Answer only the highlighted area. PREPARATORY EXERCISE [Hint ensure you use the
ID: 564300 • Letter: A
Question
Answer only the highlighted area.
PREPARATORY EXERCISE [Hint ensure you use the correct units! Reaction of ethene, C Ha, with iodine is an endothermic process which is being considered as a route to make di-iodoethane according to the following equation: At 50 °C, the equilibrium partial pressures of CHa, iodine and CHalz were found to be PcH-107.6 mmHg, P, = 1.17 mmHg and PCM.½= 1.14 mmHg. Write an expression for the equilibrium constant, K, for this reaction. Determine the value of the equilibrium constant, Kp for this reaction at 50°C. Determine the value of the equilibrium constant K for this reaction at 50 °C Evaluate the equilibrium constant for the related process: Ha ()+12) (e) Which of the following would help maximize the amount of C Hal2 produced?: (i) Decrease the volume of the reaction (i) Remove iodine from the reaction mixture i) Increase the reaction temperature (iv) Add C2Hs to the reaction mixture Additional iodine is added to this equilibrium mixture to initially raise the iodine pressure to 3.00 mbar and the system is then allowed to return to equilibrium. (f) Determine the reaction quotient Q immediately after addition of iodine (g) Based on your value of Qp, determine whether the system will shift to the left, shift to the right or is unaffected by the addition of iodine. (h) Make an ICE Table to show the initial vapor pressures of the three reagents after addition of the extra iodine and determine the new equilibrium partial pressures of each compound presentExplanation / Answer
The reaction is C2H4(g)+ I2(g)ß->C2H4I2(g)
Let P= partial pressure of given component in atm
PC2H4I2= partial pressure of C2H4I2=1.14 mm Hg = 1.14/760 atm
PC2H4= partial pressure of C2H4= 107.6 mm Hg= 107.6/760 atm
PI2= partial pressure of I2= 1.17 mm Hg= 1.17/760 atm
KP= PC2H4I2/(PC2H4*PI2)
KP= (1.14/760)/(107.6/760)*(1.17/760)=6.88/atm
KP= KC*(RT)deltan, deltan= change in no of moles during the course of reaction
Deltan= 1-2= -1
6.88 = KC*{0.0821 L.atm/mole.K)* (50+273}-1, KC= 182.44
For the reaction C2H4I2(g)ß->C2H4(g)+ I2(g)
Kp’= [C2H4][I2]/[C2H4I2]= 1/KP= 1/6.88= 0.1453 and KC’= 1/KC= 1/182.44 =0.0055
Lechatlier principle will be used to explain which of the effect increases the production of C2H4I2.
Since K= [C2H4I2]/[C2H4][I2], for increasing the production of C2H4I2, when [C2H4] is added, this will be increase K. But at any given temperature, K is constant and hence to keep K constant, [C2H4I2] has to be produced more. Hence addition of C2H4 will enable more production of C2H4I2.
When PI2= 3mbar, 1 bar= 0.9869 atm, PI2= 3 mbar= 3*10-3 bar= 3*10-3*0.9869 atm=0.003 atm
Q= Reaction coefficient =PC2H4I2/ (PC2H4)*PI2= (1.14/760)/(0.003*107.6/760)=3.53<KP, so for increasing Q, more C2H4I2 has to form. The reaction shifts to the right.
ICE Table is
S.No
PC2H4 (atm)
PI2 (atm)
PC2H4I2 (atm)
1
Initial
107.6/760=0.1415
0.003
1.14/760= 0.0015
2
Change
-P
-P
+P
3
Equilibrium
0.1415-P
0.003-P
0.0015+P
KP= (0.0015+P)/ (0.1415-P)*(0.003-P) = 6.88
When solved using excel, P= 0.000714 atm
So at equilibrium, PC2H4= 0.1415-0.000714=0.140786 atm , PI2= 0.003-0.000714= 0.002286 atm and PC2H4I2= 0.0015+0.000714= 0.002214 atm
S.No
PC2H4 (atm)
PI2 (atm)
PC2H4I2 (atm)
1
Initial
107.6/760=0.1415
0.003
1.14/760= 0.0015
2
Change
-P
-P
+P
3
Equilibrium
0.1415-P
0.003-P
0.0015+P