One litre of pure methanol at 10 degree C is to be equilibrated isothermally wit

Question

One litre of pure methanol at 10 degree C is to be equilibrated isothermally with a large volume of a gas at 60 bar total pressure consisting of 5% by mole CO, 20% CO_2 and the rest k H_2, Using the literature data provided, calculate: (a) the molar composition of the liquid phase (b) the heat liberated Data CO_2 in methanol. The table be km lists ml (NTP) CO_2 devolved in 1 g methanol (rho of 791 kg m^3) at the dated temperature and pressure. CO in methanol(at 10 degree C) H_2 in methanol(at 10 degree C)

Explanation / Answer

Using the element weights and the molar convertion (6.023X10-23 mol) the answer will be

a)Methanol (32g) 1.92X 10-21 mol at 60 bar = 1.13 X10-19 mol/atm

Gas (74g) 4.45X 10-21mol at 60 bar = 2.63 X10-19 mol/atm

Liquid phase (106g) 6.38x 10 -23 mol 3.77 X10-19 mol/atm

b) Methanol heat (238.4 kJ/mol)

Gas (-110.53 kJ/mol)

Liquid phase 341.66 kJ/mol at 60 bar = 20229.68 kJ/atm

1) Using th information given the results are and the units are g atm oC

2) a. 2160 g mL/atm b. 18.75 g mL/atm c.19 g mL/atm of CO

3) a. 1.23 g mL/atm b. 1.25 g mL/atm c. 1.31 g mL/atm of H2

(p/atm) / temp 0 25 49.8 75 6.8 0 7.22134E-06 9.38146E-06 9.27423E-06 10.7 0 1.87356E-05 0.019221725 2.54242E-05 16.5 0 2.88914E-05 6.04701E-05 4.04735E-05 22.3 0 9.34597E-05 0.000113439 0.000115478

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