CHEMICAL EQUILIBRIUM Chapter 16 ALL BOLD NUMBERED PROBLEMS

Properties of an Equilibrium Equilibrium systems are DYNAMIC (in constant motion) & REVERSIBLE

CoCl42- (aq) + 6 H2O (l)  Co(H2O)62+ (aq) Cool Heat Blue to pink CoCl42- (aq) + 6 H2O (l)  Co(H2O)62+ (aq) Cool Pink to blue Co(H2O)62+ (aq)  CoCl42- (aq) + 6 H2O (l) Heat

Chemical Equilibrium Fe3+ + SCN- FeSCN2+ Fe(H2O)63+ Fe(SCN)(H2O)52+ + SCN- + H2O colorless red-orange

Chemical Equilibrium Fe3+ + SCN- FeSCN2+ After a period of time, the concentrations of reactants and products are constant. The forward and reverse reactions continue after equilibrium is attained.

Examples of Chemical Equilibria Phase changes such as H2O(s) H2O(liq)

Examples of Chemical Equilibria Acid Base Equilibria +Na2CO3 H3O+(aq) + 2CO32-(aq) OH-(aq) + 2HCO3-(aq) +CO2 Color Red Purple Violet Blue Blue-Green Green pH 2 4 6 8 10 12

Examples of Chemical Equilibria Formation of stalactites (ceiling) and stalagmites (floor) CaCO3(s) + H2O(liq) + CO2(g) Ca2+(aq) + 2 HCO3-(aq)

Chemical Equilibria CaCO3(s) + H2O(liq) + CO2(g) Ca2+(aq) + 2 HCO3-(aq) At a given T and P of CO2, [Ca2+] and [HCO3-] can be found from the EQUILIBRIUM CONSTANT.

THE EQUILIBRIUM CONSTANT For any type of chemical equilibrium of the type a A + b B c C + d D the following is a CONSTANT (at a given T). If K is known, then we can predict concentrations of products or reactants.

Determining K 2 NOCl(g) 2 NO(g) + Cl2(g) Place 2.00 mol of NOCl in a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. Calculate K. Solution Set up a table of concentrations: [NOCl] [NO] [Cl2] Initial 2.00 0 0 Change Equilibrium 0.66

Determining K 2 NOCl(g) 2 NO(g) + Cl2(g) Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. Calculate K. Solution Set of a table of concentrations: [NOCl] [NO] [Cl2] Initial 2.00 0 0 Change Equilibrium 0.66 -0.66 +0.66 +0.33 1.34 0.33

Determining K K = [ NO ] Cl NOCl K = [ NO ] Cl NOCl (0.66) (0.33) 2 NOCl(g) 2 NO(g) + Cl2(g) [NOCl] [NO] [Cl2] Initial 2.00 0 0 Change -0.66 +0.66 +0.33 Equilibrium 1.34 0.66 0.33 K = [ NO ] 2 Cl NOCl K = [ NO ] 2 Cl NOCl (0.66) (0.33) (1.34) 0.080

Writing and Manipulating K Expressions Solids and liquids NEVER appear in equilibrium expressions. S(s) + O2(g) SO2(g) K = [SO 2 ] [O

Writing and Manipulating K Expressions Solids and liquids NEVER appear in equilibrium expressions. NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) K = [NH 4 + ][OH - ] 3

Writing and Manipulating K Expressions Changing coefficients S(s) + 3/2 O2(g) SO3(g) 2 S(s) + 3 O2(g) 2 SO3(g) K = [SO 3 ] [O 2 3/2 K new = [SO 3 ] 2 [O K new = [SO 3 ] 2 [O (K old )

Writing and Manipulating K Expressions Changing direction S(s) + O2(g) SO2(g) SO2(g) S(s) + O2(g) K = [SO 2 ] [O K new = [O 2 ] [SO K new = [O 2 ] [SO 1 old

Writing and Manipulating K Expressions Adding equations for reactions S(s) + O2(g) SO2(g) K1 = [SO2] / [O2] SO2(g) + 1/2 O2(g) SO3(g) K2 = [SO3] / [SO2][O2]1/2 NET EQUATION S(s) + 3/2 O2(g) SO3(g) K net = [SO 3 ] [O 2 3/2 1 •

Manipulating K Expressions The equilibrium expression is tied to the equation from which it is written. Changing the equation changes the expression and thus the numerical value of K. Three general cases need to be considered. 1. If the equation is multiplied by a number, “a”, then the K is raised to the “a” power. 2. If the equation is reversed, then the new K is the reciprocal of the old K. 3. If two equations are added, the new K is the product of the two old K's. Using these rules, new K's can be derived for the modified equations. SAMPLE QUESTIONS

Writing and Manipulating K Expressions Concentration Units We have been writing K in terms of mol/L. These are designated by Kc. But with gases, P = (n/V)•RT = [ ] • RT P is proportional to concentration, so we can write K in terms of P. These are designated by Kp. Kc and Kp may or may not be the same. Kp = Kc(RT)Dn

Writing and Manipulating K Expressions Practice Problems Write the equation and the Kc expression for the formation of two moles of gaseous ammonia from the elements in the standard state. Kc for this reaction at 25oC is 3.5x108. Calculate Kp for this reaction. Calculate Kc for the reaction forming the elements from one mole of ammonia gas.

The Meaning of K K = [NH ] [N ][H 3.5 x 10 We can tell if a reaction is product-favored or reactant-favored. For N2(g) + 3 H2(g) 2 NH3(g) K c = [NH 3 ] 2 [N ][H 3.5 x 10 8 Concentration of products is much greater than that of reactants at equilibrium. The reaction is strongly product-favored.

The Meaning of K For AgCl(s) Ag+(aq) + Cl-(aq) Kc = [Ag+] [Cl-] = 1.8 x 10-5 Concentration of products is much less than that of reactants at equilibrium. The reaction is strongly reactant-favored. Ag+(aq) + Cl-(aq) AgCl(s) is product-favored.

Comparing Q and K The relative magnitudes of Q and K tell us which direction the reaction will proceed to reach equilibrium. If Q<K, not at equilibrium and Reactants ------> Products. If Q=K, the system is at equilibrium. If Q>K, not at equilibrium and Products ------> Reactants.

Using Q Can tell if a reaction is at equilibrium. If not, which way it moves to approach equilibrium.

Using Q If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium? H C H—C—C—C—C—H H—C—C—C—H K = n-butane iso-butane [iso] [n] = 2.5 If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium? Which way does the reaction “shift” to approach equilibrium?

If Q = K, then system is at equilibrium. Using Q In general, all reacting chemical systems are characterized by their REACTION QUOTIENT, Q. Q = product concentrations reactant concentrations If Q = K, then system is at equilibrium. [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium? Q = conc. of iso conc. of n 0.35 0.15 2.3 Q (2.3) < K (2.5) Reaction is NOT at equilibrium, so [Iso] must become ________ and [n] must ____________. larger decrease SAMPLE QUESTIONS

Using K H2(g) + I2(g) 2 HI(g) K = [HI] [H ][I ] 55.3 PROBLEM: Place 1.00 mol each of H2 and I2 in a 1.00 L flask. Calculate the equilibrium concentrations. H2(g) + I2(g) 2 HI(g) K c = [HI] 2 [H ][I ] 55.3

H2(g) + I2(g) 2 HI(g), Kc = 55.3 Step 1. Set up table to define EQUILIBRIUM concentrations. [H2] [I2] [HI] Initial Change Equilib where x is defined as amount of H2 and I2 consumed on approaching equilibrium. 1.00 1.00 0 -x -x +2x 1.00-x 1.00-x 2x

H2(g) + I2(g) 2 HI(g), Kc = 55.3 K = [2x] [1.00 - x][1.00 x] 55.3 7 . Step 2. Put equilibrium concentrations into Kc expression. K c = [2x] 2 [1.00 - x][1.00 x] 55.3 Step 3. Solve Kc expression - take square root of both sides. 7 . 44 = 2x 1.00 - x

H2(g) + I2(g) 2 HI(g), Kc = 55.3 7 . 44 = 2x 1.00 - x Step 3. Solve Kc expression - take square root of both sides. 7 . 44 = 2x 1.00 - x x = 0.788 Therefore, at equilibrium [H2] = [I2] = 1.00 - x = 0.21 M [HI] = 2x = 1.58 M

Sample Problem At a certain temperature 8.00 atm of H2S(g) comes to equilibrium with H2(g) and S2(g). The equilibrium pressure of the sulfur gas is 0.60 atm. Write the equation for the decomposition of the hydrogen sulfide, calculate the equilibrium pressure for each gas, and calculate KP. Solution

Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g)

Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g) K c = [NO 2 ] [N O 4 0.0059 at 298 K If initial concentration of N2O4 is 0.50 M, what are the equilibrium concentrations? Step 1. Set up an equilibrium table [N2O4] [NO2] Initial 0.50 0 Change +2x Equilib 0.50 - x 2x -x

Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g) Step 2. Substitute into Kc expression and solve. K c = 0.0059 [NO 2 ] [N O 4 (2x) (0.50 - x) Rearrange: 0.0059 (0.50 - x) = 4x2 0.0029 - 0.0059x = 4x2 4x2 + 0.0059x - 0.0029 = 0 This is a QUADRATIC EQUATION ax2 + bx + c = 0 a = 4 b = 0.0059 c = - 0.0029

Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g) Solve the quadratic equation for x. ax2 + bx + c = 0 a = 4 b = 0.0059 c = - 0.0029 x = -b ± b 2 - 4ac 2a x = -0.0059 ± ( . 0059 ) 2 - 4(4)(-0.0029) 2(4) x = - 0.00074 ± 1/8(0.046)1/2 = - 0.00074 ± 0.027

Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g) = -0.0059 ± ( . 0059 ) 2 - 4(4)(-0.0029) 2(4) x = -0.00074 ± 1/8(0.046)1/2 = -0.00074 ± 0.027 x = 0.026 or -0.028 But a negative value is not reasonable. Conclusion [N2O4] = 0.50 - x = 0.47 M [NO2] = 2x = 0.052 M More Sample Problems

EQUILIBRIUM AND EXTERNAL EFFECTS 38 EQUILIBRIUM AND EXTERNAL EFFECTS Temperature, catalysts, and changes in concentration affect equilibria. The outcome is governed by LE CHATELIER’S PRINCIPLE “...if a system at equilibrium is disturbed, the system tends to shift its equilibrium position to counter the effect of the disturbance.”

EQUILIBRIUM AND EXTERNAL EFFECTS Henri Le Chatelier 1850-1936 Studied mining engineering. Interested in glass and ceramics.

Figure 16.6 50o C 0o C NO2 / N2O4 is temperature dependent.

EQUILIBRIUM AND EXTERNAL EFFECTS Temperature change ---> change in K Consider the fizz in a soft drink CO2(g) + H2O(liq) CO2(aq) + heat Decrease T. What happens to equilibrium position? To value of K? K = [CO2] / P (CO2) K increases as T goes down because [CO2] increases and P(CO2) decreases. Increase T. Now what? Equilibrium shifts left and K decreases.

Temperature Effects on Equilibrium N2O4 (colorless) + heat 2 NO2 (brown) DHo = + 57.2 kJ K c = [NO 2 ] [N O 4 Kc = 0.00077 at 273 K Kc = 0.0059 at 298 K

EQUILIBRIUM AND EXTERNAL EFFECTS Add catalyst ---> no change in K A catalyst only affects the RATE of approach to equilibrium. Catalytic exhaust system

NH3 Production Fritz Haber, 1909 N2(g) + 3 H2(g) 2 NH3(g) K = 3.5 x 108 at 298 K K = 0.16 at 723 K

EQUILIBRIUM AND EXTERNAL EFFECTS Concentration changes no change in K only the position of equilibrium changes

Le Chatelier’s Principle Adding a “reactant” to a chemical system.

Le Chatelier’s Principle Removing a “reactant” from a chemical system.

Le Chatelier’s Principle Adding a “product” to a chemical system.

Le Chatelier’s Principle Removing a “product” from a chemical system.

Figure 16.7 (a) 7 n-butane molecules (b) 7 iso-butane are added (c) What are the new equilibrium concentration?

Butane-Isobutane Equilibrium K = [isobutane] [butane] 2.5 isobutane

Butane Isobutane Assume you are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. When the system comes to equilibrium again, what are [iso] and [butane]? K = 2.5 butane isobutane

Butane Isobutane Q = [isobutane] [butane] 1.25 0.50 + 1.50 0.63 Assume you are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. When the system comes to equilibrium again, what are [iso] and [butane]? K = 2.5 Solution Calculate Q immediately after adding more butane and compare with K. Q = [isobutane] [butane] 1.25 0.50 + 1.50 0.63 Q is LESS THAN K. Therefore, the reaction will shift to the ____________. Right (products)

Butane Isobutane You are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. Solution Q is less than K, so equilibrium shifts right — away from butane and toward isobutane. Set up concentration table [butane] [isobutane] Initial Change Equilibrium 0.50 + 1.50 1.25 - x + x 2.00 - x 1.25 + x

Equilibrium has shifted toward isobutane. Butane Isobutane You are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. Solution K = 2.50 [isobutane] [butane] 1.25 + x 2.00 - x = 1.07 M At the new equilibrium position, [butane] = 0.93 M and [isobutane] = 2.32 M. Equilibrium has shifted toward isobutane.

Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g) K c = [NO 2 ] [N O 4 0.0059 at 298 K Increase P in the system by reducing the volume.

Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g) K c = [NO 2 ] [N O 4 0.0059 at 298 K Increase P in the system by reducing the volume. In gaseous system the equilibrium will shift to the side with fewer molecules (in order to reduce the P). Therefore, reaction shifts LEFT and P of NO2 decreases and P of N2O4 increases.

Sample Problem 1. The temperature is increased. Write the equation for the formation of two mole of ammonia gas from the elements in the standard state. DH for this reaction is -92kJ. Predict the change in K and the direction of shift of an equilibrium mixture when: 1. The temperature is increased. 2. The PT is increased by adding He. 3. The total volume is increased. 4.Hydrogen gas is removed. 5.Nitrogen gas is added. 6.The temperature is decreased. 7.The total volume is decreased. 3 H2 (g) + N2 (g) <--> 2 NH3 (g) + 92kJ smaller, left none, right none, left larger, right

Practice Problems 1. 2 NO (g) + O2 (g) <--> 2 NO2 (g) a) Write the Kp expression. b) Write the Kc expression. c) If Kc = 0.50 at 25oC, calculate Kp. d) Determine Kc for the following reaction: 4 NO (g) + 2 O2 (g) <--> 4 NO2 (g) e) If 2.5 moles of NO, O2, and NO2 are placed in a 5.0 L container, is the system at equilibrium? If not, will the reaction proceed to the left or to the right? f) If 2.5 moles of NO and O2 are placed in a 5.0 L container at 350oC, 1.1 moles of NO2 are present at equilibrium. Calculate Kc.

Practice Problems 2. 2.5 moles of carbon dioxide is placed in a 0.50 L flask. CO2 (g) <--> C (s) + O2 (g) Kc = 1.82 at 25oC Calculate all equilibrium concentrations. 3. Calculate all equilibrium concentrations if the initial concentration of Cl2 is .100. Cl2 (g) <--> 2 Cl (g) Kc = 0.036 4. Calculate the equilibrium pressure of NO2 if the initial pressure of N2O4 is 1.0 atm at 25o C. N2O4 (g) <--> 2 NO2 (g) Kc = 0.11

Practice Problems 5. The initial concentrations of CO2 and H2 are 0.100 M. Calculate all equilibrium concentrations. CO2 (g) + H2 (g) <--> CO (g) + H2O (g) Kc = 0.64 6. BaCO3(s) + 2H+(aq) <--> H2O(l) + CO2(g) +Ba2+(aq) Explain the effects of the following on the given equation: a) adding HCl b) adding BaCl2 c) adding H2O d) adding BaCO3 e) adding NaOH

Practice Problems Answers 1. a) b) c) 0.02 d) 0.25 e) left f) 1.6 2. 1.8, 3.2 3. 0.074, 0.052 4. 1.1 atm 5. 0.044, 0.044, 0.056, 0.056 6. a) right b) left c) none d) none e) right P2 NO2 P2 NO P O2 Kc = [NO2]2 [NO]2[O2] Kp = The End!!!

Sample Questions Write the expression. 1. 2 H2 (g) + O2 (g) <--> 2 H2O (g) Kc = [H2O]2 [H2]2[O2]

Sample Questions Write the expression. 2. 4 H2 (g) + 2 O2 (g) <--> 4 H2O (g) Kc = [H2O]4 [H2]4[O2]2

Sample Questions Write the expression. 3. 2 H2O (g) <---> 2 H2 (g) + O2 (g) Kc = [H2] 2 [O2] [H2O]2

Sample Questions 1. 2 H2 (g) + O2 (g) <--> 2 H2O (g) Kc = [H2O]2 If = 5.0 2. 4 H2 (g) + 2 O2 (g) <--> 4 H2O (g) Kc = [H2O]4 [H2]4 [O2]2 = 25 3. 2 H2O (g) <---> 2 H2 (g) + O2 (g) Kc = [H2]2 [O2] [H2O]2 = 0.20

Sample Questions Write the equation and the Kc expression for the formation of two moles of gaseous ammonia from the elements in the standard state. 3 H2 (g) + N2 (g) <--> 2 NH3 (g) Kc = [NH3] 2 [H2] 3 [N2]

Sample Questions Kc for this reaction at 25oC is 3.5x108. Calculate Kp for this reaction. 3 H2 (g) + N2 (g) <--> 2 NH3 (g) Kp = Kc (RT)Dn Kp = 3.5 x 108 (0.0821x 298)-2 Kp = 5.8 x 105

Sample Questions Calculate Kc for the reaction forming the elements from one mole of ammonia gas. NH3 (g) <--> 3/2 H2 (g) + 1/2 N2 (g) Kc = [H2] 3/2 [N2]1/2 [NH3] = 1 (3.5 x 108)1/2 = 5.3 x 10-5

The Meaning of Q Q, Reaction Quotient Same expression as the equilibrium expression. Non-equilibrium concentration numbers may be used. Used to determine if a system is at equilibrium and if not, which direction it will move to reach equilibrium.

Sample Questions 1. If [NO2] = 0.050 and [N2O4] = 0.050, is the system at equilibrium? If not, in which direction does the reaction move to come to equilibrium? N2O4 (g) <--> 2 NO2 (g) Kc = 0.36 [NO2] 2 [N2O4] (0.050) 2 (0.050) Q = = = 0.050 Q < Kc forward direction (to right, products)

Sample Questions 2. If [NO2] = 0.50 and [N2O4] = 0.40, is the system at equilibrium? If not, in which direction does the reaction move to come to equilibrium? N2O4 (g) <--> 2 NO2 (g) Kc = 0.36 [NO2] 2 [N2O4] (0.50) 2 (0.40) Q = = = 0.62 Q > Kc reverse direction (to left, reactants)

Sample Questions 3. If [NO2] = 0.50 and [N2O4] = 0.69, is the system at equilibrium? If not, in which direction does the reaction move to come to equilibrium? N2O4 (g) <--> 2 NO2 (g) Kc = 0.36 [NO2] 2 [N2O4] (0.50) 2 (0.69) Q = = = 0 .36 Q = Kc equilibrium

Determining K 2 NOCl(g) 2 NO(g) + Cl2(g) Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. Calculate K. (Review of slides 9-11.) Solution Set up a table of concentrations: [NOCl] [NO] [Cl2] Initial 2.00 0 0 Change Equilibrium 0.66 - 0.66 +0.66 + 0.33 1.34 0.33

Determining K K = [ NO ] Cl NOCl K = [ NO ] Cl NOCl (0.66) (0.33) 2 NOCl(g) 2 NO(g) + Cl2(g) [NOCl] [NO] [Cl2] Before 2.00 0 0 Change -0.66 +0.66 +0.33 Equilibrium 1.34 0.66 0.33 K = [ NO ] 2 Cl NOCl K = [ NO ] 2 Cl NOCl (0.66) (0.33) (1.34) 0.080

Sample Questions 1. PCl5 dissociates to produce PCl3 and Cl2. If 2.00 moles of PCl5 is placed in a 1.00 L flask it is found that 10.0% of the original PCl5 had dissociated, at equilibrium. Calculate Kc. PCl5 <--> PCl3 + Cl2 2.00 0 0 - 0.200 + 0.200 +0 .200 1.80 0.200 0.200 [PCl3][Cl2] [PCl5] (.200)(.200) (1.80) Kc = = = 0.0222

Sample Questions The End!!! 2. If 1.50 moles of H2O are placed in a 3.00 L flask 1.20 moles of H2O remain at equilibrium. Calculate Kc. 2 H2O <--> 2 H2 + O2 0.500 0 0 Can we use only Moles?... - 0.100 + 0.100 + 0.050 0.400 0.100 0.050 No….reason is all K’s are relative, that is since all reactions are done in different containers, moles and size of the container determine the K. [H2]2 [O2] [H2O]2 (.100)2(.050) (0.400)2 Kc = = = 0.0031 The End!!!

Sample Questions 3. When 0.020 mole N2O and 0.0560 mole O2 are placed in a 1.00 L flask it is found that there is 0.020 mole NO2 at equilibrium. Calculate Kc. 2 N2O + 3 O2 <--> 4 NO2 0.020 0.0560 0 - 0.010 - 0.015 +0 .020 0.010 0.041 0.020 [NO2]4 [N2O]2 [O2]3 (0.020)4 (.010)2(.041)3 Kc = = = 23

Sample Questions 4. Some solid ammonium hydrogen sulfide is placed in a flask at 298 K. At equilibrium, the pressure was 0.660 atm. Calculate Kp and Kc. NH4HS <--> NH3 + H2S some 0 0 - some + 0.330 + 0.330 some - some 0.330 0.330 Kp = PNH3PH2S = (0.330)2 = 0.109 Kp = Kc(RT)2 0.109 = Kc (0.0821x298)2 Kc = 0.000182

Sample Problem 2 H2S (g) <--> 2 H2 (g) + S2 (g) 8.00 0 0 - 1.20 - 1.20 + 1.20 + 0.60 6.80 1.20 0.60 PH22 PS2 PH2S2 (1.20)2 (0.60) (6.80)2 Kp = = = 0.019

Sample Problems 1. Calculate the equilibrium pressure of CO if 25 g COBr2 is placed in a 10.0 L flask at 298 K. CO (g) + Br2 (g) <--> COBr2 (g) Kp = 0.19 25 g mole 187.8 g = 0.13 mole P = nRT/V P = (0.13 mol)(0.0821 L atm/molK)(298K)/10.0 L) P = 0.32 atm

Sample Problems 1. Calculate the equilibrium pressure of CO if 25 g COBr2 is placed in a 10.0 L flask at 298 K. CO (g) + Br2 (g) <--> COBr2 (g) Kp = 0.19 0 0 0.32 + x + x - x x x 0.32 - x PCOBr2 PCOPBr2 (0.32 - x) x2 Kp = = = 0.19 x = 0.30atm = PCO

Sample Problem 2. Calculate all equilibrium concentrations if 0.500 moles of HI is placed in a 2.00 L flask. 2 HI <--> H2 + I2 Kc = 0.018 0.250 0 0 - 2x + x + x 0.250 - 2x x x [H2][I2] [HI]2 x2 (0.250 - 2x)2 Kc = = = 0.018 x = 0.026 M = [H2] = [I2] [HI] = 0.20 M

Sample Problem 3. Calculate all equilibrium concentrations when the initial concentrations of SO2 and NO2 are 0.0500. SO2 + NO2 <--> NO + SO3 Kc = 85.0 0.0500 0.0500 0 0 - x - x + x + x 0.0500 - x 0.0500 - x x x [NO][SO3] [SO2][NO2] x2 (0.0500 - x)2 Kc = = = 85.0 x = 0.0451 M = [NO] = [SO3] [SO2] = [NO2] = 0.0049 M