Question

calculate the maximum amount of work that can be obtained from the combustion of 1.00 moles of ethane, CH3CH3(g), at 25 °C and standard conditions using a thermodynamics properties chart.

Answer 1

From the thermodynamics properties chart,we know that

For C2H6 Delta H : -84.5 KJ/mole, Delta S = 229.5 J/mole.K

For CO2 Delta H : -394 KJ/mole, Delta S = 213.6 J/mole.K

For H2O Delta H : -286 KJ/mole, Delta S = 69.96 J/mole.K

For O2 Delta H :        0 KJ/mole, Delta S = 205 J/mole.K

Equation for the combustion of ethane, CH3CH3(g) is

2C2H6(g) +7O2(g) --> 4CO2(g) +6H2O(l)

We know that the maximum amount of work = Delta G (Gibbs free energy)

G = H - T S---(1)

Let us calculate first H = H products -H reactants = (4 x -394 + 6 x -286)-(2x-84.5 + 0)kj/mole = -3123 kj/mole

S = 4x213.6 + 6x69.96 - ( 2x229.5 + 7x205 )j/mole,K 1894 =-1054.28 j/mole.k

    G = H -TS

             = -3123 x - 298 x (-1.5428 ) = -1435.812 kj

As two moles of ethane will react in the above equation G = -717.906 kj