Then, we could convert to (gaseous) moles via: n = PV/(RT).] The initial concentrations are: 6 moles [H 2 ] o = 5 L = 1.20 M and [CO] o = 3 mole 5 L = 0.60 M. Thus, we calculate Q (i.e., Q C ). Q = [H 2 ] 2 o [CO] o = (1.20)2 (0.60) = So, Q > K C and the reaction proceeds toward reactants to establish equilibrium. We set up our table, let x = the increase in [CO] to establish equilibrium, the change in [CO] is -x. The completed table is below. CH 3 OH(l) 2 H 2 (g) + CO(g) [initial] 2.00 moles * 1.20 M 0.60 M [change] ** x -x [equil] ** x x The * means that only the gaseous species (i.e., only those that appear in the K eq expression) should enter the table; and then, only in units of moles/l. Obviously, the moles of liquid that are produced are related stoichiometrically to the x changes of the gaseous species. More about that later. Let's find the equilibrium concentration of H 2 (g) and of CO(g). K C = = [H 2 ] 2 eq [CO] eq = (1.20-2x)2 ( x). We can solve the above algebraic equation exactly, by suitable factoring. We have: = (2) 2 ( x) 3. Thus, = ( x) = x = M. [H 2 ] eq = x = M (moles H 2 ) eq = (0.648 M)(5 L) = moles ; and 14

15 and [CO] eq = x = M (moles CO) eq = (0.324 M)(5 L) = moles Now:from the table, moles of CO(g) reacted = x x 5 L = (0.276 M)(5 L) = moles of CO(g) reacted. The stoichiometry of the reaction tells us that the moles of CH 3 OH(l) produced = moles of CO(g) reacted = moles. At equilibrium: moles CH 3 OH(l) in container = ( )moles = moles. Example # 2: Consider the reaction for the dehydration of CuSO 4 5H 2 O(s) (Blue Vitriol) at 500 K: CuSO 4 5H 2 O(s) CuSO 4 (s) + 5 H 2 O(g) ; K P = 3.60 x (a) At 500 K, calculate the equilibrium pressure of H 2 O(g) (in torr) and the minimum number of grams of CuSO 4 5H 2 O(s) (M.W. = g/mole) that must be placed in a previously empty 5.00 L container, in order to achieve this H 2 O(g) pressure. [As usual, the volume of both solids can be assumed to be negligible - when compared to the volume of the 5.00 L container.] (b) If the volume of the container is increased gradually, at constant temperature, what will happen to the established equilibrium pressure of H 2 O(g)? Does this agree with LeChâtelier? Solution: If we write down the equilibrium expression, we have: K P = P 5 H 2 O,eq. (a) Since the law of chemical equilibrium depends on only one partial pressure, we can solve directly: Thus: P H2 O,eq = 5 K P = x 10-5 = atm. = torr. The moles of H 2 O(g) that must be produced to achieve equilibrium under these conditions is moles of H 2 O(g) at equilibrium = moles of H 2 O(g) at equilibrium. P H2 O,eq V RT = ( atm.)(5.00 L) L atm. = mole K (500 K) grams of Blue Vitriol needed = ( moles H 2 O) 1 mole Blue Vitriol 5 moles H 2 O g Blue Vitriol 1 mole Blue Vitriol = g. Thus, a little bit more that g of Blue Vitriol must initially be placed in the container, at these conditions; since, at equilibrium, there must be some Blue Vitriol in the container also. The actual amount of Blue Vitriol that must remain is not critical since solids do not appear in the K eq expression. 15

16 (b) At constant temperature, K P does not change. Thus, since P H2 O,eq = 5 K P, the equilibrium pressure of H 2 O(g) will not change. In other words, as long as the volume increase is gradual, the adjustment is instantaneous and the pressure will never vary much from its previously established value. Compensation, in this case, is complete! This does agree with LeChâtelier. According to LeChâtelier, as the volume is increased (at constant temperature) the reaction will shift to the side with more moles of gas, to re-establish equilibrium. In this case, more moles of products (i.e., more H 2 O(g)) will be produced. Why doesn't the pressure of H 2 O(g) increase? Because the mole increase exactly compensates the volume increase. n H2 Thus, since P H2 O = O RT V = [H 2 O] eq RT. n H 2 O and V increase by the same amount - thus P H2 O and [H 2O] does not change. What will happen as V is increased gradually, at constant temperature, from an initial volume, V o (5 L, here) and becomes very large? The pressure of H 2 O(g) will stay constant until the Blue Vitriol runs out! Now, the equilibrium is destroyed! Then, since no more moles of H 2 O(g) can be produced, we have volume increase at constant temperature and (now) constant moles of H 2 O(g). This is Boyle's Law - pressure will decrease as volume increases, i.e., P α 1 V, a hyperbolic dependence. The graph below, where we plot P H 2 O versus Volume (V). V o denotes the initial volume of equilibration (5 L, here) and V* denotes the volume at which the Blue Vitriol runs out and hence the onset of Boyle's Law. P water atm. Boyle's Law V 0 V* V 16

17 Chemistry SEAS LeChâtelier's Principle - Return to Equilibrium LeChâtelier: when a small disturbance or stress is applied to an equilibrated system, the system will attempt to restore equilibrium in a fashion to minimize the effect of the stress. Stress Direction of Approach to Equilibrium Reason A. Temperature Increase in temperature favors the endothermic Van't Hoff change at reaction. Decrease in temperature favors the constant exothermic reaction. pressure B. Pressure Decrease in pressure favors the side of Q versus K eq change at the reaction with the most moles of gas. constant Increase in pressure favors the side of temperature, by the reaction with the least moles of gas. increasing/ decreasing volume C. Pressure No Effect Ideal gas ; change, at Q versus K eq constant volume and constant temperature, by adding an inert gas D. Addition of an This stress is equivalent to an increase in Q versus K eq inert gas, at volume, at constant temperature. The side of constant total the reaction with the most moles of gas is favored, pressure and in order to restore equilibrium constant temperature E. Pressure or Removing product - restoration is toward products. Q versus K eq Concentration Removing reactant - restoration is toward reactants. change, at Adding reactant - restoration is toward products. constant Adding product - restoration is toward reactants. temperature and constant volume by adding or removing reactant/product gaseous species 17

18 Stress Direction of Approach to Equilibrium Reason F. Adding or removing No effect - only if Q versus K eq pure solid or pure enough is present. liquid, that is a reactant or product, at constant temperature and constant volume. G. Adding a catalyst No Effect Kinetics 18