Question

Methanol is obtained from carbon monoxide and hydrogen, according to:

CO(g) + H2 - CH3OH (9)

Suppose 100 m³/h of reagent mixture is fed in stoichiometric amounts, at 35°C and 1 atm. If 17.2 kW of heat is removed from the reactor, determine the conversion rate (using four decimal places) of the reaction when the outlet temperature is 135°C.

Answer 1

The reaction occurring is

$CO + 2H_{2} \rightarrow CH_{3}OH$

Volumetric flowrate = 100 m^{​​​​​​3}/h

T = 35°C = 308 K

P = 1 atm = 1.013×10⁵ Pa

n = PV/(RT)

n = (1.013×10⁵×100)/(8.314×308)

n = 3955.931 mol/h

Stiochiometric amount is supplied

Moles of CO = 3955.931(1/3)= 1318.643 mol/h

= 0.36628 mol/s

Moles of H2 = 2637.2873 mol/h = 0.73257 mol/s

At T = 25°C , from handbook

∆Hf(CO)(g)= -110.53 KJ/mol

∆Hf(CH3OH)(g)= -205 KJ/mol

∆Hf(H2)(g)= 0

∆Hr = ∆H_products - ∆H_reactants

∆Hr = -205 -(-110.53)

∆Hr (25°C)= -94.47 KJ/mol

From handbook at average temperature of (25+35)/2 = 30°C

Cp(CO)= 29.1545 J/mol K

Cp(Methanol)= 44.3938 J/mol K

Cp(H2)(g)= 28.859 J/mol K

∆Cp =(nCp) _products -(nCp)_reactants

∆Cp = (44.3938) -(2×28.859) - 29.1545= -42.478

Kirchoffs law

∆H(35°C) = ∆H(25°C)+ ∆Cp(35-25)

∆H(35°C)= -94.47(1000)+ (-42.478)(35-25)= -94.894 KJ/mol

Let x be amount of methanol reacted

Outlet temperature = 135°C

Enthaply reference = 25°C

Reactant average temperature = (35+25)/2 = 30°C

Product average temperature = (35+135)/2 = 85°C

All cp values are taken at average temperature from handbook

Component	inlet moles (mol/s)	Cp	Hi	outlet moles (mol/s)	Cp	Ho
CO	0.36628	0.0291545	0.1067	0.36628-x	0.029237
H2	0.73256	0.028859	0.2114	0.73256-(2x)	0.029041
Methanol	0	0.0443938	-	x	0.048496
Total			0.3181

Q = ∆H_r(x)+ Ho + Hi

Q = 17.2 KW

- 17.2 = -94.894(x)+ 0.3181 +

((0.36628-x)(0.029237)+ (0.73256-2x)(0.029041)+ (0.048496(x))(135-35)

x = 0.209730 mol/s

Initial moles = 0.36628 mol/s

Conversion =

( (0.36628-0.209730)/(0.36628))(100) = 42.74052 %

Please upvote if helpful