Question

2. We wish to produce formaldehyde by the gas-phase pyrolysis of methanol, CH,OH (8)=CH,0 (8)+H, () The following table provides AH 295, AGM298 and C/R = A+BT +CT2 + DT ?, T'in K, for the species involved in the pyrolysis reaction. Species AH 29(J/mol) AG298 (J/mol) A 10B 10°C 10 D 0 CH,OH CH,0 H -200660 -108570 -161960 -102530 2.211 2.264 3.249 12.216 7.022 0.422 -3.450 -1.877 0 0.083 0 0 0 (a) (15 pts) Calculate heat of reaction and equilibrium constant at 298 K. Should we expect a substantial production of formaldehyde by methanol pyrolysis at 298 K? Briefly explain your expectation (6) (25 pts) Calculate the heat of reaction AH; and equilibrium constant K, at 450°C. (c) (15 pts) If nearly pure methanol is fed into a catalytic reactor which is controlled at 450°C and 2 bar, determine the maximum (equilibrium) conversion of methanol?

Answer 1

A)

The reaction occuring is

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

From given data

At T = 298 K

∆Hf(CH3OH) (g) = -200660 J/mol

∆Hf(HCHO) (g) = -108570 J/mol

∆Hf(H2) (g) = 0

∆H_r = ∆Hf_products - ∆Hf_reactants

∆H_r(298 K) = (-108570) -(-200660)

= 92090 J/mol

= 92.090 KJ/mol

Heat of reaction at 298 K = +92.090 KJ/mol

At T = 298 K

∆Gf(CH3OH) (g) = -161960 J/mol

∆Gf(HCHO) (g) = -102530 J/mol

∆Gf(H2) (g) = 0

∆G = ∆G(products) - ∆G(reactants)

∆G = (-102530) -(-161960) = 59430 J/mol

AG -In(K) RT

$\frac{59430}{(8.314)(298)}= -ln(K)$

K = 3.8237×10^-11

At this conditions ∆G is positive and equilibrium constant is less than 1 , so substantial production of formaldehyde cannot be achieved

At this conditions , reverse reaction is favoured and hence formaldehyde production is not favoured  

B)

At T = 298 K

∆G = ∆H - T∆S

59430 = 92090 -(298×∆S)

∆S(298 K) = 109.597 J/mol K

T = 450°C = 723 K

Average temperature = (298+723) /2 = 510.5 K

Cp equation of all components is given

General equation (Cp/R) = A + BT+ CT² + D/T²

T is in K

At T = 510.5 K

Cp(CH3OH) (g) = 62.7554 J/mol K

Cp(HCHO) (g) = 44.5594 J/mol K

Cp(H2) (g) = 29.0680 J/mol K

According to kirchoffs law

∆H(723 K) = ∆H(298 K) + ∆Cp(723 - 298)

∆Cp = (nC_p) _products -(nC_p) _{​​​​​​reactants}

∆Cp = (29.0680) +(44.5594) -(62.7554) =

10.872 J/mol K

∆H(510.5 K) = 92090 + (10.872) (723 -298)

∆H( K) = 96710.6 J/mol

Heat of reaction at 723 K (450°C)

= 96710.6 J/mol = 96.7106 KJ/mol

∆S(723 K) = C_pavg ln(T/298)

C_pavg according to Stiochiometry = (62.7554) (0.33) + (44.5594) (0.33) +(29.0680) (0.33) = 45.006 J/mol K

∆S(723 K) = 45.006 ln(723/298)

= 39.8895 J/mol K

∆S(total) = ∆S(298 K) + ∆S(510.5 K)

∆S(total) = 109.597 + 39.8895

= 149.486 J/mol K

∆G = ∆H - T∆S

∆G = 94400.3 - (723 × 149.486)

∆G = -13678.078 J/mol

AG -In(K) RT

- 13678.078 (8.314) (723) = -In(K)

K = 9.73278

C)

T = 450°C , P= 2 bar

Equilibrium constant is independent of pressure and therefore K = 9.73278

Feed is pure methanol

Basis : 1 mole methanol

$9.73278 = \frac{\epsilon ^{2}}{(1-\epsilon )}$

$\epsilon$ = 0.91414

Methanol reacted = 0.91414 moles

Unreacted methanol = 1-0.91414 = 0.08586 moles

Conversion = 1-(0.08586/1) = 0.91414

Equilibrium Conversion of Methanol = 91.414 %

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