Question

A prominent chemical manufacturer finds itself producing large amounts of chloroethane (CH₃CH₂Cl), a toxic byproduct of one of its processes. Using water, a clever chemist proposes to convert this compound into ethanol and hydrochloric acid (converting the toxic waste into two value-added products) using the following reaction:

CH₃CH₂Cl_(g) + H₂O_(g) ↔ CH₃CH₂OH_(g) + HCl_(g)      (100 °C)

Use the data in Appendix A. (Below)

a. Calculate the value of D_rHº

b. Calculate D_rSº, D_surroundingsSº   and   D_universeSº a. Calculate the value of D_rHº

c. Calculate D_rGº for this reaction at 100 °C (Assume that DSº and DHº values do not vary with temperature)

d. Is this reaction spontaneous at 100^oC? What effect would lowering the temperature to 50^oC have on the position of the equilibrium?

e. Find the K_Pvalue of the reaction. Write the equilibrium constant expression for the reaction.

f. If a closed vessel containing 0.010 atm of chloroethane is reacted with 1.0 atm of water vapor at 100 °C, what will be the partial pressure of chloroethane in the vessel when equilibrium is reached?

Appendix A: Thermodynamic parameters for selected molecules

molecule	DH0f (kJ mol-1)	Sf0 (J K-1 mol-1)
C2H5OH (g)	-234.8	281.6
C2H5Cl (g)	-261.9	268.4
HCl (g)	-92.31	186.91
H2(g)	0	130.6
H2O(l)	-285.8	69.9
H2O(g)	-241.8	188.7

Answer 1

The given chemical transformation,

CH₃CH₂Cl (g) + H₂O(g) ? CH₃CH₂OH(g) + HCl(g)      (100 °C)

a)Calculation of the value of ?rHº

Formula,

?rHº = ??rHº(Products) - ? ?Hº(Reactants)

?rHº = [?rHº(CH₃CH₂OH(g)) + ?rHº (HCl(g))] –[ ?rHº(CH₃CH₂Cl (g)) + ?rHº(H₂O(g))]

Substituting all the known values from Appendix A,

?rHº = [(-234.8) + (-92.31)] –[ (-261.90) + (-241.80)]

?rHº = [-327.11]-[-503.70]

?rHº = -176.59 kJ

?rHº = -176.6 kJ

?rHº = -176.6 x 1000 J

?rHº = -176600 J.

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b. Calculation of ?rSº, ?surroundingsSº   and   ?universeSº.

i) Calculation of ?rSº,

Formula,

?rSº = ??rSº(Products) - ? ?Sº(Reactants)

?rSº = [?rSº(CH₃CH₂OH(g)) + ?rSº (HCl(g))] –[ ?rSº(CH₃CH₂Cl (g)) + ?rSº(H₂O(g))]

Substituting all the known values from Appendix A,

?rSº = [(281.6) + (186.91)] –[ 268.4 + 188.7]

?rSº = [468.51]-[457.1]

?rSº = 11.41 J.K^-1.

ii) Calculating ?surroundingsSº and ?universeSº

So far we got, ?rHº = -176600 J and ?rSº = 11.41 J.K^-1

i.e. Standard enthalpy change is –ve and standard entropy change is +ve and hence ?rGº will be –ve at all temperature and reaction will be spontaneous.

For non-isolated spontaneous process ,

?rSº(total)= ?rSº(system) + ?rSº(surrounding) > o

Hence, we must have, ?rSº(surrounding) > o.

And also ?universeSº > 0.

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c. Calculation of ?rGº for this reaction at 100 °C

We have, ?rHº = -176600 J and ?rSº = 11.41 J.K^-1, T =100 °C = 373.15 K

Formula,

?rGº = ?rHº -T. ?rSº……………..(1)

?rGº = -176600 – 373.15 X 11.41

?rGº = =-176600 – 4257.64

?rGº = - 180857.64 J

?rGº = -180.857 kJ

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d)

i) At 100 ^oC we have, ?rGº = -180.857 kJ i.e. –ve . Yes, this reaction spontaneous at 100 ^oC .

Lowering the temperature to 50 ^oC will have no effect on the position of the equilibrium as we can see from eq.1 that at all temperature ?rGº will be –ve as ?rHº is -ve and ?rSº is +ve .

e.   Calculation for K_P value of the reaction.

?rGº , and Kp related by equation,

?rGº = -RT ln(Kp)

Substitute known values, ?rGº = -180857.64 J, T = 373.15 K, R= 8.314 J.K^-1.mol^-1.

-180857.64 = -8.314 x 373.15 x ln(Kp)

ln(Kp) = -180857.64 / -31102.37

ln(Kp) =58.3

Kp =e^58.3

Kp =2.07 x 10²⁵ .

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