Question

Oxidation of propane is represented as
`C_(3)H_(8_((g))) + 50_(2_((g))) to 3CO_(2_((g)))+ 4H_(2)O_((g)) DeltaH^(@) =- 2043 kJ`
How much pressure volume work is done and what is the valune of `DeltaU` at constant pressure of 1 atm when the volume change is + 22.4 L

Answer 1

Correct Answer - `W= - 2.27 kJ ,Delta U =- 2045. 27 kJ`
Given : `Delta H^(@) =- 2043 kJ`
Change in volume `= Delta V = + 22.4 L`
`Delta U = ?`
`C_(3)H_(8_((g)) + 5O _(2_((g))) to 3CO_(2_((g))) + 4H_(2)O_((g))`
`Delta n =(n_(2))_("gaseous products ") - (n_(1))_("gaseous reactants ")`
`= (3 + 4) - (1+5)`
`= 1"mol" `
Since there is the an increase in number of moles the work will be of expansion .
`W=- P xx Delta V`
`=- 1 xx 22.4 L atm`
`=-1 xx 22.4 xx 101.3 J`
`=- 2270 J`
`=- 2.27 kJ`
`Delta H = Delta U + PDelta V`
`:. Delta U = Delta H - P Delta V`
`=- 2043 - (2.27 )`
`=- 2045 .27 kJ`