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Wright CHM 1220 - CHM 1220 SI_Chemical Equilibrium_Rev 2_ANSWERS

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Chapter 14 SI Activity: Chemical Equilibrium CHM 1220Problems1. The equilibrium constant Kp for the reaction PCl5 (g)  PCl3 (g) + Cl2 (g) is 3.81 × 102 at 600 K and 2.69 × 103 at 700 K.Ⓐ Without doing any calculations, predict whether this reaction is endothermic or exothermic.Ⓑ Verify your answer in part Ⓐ by calculating the change in enthalpy.Ⓒ Predict how the equilibrium would shift under the following stresses: (i) an increase in volume, (ii) addition of an inert gas, and (iii) addition of a catalyst.Ⓓ Calculate Kc for the reaction at 600 K.Ⓔ If the initial concentrations of the species are [PCl5] = 0.500 M, [PCl3] = 0.150 M, and [Cl2] = 0.600 at 600 K, in which direction does the reaction proceed toward equilibrium? What are the equilibrium concentrations? lnlnKKHRTTRKKHTT212121211111=− −−=−∆∆¥¥=−−=−()−⋅∆HRKKTT¥ln .l21211118 31447Jmol Knn...26910381101700160068 3321××−= +− K K kJmolANSWERSQuestions? Comments? Suggestions? Email me at [email protected] the reaction becomes increasingly product-favored as the temperature increases, the reaction must be en-dothermic by Le Châtlier’s principle. (We can imagine “heat” as a reactant for endothermic reactions, so heating the reaction vessel would be like adding reactant, which forces the reaction toward the right.)(i) Rxn shifts toward side with more molecules of gas to fill the available space. (ii) Adding inert gas increases the pressure, which causes the rxn to shift toward the side with fewer moles of gas. (iii) Cata-lysts affect the rates of the forward and reverse rxns equally, so there is no net effect on the equilibrium.Need to find ΔH for rxn given two Ks and two Ts:Use relationship between Kc and Kp:(i) (iii)(ii) Shifts Right (Products)Shifts Left (Reactants)No effectSince Q (0.180) is smaller than K (7.74), the rxn will proceed to the right (products).KKRTKKRTnnpccpL atmmol K =()=()=×()(∆−∆⋅⋅3811000820572..))()=→−−()600 7 73824 77421 K . .Need Q before determining rxn direction:Then, need ICE table to find concentrations:Use values from ICE table in equilibrium expression:Rewrite expression as a standard-form quadratic eqn:Solve using the quadratic formula (work not shown):Calculate equilibrium concentrations:Comparing Q to K:Q =[][][]=[][][]=PClClPCl M M M3250 600 0 1500 5000 180....PCl5 (g)PCl3 (g) + Cl2 (g)I 0.500 M 0.150 M 0.600 MC – x + x + xE 0.500 – x 0.150 + x 0.600 + xKxxxcPClClPCl=[][][]=+[]+[]−[]3250 600 0 1500 500...Kx xxxxxc0 500 0 600 0 1503 869 7 738 0 0900 0 750..... ..−[]=+[]+[]−=++22203779 8 488=− ++..xx8.912x = −Reject because wouldcause a negative concentration0 4240or .x =  [ ][ ][ ]532PCl 0.500 0.500 0.4240PCl 0.150 0.10.0759 M0.574 50 0.4240Cl 0.600 0.600 0.M1.02 4 M240xxx= −= − == += + == += + =2. The equilibrium concentrations in a gas mixture at a 298 K for the reaction H2 + I2  2 HI are [H2 ] = 0.13 M, [I2 ] = 0.70 M, and [HI] = 2.1. What equilibrium concentrations are obtained at the same temperature when 0.20 mol of HI is injected into an empty 500.0 mL container? You may assume all three gases behave ideally. Rev. 2 – Fall 2017 – J. Stevens3. For a particular biochemical reaction of the form A  2 B at 37 °C (normal body temperature), the enthalpy change is +20. kJ/mol and the entropy change is +34 J/mol ∙ K. Assuming that these quantities are independent of temperature, calculate the temperature conditions under which the reaction is product-favored. Then, cal-culate the equilibrium concentration of the product if the reaction were to take place at 37 °C with an initial reactant concentration of 2.00 M.Product-favored when K ≥ 1 (so when ln K ≥ 0):Calculate Kc from data given:Calculate [HI] injected:Use ICE table to find new concentrations:Use the given data to find K:Next, set up an ICE table:Use values from ICE table in equilibrium expression:020 104≤=−+−≤−≥≥=×ln.KHRTSRSRHRTTSRHRTHS∆∆∆∆∆∆∆∆¥¥¥¥¥¥¥¥ JmoolJmol K K34590⋅≥TA2 BI 2.00 M 0.00 MC – x + 2 xE 2.00 – x + 2 xlnln..KHRTSRK=− +=−×()()()⋅∆∆¥¥20 108 31447 3354 KJmolJmol K++()()=− ⇒= ⇒=⋅−348 314473 09118 003 09118 JmolJmol K.ln...KKeK445448Rewrite expression as a standard-form quadratic eqn:KxxKxxx=[][]=[]−[]−[]=[]−−BA222220020020 090897 0 045448 4....xx20=Solve using the quadratic formula (work not shown):Calculate equilibrium concentration:0.156x = −Reject because wouldcause a negative concentration0 1452or .x =  [ ]( )B 2 2 0.1452 0.29 Mx= = =KcHIHI=[][][]=[][][]=22222101307048 46154....HI mol mL 1 L mL M[]===nV0205001000040..H2 (g) + I2 (g) 2 HI (g)I 0 0 0.40 MC + x + x – 2 xE x x 0.40 – 2 xUse values from ICE table in equilibrium expression:KxxxcHIHI=[][][]=−[][][]22220402.Rewrite expression as a standard-form quadratic eqn:Kx xxxxxxc[]=−+=−+=−−222201616448 46154 0161640016 16 44....... ..46154Solve using the quadratic formula (work not shown):0.08062x = −Reject because wouldcause a negative concentrationo 0.04464r x =     Calculate equilibrium concentrations:HIHI M M220 0450310402 04020


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Wright CHM 1220 - CHM 1220 SI_Chemical Equilibrium_Rev 2_ANSWERS

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