ide. SO2Cl2, is used as a reagent in the syn- 56 hesis of organic compounds. Whe
ID: 1029875 • Letter: I
Question
ide. SO2Cl2, is used as a reagent in the syn- 56 hesis of organic compounds. When heated to a suffi tly high temperature, it d sO2C12(g) SO2(g) + Cl2(g) a) A ecomposes to SO2 and Cl2 Kc = 0.045 at 375 °C 10.0-L flask containing 6.70 g of SO2Cl2 is heated to 375 °C. What is the concentration of each of the compounds in the system when equilibrium is achieved? What fraction of SO2Cl2 has dissociated? (b) What are the concentrations of SO2CI2, SO2, and Cl2 at equilibrium in the 10.0-L flask at 375 °C if you begin with a mixture of SO2Cl2 (6.70 g) and Ch (0.10 atm)? What fraction of SO2Cl2 has dissociated? c) Compare the fractions of SO2Cl2 in parts (a) and (b) Do they agree with your expectations based on Le Chatelier's principle? mori (Hb) can form a complex with both O and CO. action Hl eHbCO(aq)+ O2(g)Explanation / Answer
SO2Cl2(g)ß->SO2(g)+Cl2(g)
Moles of SO2Cl2= mass/molar mass, molar mass of SO2Cl2= 6.7/135=0.0496
Concentration of SO2Cl2= moles/Volume in Liters= 0.0496/10=0.00496M
Reaction coefficient, Q= [SO2][Cl2]/[SO2Cl2]= 0*0/0.00496= 0 <KC, equilibrium constant. So the reaction proceeds in the forward direction so as to increase Q.
Preparing the ICE table
Component initial concentration(M) Change (M) Eq. concentration(M)
SO2Cl2 0.00496 -x 0.00496-x
SO2 0 x x
Cl2 0 x x
KC= [SO2][Cl2]/[SO2Cl2]= x2/(0.00496-x)= 0.0451
When solved using excel, x=0.00451
At equilibrium, [SO2]=[Cl2]=0.00451, [SO2Cl2]=0.00496-0.0045=0.00045M9
Fraction of dissociation = (change in concentration/initial concentration)=100*0.0045/0.00496= 0.9073
b) moles of Cl2 used can be calculated from gas law, n= PV/RT
P= 0.1 atm, V= 10L, R =0.0821 L.atm/mole.K, T= 375 deg.c =375+273= 648K
No of moles of chlorine =0.1*10/(0.0821*648)= 0.019
Concentration of Cl2= 0.019/10=0.0019M
Q=[SO2][Cl2]/[SO2Cl2]= 0*0.0019/0.00496=0<KC, so the reaction proceeds in the forward direction for decomposition of SO2Cl2
Preparing the ICE table
Component initial concentration(M) Change (M) Eq. concentration(M)
SO2Cl2 0.00496 -x 0.00496-x
SO2 0 x x
Cl2 0.0019 x 0.0019+ x
Kc= x*(0.0019+x)/(0.00496-x)= 0.045
When solved for x, x= 0.00243,
So at equilibrium, [SO2Cl2]= 0.00253M, [SO2]=0.00243M and [Cl2]=0.0019+0.00243=0.00433M
Fraction of SO2Cl2 dissociated= 0.00243/0.00496=0.49
Since there is increase in no of moles of products due to addition of Cl2,the reaction shifts in a direction where there is decrease in no of moles in accordance to Lechatlier principle ( favouring reactant SO2Cl2) and this reduces the % dissociation of SO2Cl2.
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