Answers: Given: No. [COCl 2 ] = K c [CO][Cl 2 ], but there are many possible values for [CO]=[Cl 2 ]

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1 Chemical Equilibrium What are the concentrations of reactants and products at equilibrium? How do changes in pressure, volume, temperature, concentration and the use of catalysts affect the equilibrium state? The state of a system at equilibrium is dynamic The rate of the forward reaction is equal to the rate of the reverse reaction. The system hasn t stopped reacting. In chemical kinetics, the focus is on the portion of the reaction prior to t e (the time at which equilibrium is reached). In equilibrium, the focus is on time after which t e occurs. The addition of radioactive NaCl to a saturated solution. Experimental results show that for a given reaction equation at a particular temperature: aa + bb + gg + hh + K c = [G] g [H] h /[A] a [B] b This is known as the equilibrium constant expression and K c is called the concentration equilibrium constant (given as a dimensionless quantity) Question 1: CO (g) + Cl 2(g) COCl 2(g) K c = 1.2x10 3 at 395 o C If [CO] equals [Cl 2 ] at equilibrium for the reaction, is there just one possible value of [COCl 2 ]? Explain. Question 2: Suppose [O 2 ] is fixed at a certain constant value when equilibrium is reached in the following reversible reaction: 2SO 2(g) + O 2(g) 2SO 3(g) K c = 1.00x10 2 Do [SO 2 ] and [SO 3 ] have unique values? Does the ratio [SO 2 ]/[SO 3 ] have a unique value? Describe the equilibrium state when [O 2 ] = 1.00M. Answers: Question 1: CO (g) + Cl 2(g) COCl 2(g) K c = 1.2x10 3 at 395 o C If [CO] equals [Cl 2 ] at equilibrium for the reaction, is there just one possible value of [COCl 2 ]? Explain. No. [COCl 2 ] = K c [CO][Cl 2 ], but there are many possible values for [CO]=[Cl 2 ] Question 2: Suppose [O 2 ] is fixed at a certain constant value when equilibrium is reached in the following reversible reaction: 2SO 2(g) + O 2(g) 2SO 3(g) K c = 1.00x10 2 Do [SO 2 ] and [SO 3 ] have unique values? Does the ratio [SO 2 ]/[SO 3 ] have a unique value? Describe the equilibrium state when [O 2 ] = 1.00M. [SO 2 ] and [SO 3 ] do not have unique values, but the following ratios do: [SO 3 ] 2 /[SO 2 ] 2 ; [SO 2 ] 2 /[SO 3 ] 2 ; [SO 3 ]/[SO 2 ], [SO 2 ]/[SO 3 ]. If [O 2 ] = 1.00M 100 = [SO 3 ] 2 /([O 2 ][SO 2 ] 2 ) 100 = ([SO 3 ]/[SO 2 ]) 2 10 = [SO 3 ]/[SO 2 ] [SO 3 ] = 10[SO 2 ] 1

2 Given Equilibrium From the Perspective of Kinetics and Thermodynamics 2HI (g) H 2(g) + I 2(g) And the rate laws are known to be Rate f = k f [HI] 2 Rate r = k r [H 2 ][I 2 ] At equilibrium Rate f = Rate r If the same rate laws govern the reaction at equilibrium: k f [HI] 2 = k r [H 2 ][I 2 ] k f /k r = [H 2 ][I 2 ]/[HI] 2 = K c (the concentration units in k f /k r will match those of K c only if they too are based on the stoichiometric coefficients of the equations) (Sometimes this requires substitution of values into the rate determining step) The relationship between K c (based on concentration) and K p (based on partial pressure) Using the ideal gas law equation PV = nrt and remembering Molarity (M) = n/v Solving for the partial pressure of a substance P, P = [Concentration]RT, which differs by a factor RT Substituting this into the equilibrium expression gives: K p = K c (RT) n where n n is the change in moles of gas (products reactants) from the balanced chemical equation. Modifying Equilibrium Expressions Sometimes it is desirable to modify existing equilibrium expressions to represent a new set of conditions or to represent a new reaction. Reversing a reaction: When a chemical reaction with the equilibrium constant, K c, is reversed, the reverse reaction has the equilibrium constant 1/K c Multiplying the coefficients through by some number in a chemical reaction: If the coefficients are multiplied through by a common factor, n, the original K c is raised to the power n to produce the new equilibrium constant. (This brings up the necessity of noting the balanced chemical equation when citing a value for K c. K c is also temperature dependent.) Adding Chemical Equations: When equations are added to achieve an overall equation, the equilibrium constant for each reaction is multiplied to obtain the overall equilibrium constant. Solids and Pure Liquids: Since the concentrations of pure solids and liquids are constant during a reaction (i.e. as pure substances, their concentration does not change with respect to themselves), those terms are not included in the equilibrium expression. Question 3: Given K c = 1.8x10-6 for the reaction 2NO (g) + O 2(g) 2NO 2(g) at 457K, derive the value of K p at 457K for the reaction NO 2(g) NO (g) + ½ O 2(g) Question 4: The equilibrium constant K c for the reaction ½ N 2(g) + 3/2 H 2 O (g) NH 3(g) + ¾ O 2(g) at 900K is 1.97x Calculate K c at 900K for the reaction 4NH 3(g) + 3O 2(g) 2N 2(g) + 6H 2 O (g) Question 5: The reaction of steam with iron is an old method of producing hydrogen gas: 3Fe (s) + 4H 2 O (g) Fe 3 O 4(s) + 4H 2(g) Write the equilibrium constant expression K c and K p for this reaction. Answers Question 3: Given K c = 1.8x10-6 for the reaction 2NO (g) + O 2(g) 2NO 2(g) at 457K, derive the value of K p at 457K for the reaction NO 2(g) NO (g) + ½ O 2(g) K p = 4564 = 4.6x10 3 K p = (1/1.8x10-6 ).5 (.0821x900).5 Question 4: The equilibrium constant K c for the reaction ½ N 2(g) + 3/2 H 2 O (g) NH 3(g) + ¾ O 2(g) at 900K is 1.97x Calculate K c at 900K for the reaction 4NH 3(g) + 3O 2(g) 2N 2(g) + 6H 2 O (g) K c = 6.64x10 78 (1/1.97x10-20 ) 4 Question 5: The reaction of steam with iron is an old method of producing hydrogen gas: 3Fe (s) + 4H 2 O (g) Fe 3 O 4(s) + 4H 2(g) Write the equilibrium constant expression K c and K p for this reaction. K c = [H 2 ] 4 /[H 2 O] 4 ; K p = P H24 /P H2 O 4 2

3 ICE Diagram Equilibrium Expression: Using equilibrium concentrations to experimentally determine K at a given temperature M (See next slide for results).2076m K c = [0.0276] 2 / ([0.0037][0.0037]) = 56 This will be true at this temperature regardless of the initial concentrations M Interpretation of Equilibrium Constants: A very large K c or K p (K value with double digit powers of ten) signifies the reaction goes to completion A very small K c or K p (K value with double digit negative powers of ten) signifies the forward reaction only occurs to a slight extent. Remember: Kinetics determines when equilibrium is reached. H 2(g) + O 2(g) H 2 O (g) Although K p at 298K is very large, it takes an extremely long time to occur without the addition of energy or a catalyst. The reaction is said to be thermodynamically favorable,, but is kinetically controlled. The reaction quotient (a.k.a. the mass-action action expression) gives the current state of a chemical system and identifies the direction that the reaction will go to achieve equilibrium. aa + bb + gg + hh + Q c = [G] g [H] h /[A] a [B] b Or Q P = P Gg P Hh /P Aa P Bb This looks exactly like the equilibrium expression, but applies to any current state of the system. By comparing Q with K, the progress of the reaction can be determined Example 1: 2HI (g) H 2(g) + I 2(g) K c = 1.84x10-2 In which direction would a net change occur if the initial conditions in the reaction were [HI] = 1.00M and [H 2 ] = [I 2 ] = 0.100M? Example 2: For the reaction H 2 S (g) + I 2(s) 2HI (g) + S (s), K p = 1.34x10-5 at 60 o C. Initially, we bring together the following species: H 2 S (g) at a partial pressure of 0.010atm, HI (g) at atm, I 2(s), and S (s). When equilibrium is established, which gas will show an increase/decrease in its partial pressure? Which solid will increase/decrease in its amount? 3

4 Answers: Example 1: 2HI (g) H 2(g) + I 2(g) K c = 1.84x10-2 In which direction would a net change occur if the initial conditions in the reaction were [HI] = 1.00M and [H 2 ] = [I 2 ] = 0.100M? Q c = [0.100][0.100]/[1.00] 2 Q c = 1.0x10-2 which is less than K c. The reaction will proceed forward towards products. Example 2: For the reaction H 2 S (g) + I 2(s) 2HI (g) + S (s), K p = 1.34x10-5 at 60 o C. Initially, we bring together the following species: H 2 S (g) at a partial pressure of 0.010atm, HI (g) at atm, I 2(s), and S (s). When equilibrium is established, which gas will show an increase/decrease in its partial pressure? Which solid will increase/decrease in its amount? Q p = /0.010 = 1.0x10-4 which is larger than K p. The pressure of H 2 S will increase/ HI will decrease, I 2 will increase/ S will decrease. Le Châtelier s Principle Henri Le Châtelier 1888: When a change (that is, a change in concentration, temperature, pressure or volume) is imposed on a system at equilibrium, the system responds by attaining a new equilibrium condition that minimizes the impact of the imposed change. The use of a catalyst may speed up a reaction, but it speeds it up in both the forward and reverse direction, therefore a catalyst has no effect on the equilibrium state of the system. Example problems involving establishing the equilibrium constant, K, and those using K to establish equilibrium concentrations or partial pressures. Example moles of I 2 and 4.0 moles of Br 2 are placed in a 2.0L reactor at 150 o C, and reaction occurs until equilibrium is reached: I 2(g) + Br 2(g) 2IBr (g) Analysis then shows that the reaction contains 3.2 moles of IBr. What is the value of the equilibrium constant K c for the reaction? Answer Example moles of I 2 and 4.0 moles of Br 2 are placed in a 2.0L reactor at 150 o C, and reaction occurs until equilibrium is reached: I 2(g) + Br 2(g) 2IBr (g) Analysis then shows that the reaction contains 3.2 moles of IBr. What is the value of the equilibrium constant K c for the reaction? 3.2 moles of IBr imply that 3.2(1/2) = 1.6mol of I 2 and Br 2 were used, leaving =.40mol I 2 and = 2.4mol of Br 2 left K c = [IBr] 2 /[I 2 ][Br 2 ] = [1.6] 2 /[.20][1.2] = 1.07x10 1 Note that you get the same answer in moles or molarity. 4

5 Example 2 Some hydrogen and iodine are mixed at 400 o C in a 1.00 liter container, and when equilibrium is established the following concentrations are present: [HI] = 0.490M, [H 2 ] = 0.080M, and [I 2 ] = 0.060M. If an additional moles of HI are added, what concentrations will be present when the new equilibrium is established? Answer: Example 2 Some hydrogen and iodine are mixed at 400 o C in a 1.00 liter container, and when equilibrium is established the following concentrations are present: [HI] = 0.490M, [H 2 ] = 0.080M, and [I 2 ] = 0.060M. If an additional moles of HI are added, what concentrations will be present when the new equilibrium is established? H 2 + I 2 2HI K c = [0.490] 2 /[0.080][0.060] = = [ X] 2 /[0.080+X][.060+X] X =.0329M [HI] =.7242M [H 2 ] =.1129M [I 2 ] =.0929M Example 3 For the reaction at equilibrium 2NaHCO 3(s) Na 2 CO 3(s) + H 2 O (g) + CO 2(g) H o = 128kJ state the effects (increase, decrease, or no change) of the following stresses on the number of moles of sodium carbonate (Na 2 CO 3 ) at equilibrium in a closed container. (Note that Na 2 CO 3 is a solid; its concentration will remain constant, but the amount can change.) a. Removing CO 2(g) b. Adding H 2 O (g) c. Raising the temperature d. Adding NaHCO 3(s) Answer Example 3 For the reaction at equilibrium 2NaHCO 3(s) Na 2 CO 3(s) + H 2 O (g) + CO 2(g) H o = 128kJ state the effects (increase, decrease, or no change) of the following stresses on the number of moles of sodium carbonate (Na 2 CO 3 ) at equilibrium in a closed container. (Note that Na 2 CO 3 is a solid; its concentration will remain constant, but the amount can change.) a. Removing CO 2(g) b. Adding H 2 O (g) c. Raising the temperature d. Adding NaHCO 3(s) a. Increases moles of Na 2 CO 3 b. Decreases moles of Na 2 CO 3 c. Increases moles of Na 2 CO 3 (Endothermic) d. No effect 5

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