2Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition6.2 The Equilibrium Constant6.3 Equilibrium Expressions Involving Pressures6.4 The Concept of Activity6.5 Heterogeneous Equilibria6.6 Applications of the Equilibrium Constant6.7 Solving Equilibrium Problems6.8 LeChatelier’s Principle6.9 Equilibria Involving Real Gases
3Nitrogen dioxide shown immediately after expanding
4Figure 6.1: Reaction of 2NO2(g) and N2O4(g) over time in a closed vessel
6Reaching Equilibrium on the Macroscopic and Molecular Level N2O4 (g) NO2 (g)Colorless Brown
7The State of Equilibrium For the Nitrogen dioxide - dinitrogen tetroxide equilibrium:N2O4 (g, colorless) = 2 NO2 (g, brown)At equilibrium: ratefwd = raterevratefwd = kfwd[N2O4] raterev = krev[NO2]2kfwd [NO2]2krev [N2O4]= = Keqkfwd[N2O4] = krev[NO2]21) Small kN2 (g) + O2 (g) NO(g) K = 1 x2) Large k CO(g) + O2 (g) CO2 (g) K = 2.2 x 10223) Intermediate k BrCl(g) Br2 (g) + Cl2 (g) K = 5
8Writing the Reaction Quotient or Mass-Action Expression Q = mass-action expression or reaction quotientProduct of the Product ConcentrationsQ =Product of the Reactant ConcentrationsFor the general reaction:a A + bB cC + dDQ =[C]c [D]d[A]a [B]bExample: The Haber process for ammonia production:N2 (g) + 3 H2 (g) NH3 (g)[NH3]2Q =[N2][H2]3
10Reaction Direction and the Relative Sizes of Q and K
11Initial and Equilibrium Concentrations for the N2O4-NO2 System at 100°C Initial Equilibrium Ratio[N2O4][NO2][N2O4][NO2][NO2]2[N2O4]
12Figure 6.2: Changes in concentration with time for the reaction H2O(g) + CO(g) H2 (g) + CO2 (g)
13Molecular model: When equilibrium is reached, how many molecules of H2O, CO, H2, and CO2 are present?
14Figure 6.3: H2O and CO are mixed in equal numbers H2O(g) + CO(g) H2 (g) + CO2 (g)
15Figure 6.4: Changes with time in the rates of forward and reverse reactions H2O(g) + CO(g) H2 (g) + CO2 (g)
16Figure 6.5: Concentration profile for the reaction
17Like Example 6.1 (P 201) - IThe following equilibrium concentrations were observed for theReaction between CO and H2 to form CH4 and H2O at 927oC.CO(g) + 3 H2 (g) = CH4 (g) + H2O(g)[CO] = mol/L [CH4] = mol/L[H2] = mol/L [H2O] = mol/LCalculate the value of K at 927oC for this reaction.Calculate the value of the equilibrium constant at 927oC for:H2O(g) + CH4 (g) = CO(g) + 3 H2 (g)1/3 CO(g) + H2 (g) = 1/3 CH4 (g) + 1/3 H2O(g)Solution:a) Given the equation above:[CH4][H2O](0.387 mol/L)(0.387 mol/L)K = = = _______ L2/mol2[CO][H2]3(0.613 mol/L)(1.839 mol/L)3
18Like Example 6.1 (P 201) - IIb)Calculate the value of the equilibrium constant at 927oC for:H2O(g) + CH4 (g) = CO(g) + 3 H2 (g)[CO][H2]3(0.613 mol/L)(1.839 mol/L)3K = = = _____ mol2/L2[H2O][CH4](0.387 mol/L)(0.387 mol/L)This is the reciprocal of K:1K1= = ____________ mol2/L2L2/mol2Calculate the value of the equilibrium constant at 927oC for:1/3 CO(g) + H2 (g) = 1/3 CH4 (g) + 1/3 H2O(g)[H2O]1/3[CH4]1/3(0.387mol/L)1/3(0.387 mol/L)1/3K = =[CO]1/3[H2](0.613 mol/L)1/3(1.839 mol/L)(0.729) (0.729)K = = L2/3/mol2/3 = (0.0393L2/mol2)1/3(0.850)(1.839)
19Summary: Some Characteristics of the Equilibrium Expression The equilibrium expression for a reaction written in reverse is thereciprocal of that for the original reaction.When the balanced equation for a reaction is multiplied by afactor n, the equilibrium expression for the new reaction is theoriginal expression raised to the nth power. Thus Knew = (Koriginal)nThe apparent units for K are determined by the powers of thevarious concentration terms. The (apparent) units for K thereforedepend on the reaction being considered. We will have more to sayabout the units for K in section 6.4.
21Expressing K with Pressure Units VFor gases, PV=nRT can be rearranged to give: P = RTn PV RTnVor: =Since = Molarity, and R is a constant if wekeep the temperature constant thenthe molar concentration is directlyproportional to the pressure.Therefore for an equilibrium between gaseous compounds we canexpress the reaction quotient in terms of partial pressures.For: NO(g) + O2 (g) NO2 (g)If there is no change in the number of moles ofreactants and products then n = 0 thenKc = Kp , or if there is a change in the number ofmoles of reactants or products then:P 2NO2Qp =P 2NO x PO2Kp = Kc(RT) ngas
22( )l ( )m ( )j ( )k The Concept of Activity Pi Preference Activity (ith component) = ai =Pi= partial pressure of the ith gaseous componentPreference= 1 atm (exactly)For the reaction: jA(g) + kB(g) == lC(g) + mD(g)( )lPCPref( )mPDPref(aC)l (aD)m(aA)J (aB)kPClPDmPAjPBkK = = =( )j( )kPAPrefPBPrefBecause of the difference in reference states between concentrationunits and pressure units, unless the numbers of moles of reactant andproducts are the same, then: K = Kp
23Figure 6.6: Position of the equilibrium CaCO3 (s) CaO(s) + CO2 (g) Keq = [CO2]Kp = PCO2
24Writing the Reaction Quotient from the Balanced Equation Problem: Write the reaction quotient for each of the following reactions:(a) The thermal decomposition of potassium chlorate:KClO3 (s) = KCl(s) + O2 (g)(b) The combustion of butane in oxygen:C4H10 (g) + O2 (g) = CO2 (g) + H2O(g)Plan: We first balance the equations, then construct the reaction quotientas described by equation 17.4.Solution:(a) 2 KClO3 (s) KCl(s) + 3 O2 (g) Qc = = [O2]3[KCl]2[O2]3[KClO3]2(b) 2 C4H10 (g) + 13 O2 (g) CO2 (g) + 10 H2O(g)[CO2]8 [H2O]10Qc =[C4H10]2 [O2]13
25Writing the Reaction Quotient for an Overall Reaction–I Problem: Oxygen gas combines with nitrogen gas in the internalcombustion engine to produce nitric oxide, which when out in theatmosphere combines with additional oxygen to form nitrogen dioxide.(1) N2 (g) + O2 (g) NO(g) Kc1 = 4.3 x 10-25(2) 2 NO(g) + O2 (g) NO2 (g) Kc2 = 6.4 x 109(a) Show that the overall Qc for this reaction sequence is the same as theproduct of the Qc’s for the individual reactions.(b) Calculate Kc for the overall reaction.Plan: We first write the overall reaction by adding the two reactionstogether and write the Qc. We then multiply the individual Kc’s for thetotal K.(1) N2 (g) + O2 (g) NO(g)(2) NO(g) + O2 (g) NO2 (g)overall: N2 (g) + 2 O2 (g) NO2 (g)
26Writing the Reaction Quotient for an Overall Reaction–II (a) cont.[NO2]2Qc (overall) =[N2][O2]2For the individual steps:[NO]2(1) N2 (g) + O2 (g) NO(g) Qc1 =[N2] [O2][NO2]2(2) 2 NO(g) + O2 (g) NO2 (g) Qc2 =[NO]2 [O2][NO]2[NO2]2[NO2]2Qc1 x Qc2 = x =The same![N2] [O2][NO]2 [O2][N2][O2]2(b) K = Kc1 x Kc2 = (4.3 x 10-25)(6.4 x 109) = ______________
27The Form of Q for a Forward and Reverse Reaction The production of sulfuric acid depends upon the conversion of sulfurdioxide to sulfuric trioxide before the sulfur trioxide is reacted withwater to make the sulfuric acid.2 SO2 (g) + O2 (g) SO3 (g)[SO3]2Qc(fwd) =[SO2]2[O2]For the reverse reaction:2 SO3 (g) SO2 (g) + O2 (g)Qc(rev) = =[SO2]2[O2][SO3]21Qc(fwd)at 1000K Kc(rev) = 26111and: Kc(fwd) = = = _____________Kc(rev)261
28Ways of Expressing the Reaction Quotient, Q Form of Chemical Equation Form of Q Value of K[B]eq[B]Reference reaction: A B Q(ref) = K(ref) =Reverse reaction: B A Q = = K =Reaction as sum of two steps:[A][A]eq[A]1K(ref)Q(ref) [B][A] [C][C] [B]Q1 = ; Q2 =(1) A CQoverall = Q1 x Q2 = Q(ref) Koverall = K1 x K2= x =(2) C B[C] [B] [B]= K(ref)[A] [C] [A]Coefficients multiplied by n Q = Qn(ref) K = Kn(ref)Reaction with pure solid or Q’ = Q(ref)[A] = [B] K’ = K(ref)[A] = [B]liquid component, such as A(s)
29Example 6.2 (P 209) - IFor the synthesis of ammonia at 500oC, the equilibrium constant is6.0 x 10-2 L2/mol2. Predict the direction in which the system willshift to reach equilibrium in each of the following cases.[NH3]0 = 1.0 x 10-3 M; [N2]0= 1.0 x 10-5 M; [H2]0=2.0 x 10-3 M[NH3]0 = 2.00 x 10-4 M; [N2]0= 1.50 x 10-5 M; [H2]0= 3.54 x 10-1 M[NH3]0 = 1.0 x 10-4 M; [N2]0= 5.0 M; [H2]0= 1.0 x 10-2 MSolutiona) First we calculate the Q:[NH3]02[N2]0[H2]03(1.0 x 10-3 mol/L)2Q = == ______________ L2/mol2(1.0 x 10-5 mol/L)(2.0 x 10-3 mol/L)3Since K = 6.0 x 10-2 L2/mol2, Q is much greater than K. For thesystem to attain equilibrium, the concentrations of the productsmust be decreased and the concentrations of the reactantsincreased. The system will shift to the left:
30Example 6.2 (P 209) - II b) We calculate the value of Q: [NH3]02 (2.00 x 10-4 mol/L)2Q = == _______________ L2/mol2(1.50 x 10-5 mol/L) (3.54 x 10-1 mol/L)3)In this case Q = K, so the system is at equilibrium. No shift will occur.c) The value of Q is:[NH3]02[N2]0[H2]03(1.0 x 10-4 mol/L)2Q = == ________________ L2/mol2(5.0 mol/L) (1.0 x 10-2 mol/L)3Here Q is less than K, so the system will shift to the right, attainingequilibrium by increasing the concentration of the product anddecreasing the concentrations of the reactants. More Ammonia!
31Like Example 6.3 (P ) - ILook at the equilibrium example for the formation of Hydrogen Chloridegas from Hydrogen gas and Chlorine gas. Initially mol of H2, and4.000 mol of Cl2, are added to mol of gaseous HCl in a literflask.H2 (g) + Cl2 (g) HCl(g)K = 2.76 x 102 = [Cl2] = [H2] = 4.000mol/2.000L = 2.000M[HCl] = mol/2.000L = 1.000MInitial Concentration Change Equilibrium Conc.(mol/L) (mol/L) (mol/L)[H2]o = 2.000M x [H2] = x[Cl2]o = 2.000M x [Cl2] = x[HCl]o = 1.000M x [HCl] = x[HCl]2[H2] [Cl2]
32Like Example 6.3 (P210-211) - II [HCl]2 [H2] [Cl2] (1.000 + 2x)2 (2.000 –x)(2.000 – x)( x)2(2.000 – x)2K = 2.76 x 102 = = =Take the square root of each side:16.61 =( x)(2.000 – x)33.22 – 16.61x = x Therefore: [H2] = M32.22 = 18.61x [Cl2] = Mx = [HCl] = MCheck:= = OK![HCl]2[H2] [Cl2](4.462)2(0.269)(0.269)
33Summary: Solving Equilibrium Problems Write the balanced equation for the reaction.Write the equilibrium expression using the law of mass action.List the initial concentrations.Calculate Q and determine the direction of the shift to equilibrium.Determine the change needed to reach equilibrium, and define theequilibrium concentrations by applying the change to theinitial concentrations.Substitute the equilibrium concentrations into the equilibriumexpression, and solve for the unknown.Check your calculated equilibrium concentrations by making sure thatthey give the correct value of K.
34Determining Equilibrium Concentrations from K–I Problem: One laboratory method of making methane is from carbondisulfide reacting with hydrogen gas, and K this reaction at 900°C is 27.8.CS2 (g) + 4 H2 (g) CH4 (g) + 2 H2 S(g)At equilibrium the reaction mixture in a 4.70 L flask contains molCS2, 1.10 mol of H2, and 0.45 mol of H2S, how much methane wasformed?Plan: Write the reaction quotient, and calculate the equilibriumconcentrations from the moles given and the volume of the container.Use the reaction quotient and solve for the concentration of methane.Solution:CS2 (g) + 4 H2 (g) CH4 (g) + 2 H2 S(g)[CS2] =0.250 mol4.70 L[CH4] [H2S]2K = = 27.8[CS2] [H2]4[CS2] = ____________ mol/L
35Determining Equilibrium Concentrations from K–II Solution cont.1.10 mol0.450 mol[H2] = = ________ mol/L[H2S] = = _________ mol/L4.70 L4.70 LKc [CS2] [H2]4(27.8)( )( )4[CH4] = =[H2S]2( )2[CH4] = = mol/L = MCheck: Substitute the concentrations back into the equation for K andmake sure that you get the correct value of K( M)( M)2[CH4] [H2S]2K = = =[CS2] [H2]4( M)( M)4OK!
36Determining Equilibrium Concentrations from Initial Concentrations and K –IProblem: Given the that the reaction to form HF from molecularhydrogen and fluorine has a reaction quotient of 115 at a certaintemperature. If mol of each component is added to a L flask,calculate the equilibrium concentrations of each species.H2 (g) + F2 (g) HF(g)Plan: Calculate the concentrations of each component, and then figurethe changes, and solve the equilibrium equation to find the resultantconcentrations.Solution:3.000 mol[H2] = = M1.500 LK = = 115[HF]2[H2] [F2]3.000 mol[F2] = = M1.500 L3.000 mol[HF] = = M1.500 L
37Determining Equilibrium Concentrations from Initial Concentrations and K–II Concentration (M) H F HFInitialChange x x xFinal x x xK = = = =[HF]2[H2][F2]( x)2( x)2( x)( x)( x)2Taking the square root of each side we get:( x)(115)1/2 = =x = 1.528( x)K = =[HF]2[H2][F2](5.056 M)2[H2] = = M(0.472 M)(0.472 M)[F2] = = Mcheck:K = 115[HF] = (1.528) = M
38Calculating K from Concentration Data–I Problem: Hydrogen iodide decomposes at moderate temperatures by thereaction below:When 4.00 mol HI was placed in a 5.00 L vessel at 458°C, the equilibriummixture was found to contain mol I2. What is the value of Kc ?Plan: First we calculate the molar concentrations, and then put them intothe equilibrium expression to find it’s value.Solution: To calculate the concentrations of HI and I2, we divide theamounts of these compounds by the volume of the vessel.2 HI(g) H2 (g) + I2 (g)4.00 mol5.00 LStarting conc. of HI = = M0.442 mol5.00 LEquilibrium conc. of I2 = = MConc. (M) HI(g) H2 (g) I2 (g)StartingChange x x xEquilibrium x x x =
39Calculating K from Concentration Data–II [HI] = M = ( x ) M = M[H2] = x = M = [I2][H2] [I2]( )(0.0884)Kc = = = ___________(0.623)2[HI]2Therefore the equilibrium constant for the decomposition of HydrogenIodide at 458°C is only meaning that the decomposition does notproceed very far under these temperature conditions. We were given theinitial concentrations, and that of one at equilibrium, and found theothers that were needed to calculate the equilibrium constant.
40Using the Quadratic Formula to Solve for the Unknown Given the Reaction between CO and H2O:Concentration (M) CO(g) H2O(g) CO2(g) H2(g)InitialChange x x x xEquilibrium x x x x[CO2][H2](x) (x)x2Qc = = = = 1.56[CO][H2O](2.00-x)(1.00-x)x xWe rearrange the equation: x x = 0ax2 + bx + c = 0quadratic equation:x =- b b2 - 4ac2a[CO] = 1.27 M[H2O] = 0.27 M[CO2] = 0.73 M[H2] = 0.73 M(-4.68)2 - 4(0.56)(3.12)x = = 7.6 Mand 0.73 M2(0.56)
41Predicting Reaction Direction and Calculating Equilibrium Concentrations –IProblem: Two components of natural gas can react according to thefollowing chemical equation:CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)In an experiment, 1.00 mol CH4, 1.00 mol CS2, 2.00 mol H2S, and2.00 mol H2 are mixed in a 250 mL vessel at 960°C. At this temperature,K = (a) In which direction will the reaction go?(b) If [CH4] = 5.56 M at equilibrium, what are the concentrations of theother substances?Plan: The find the direction, we calculate Qc using the calculatedconcentrations from the data given, and compare it with Kc. (b) Basedupon (a), we determine the sign of each component for the reaction tableand then use the given [CH4] at equilibrium to determine the others.Solution:[H2S] = 8.00 M, [CS2] = 4.00 Mand [H2 ] = 8.00 M1.00 mol0.250 L[CH4] = = M
42Predicting Reaction Direction and Calculating Equilibrium Concentrations –II[CS2] [H2]44.00 x (8.00)4Q = = = 64.0[CH4] [H2S]24.00 x (8.00)2Comparing Q and K: Q > K (64.0 > 0.036, so the reaction goes tothe left. Therefore, reactants increase and products decrease theirconcentrations.(b) Setting up the reaction table, with x = [CS2] that reacts, which equalsthe [CH4] that forms.Concentration (M) CH4 (g) H2S(g) CS2(g) H2(g)InitialChange x x x xEquilibrium x x x xSolving for x at equilibrium: [CH4] = 5.56 M = 4.00 M + xx = 1.56 M
43Predicting Reaction Direction and Calculating Equilibrium Concentrations –IIIx = 1.56 M = [CH4]Therefore:[H2S] = 8.00 M + 2x = 8.00 M + 2(1.56 M) = M[CS2] = 4.00 M - x = 4.00 M M = M[H2] = 8.00 M - 4x = 8.00 M - 4(1.56 M) = 1.76 M[CH4] = 1.56 M
44Le Chatelier’s Principle “If a change in conditions (a “stress”) is imposed on a systemat equilibrium, the equilibrium position will shift in a directionthat tends to reduce that change in conditions.”A + B C + D + EnergyFor example: In the reaction above, if more A or B is added youwill force the reaction to produce more product, if they are removed,it will force the equilibrium to form more reactants. If C or D isadded you will force the reaction to form more reactants, if they areRemoved from the reaction mixture, it will force the equilibrium toForm more products. If it is heated, you will get more reactants,and if cooled, more products.
45Henri Louis Le Chatelier Source: Photo Researchers
49The Effect of a Change in Concentration–I Given an equilibrium equation such as :CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g)If one adds ammonia to the reaction mixture at equilibrium, it will forcethe reaction to go to the right producing more product. Likewise, if onetakes ammonia from the equilibrium mixture, it will force the reactionback to produce more reactants by recombining H2 and HCN to givemore of the initial reactants, CH4 and NH3.CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g)Add NH3Forces equilibrium toproduce more product.CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g)Remove NH3Forces the reaction equilibrium to go backto the left and produce more of the reactants.
50The Effect of a Change in Concentration–II CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g)If to this same equilibrium mixture one decides to add one of theproducts to the equilibrium mixture, it will force the equilibrium backtoward the reactant side and increase the concentrations of reactants.Likewise, if one takes away some of the hydrogen or hydrogen cyanidefrom the product side, it will force the equilibrium to replace it.CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g)Forces equilibrium to gotoward the reactant direction.Add H2CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g)Remove HCNForces equilibrium to make moreproduce and replace the lost HCN.
51The Effect of a Change in Pressure (Volume) Pressure changes are mainly involving gases as liquids and solidsare nearly incompressible. For gases, pressure changes can occur inthree ways:Changing the concentration of a gaseous componentAdding an inert gas (one that does not take part in the reaction)Changing the volume of the reaction vesselWhen a system at equilibrium that contains a gas undergoes a changein pressure as a result of a change in volume, the equilibrium positionshifts to reduce the effect of the change.If the volume is lower (pressure is higher), the total number of gasmolecules decrease.If the volume is higher (pressure is lower), the total number of gasmolecules increases.
52Figure 6.8: A mixture of NH3(g), N2(g), and H2(g) at equilibrium N2 (g) + 3 H2 (g) NH3 (g)
53Figure 6.9: Brown NO2(g) and colorless N2O4(g) at equilibrium in a syringe 2 NO2 (g) N2O4 (g)Brown ColorlessSource: Ken O’Donoghue
54The Effect of a Change in Temperature Only temperature changes will alter the equilibrium constant, and that iswhy we always specify the temperature when giving the value of Kc.The best way to look at temperature effects is to realize that temperatureis a component of the equation, the same as a reactant, or product. Forexample, if you have an exothermic reaction, heat (energy) is on theproduct side of the equation, but if it is an endothermic reaction, it willbe on the reactant side of the equation.O2 (g) + 2 H2 (g) H2O(g) + Energy = ExothermicElectrical energy + 2 H2O(g) H2 (g) + O2 (g) = EndothermicA temperature increase favors the endothermic direction and atemperature decrease favors the exothermic direction.A temperature rise will increase Kc for a system with a positive H0rxnA temperature rise will decrease Kc for a system with a negative H0rxn
58Table 6.4 (P 222) Shifts in the Equilibrium Position for the Reaction N2O4(g) NO2 (g)Change ShiftAddition of N2O4 (g) More ProductsAddition of NO2 (g) More ReactantsRemoval of N2O4 (g) More ReactantsRemoval of NO2 (g) More ProductsAddition of He(g) NoneDecrease in Container Volume More ReactantsIncrease in Container Volume More ProductsIncrease in Temperature More ProductsDecrease in Temperature More Reactants
59Table 6.5 (P 223) Values of Kpobs at 723 K for the Reaction N2 (g) H2 (g) NH3 (g)as a Function of Total Pressure (at equilibrium)Total Pressure Kpobs(atm) (atm-2)x 10-5x 10-5x 10-5x 10-5x 10-4x 10-4
60Percent Yield of Ammonia vs. Temperature (°C) at five different operating pressures.
62Predicting the Effect of a Change in Concentration on the Position of the Equilibrium Problem: Carbon will react with water to yield carbon monoxide andand hydrogen, in a reaction called the water gas reaction that was usedto convert coal into a fuel that can be used by industry.C(s) + H2O (g) CO(g) + H2 (g)What happens to:(a) [CO] if C is added? (c) [H2O] if H2 is added?(b) [CO] if H2O is added? (d) [H2O] if CO is removed?Plan: We either write the reaction quotient to see how equilibrium willbe effected, or look at the equation, and predict the change in directionof the reaction, and the effect of the material desired.Solution:(a) No change, as carbon is a solid, and not involved in theequilibrium, as long as some carbon is present to allow the reaction.(b) The reaction moves to the product side, and [CO] increases.(c) The reaction moves to the reactant side, and [H2O] increases.(d) The reaction moves to the product side, and [H2O] decreases.
63Predicting the Effect of Temperature and Pressure Problem: How would you change the volume (pressure) or temperaturein the following reactions to increase the chemical yield of the products?(a) 2 SO2 (g) + O2 (g) SO3 (g); H0 = 197 kJ(b) CO(g) + 2 H2 (g) CH3OH(g); H0 = kJ(c) C(s) + CO2 (g) CO(g); H0 = kJ(d) N2(g) + 3 H2(g) NH3(g); H0 = kJPlan: For the impact of volume (pressure), we examine the reaction forthe side with the most gaseous molecules formed. For temperature, wesee if the reaction is exothermic, or endothermic. An increase in volume(pressure) will force a reaction toward fewer gas molecules.Solution: To get a higher yield of the products you should:(a) Increase the pressure, and increase the temperature.(b) Increase the pressure, and decrease the temperature.(c) Decrease the pressure, and increase in thetemperature will increase the product yield.(d) Increase the pressure, and decrease the temperature.
64Effect of Various Disturbances on an Equilibrium System Disturbance Net Direction of Reaction Effect on Value of KConcentrationIncrease [reactant] Toward formation of product NoneDecrease [reactant] Toward formation of reactant NonePressure (volume)Increase P Toward formation of loweramount (mol) of gas NoneDecrease P Toward formation of higherTemperatureIncrease T Toward absorption of heat Increases if H0rxn> 0Decreases if H0rxn< 0Decrease T Toward release of heat Increases if H0rxn< 0Decreases if H0rxn> 0Catalyst added None; rates of forward and reversereactions increase equally None