Question

2. The following synthesis reaction of methanol from carbon monoxide (CO) is an important part of the so-called CI chemistry, CO(g) + 2H, (8) catalya >CH,OH (8) because CO can be obtained from coal and methanol can easily go on to form heavier hydro-carbons and carbonhydrates. (a) Calculate the heat of reaction and equilibrium constant at 150 °C. (b) Determine the equilibrium composition at 150 °C and 5 bar if a feed mixture of CO and Hz in a 1:1 molar ratio is considered. (C) To increase the driving force for methanol formation, circle the correct answers from the following options (1) increase or decrease system temperature? (ii) increase or decrease system pressure? (iii) increase or decrease the feed rate of Hz? Please use Tables C.1 and C.4 for the needed AH%295, AG 29 and C/R = A+BT +CT? + DT-² for the species involved in the above reaction.

Answer 1

The reaction occuring is

$CO(g) + 2H_{2}(g) \rightleftharpoons CH_{3}OH(g)$

Reaction occurs at 150°C

From handbook at T = 25°C

∆H(CH3OH) (g) = -205 KJ/mol

∆H(CO) (g) = -110. 53 KJ/mol

∆H(H2) (g) = 0

∆H_r = ∆H_products - ∆H_reactants

∆H_r(25°C) = (-205) -(-110.53) = -94. 47 KJ/mol

According to kirchoffs law

∆H(150°C) = ∆H(25°C) + ∆Cp(150-25)

∆Cp = (nCp) _products -(nCp) _reactants

Cp is taken from handbook at average temperature of (25+150) /2 = 87.5°C

From handbook at T =87.5°C

Cp(CH3OH) (g) = 0.048694 KJ/mol K

Cp(H2) (g) = 0.029050 KJ/mol K

Cp(CO) (g) = 0.029240 KJ/mol K

∆Cp = (0.048694) -(2×0.029050) -(0.029240) = -0. 038646 KJ/mol K

∆H(150°C) = -94. 47+ (-0.038646) (150-25)

∆H(150°C) = -99. 300 KJ/mol

Heat of reaction at 150°C = -99. 30075 KJ/mol

At P = 1 atm and T = 25°C from handbook

S(CH3OH) (g)= 0.2399 KJ/mol K

S(CO) (g) = 0.197660 KJ/mol K

S(H2) (g) = 0.13068 KJ/mol K

∆S = ∆S_products - ∆S_reactants

∆S(25°C ) = (0.2399) -(2×0.13068) -(0.197660) = -0. 21912 KJ/mol K

∆S(423.15 K) = C_pavg ln(T/298.15 K)

C_pavg = (0.029050) (0.5) + (0.048694) (0.25) +(0.029240) (0.25) = 0.0340 KJ/mol K

∆S(150°C) = 0.0340 ln(423.15/298.15)

∆S(150°C) = 0.011904 KJ/mol K

∆S(T) = ∆S(25°C) + ∆S(150°C)

∆S(Total) = -0. 21912+ 0.011904 =

-0. 20721 KJ/mol K

∆G = ∆H -T∆S

T = 150°C = 423.15 K

∆G = (-99.30075) -(423.15×-0.20721)

∆G = -11. 61747 KJ/mol = -11617. 477 J/mol

$\frac{\Delta G}{RT}= -ln(K)$

$\frac{-11617.477}{(8.314)(423.15)}= -ln(K)$

K = 27.17319

Equilibrium constant = 27.17319

B)

Equilibrium constant is independent of pressure hence

Equilibrium constant at 150°C , 1 atm = equilibrium constant at 150°C ,5 bar

K = 27.17319

Feed ratio = 1:1 (CO: H2)

Stiochiometry ratio = (H2:CO) = 2:1

Basis : 2 mole feed

$27.17319 = \frac{\epsilon }{(1-\epsilon )(1-2\epsilon )^{2}}$

$\epsilon$ = extent of reaction

Solving we get

$\epsilon$ = 0.41861

Equilibrium product analysis

Component	mol	mol%
CO	1-0.41861 = 0.58139	50
H2	1-(2×0.41861) = 0.16278	14
CH3OH	0.41861	36
Total	1.16278	100

C)

I) The reaction is exothermic and according to lechartier principle decreasing the system temperature will favour methanol formation as system tries to eliminate the effect of change in equilibrium

ii) There is decrease in number of moles of reaction ∆n = -2

So increasing the system pressure increases the methanol formation driving force

iii)

In this case H2 is limiting reactant

For every mole of CO 2 moles of H2 is needed

When H2 is increased the system tries to reduce the concentration of H2 by favouring forward reaction

So increasing feed rate of H2 will increase the methanol formation

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