Question

n-Butane is converted to isobutane in a continuous isomerization reactor that operates isothermally at 149.0°C. The feed to the reactor contains 96.00 mole%n-butane, 2.00% isobutane, and 2.00% HCl at 149.0°C, and a 35.00% conversion of n-butane is achieved. In the following calculations, one should use at least five significant figures. Physical Property Tables Feed and Product Composition Taking a basis of 1 mol of feed gas, calculate the moles of each component of the feed and product mixtures and the extent of reaction, Basis = 1 mol total nin (mol) Nout(mol) n-butane i i iso-butane i HCI i i

also make an enthalpy table.

Answer 1

Ans

Basis = 1 mol of feed gas

Moles of butane in reactants nb = 0.96000 mol

Moles of isobutane in reactants ni = 0.020000 mol

Moles of HCl in reactants nh = 0.020000 mol

Conversion of butane X = 35%

Moles of butane in product n1 = (1-X)*moles of butane in reactants

= (1-0.35)*0.96

= 0.62400 mol

Moles of isobutane in product n2 = moles of isobutane in reactants + X*moles of butane in reactants

= 0.02 + 0.35*0.96

= 0.35600 mol

Moles of HCl in product n3 = moles of HCl in reactants

= 0.020000 mol

Extent of reaction = (n1 - nb) / stoichiometric coefficient of butane

= (0.624 - 0.96)/(-1)

= 0.33600 mol

Part b

Inlet temperature = 149°C

For isothermally process

Outlet temperature = 149°C

Reference temperature = 25°C

Enthalpy of butane at inlet = enthalpy of butane at outlet

= 92.3*10^-3 * (149-25) + (27.88/2)*10^-5 * (149²-25²)

= 14.453 kJ/mol

Enthalpy of isobutane at inlet = enthalpy of isobutane at outlet

= 89.46*10^-3 * (149-25) + (30.13/2)*10^-5 * (149²-25²)

= 14.343 kJ/mol

Enthalpy of HCl at inlet = enthalpy of HCl at outlet

= 29.13*10^-3 * (149-25) - (0.134/2)*10^-5 * (149²-25²)

= 3.598 kJ/mol