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For his 18th birthday in February Peter ...

For his 18th birthday in February Peter plants to turn a hut in the garden of his parents into a swimming pool with an artifical beach. In order to estimate the consts for heating the water and the house , peter obtains the data for the natural gas combustion and its price.
Write down the chemical equations for the complete conbustion of the main components of natural gas, methane and ethane , given in table-1. Assume tht nitrogen is inert under the chosen conditions. Calculate the reaction enthalpy , the reaction entropy , and the Gibbs energy under standard conditons `(1.013 xx 10^(5) Pa , 25.0^(@)C)` for the combustion of methane and ethane according to the equations above assuming that all products are gaseous .
The thermodynamic properties and the composition of natural gas can be found in table 1.

Text Solution

Verified by Experts

The correct Answer is:
Amount of methane and ethane in `1 m^(3)` natural gas :
`m= gt xx V=0.740 g dm^(-3) xx 1000 dm^(3) =740 g`
`M_(av) = sum_(i) xx (i) M(ii)=(0.0024 xx44.01 g "mol"^(-1)) + (0.0134 xx28.02 g "mol"^(-1))+(0.9732 xx 16.05 g "mol"^(-1)) + (0.011 xx 30.08 g "mol"^(-1))=16.43 g "mol"^(-1)`
`n_("tol") =m(Mav)^(-1) = 740 gxx (16.43 g//mol)^(-1)=45.83 "mol"`
n(i) = x(i) . n_("tot")
`n(Ch_(4)) = x(CH_(4)) xx n_("tot") =09732xx 45.04 "mol"=0.495 "mol"`
`n(C_(2)H_(6))=x(C_(2)H_(6)xxn_("tot")=0.0110 xx 45.04 "mol" =0.495 "mol"`
(b) Energy of combustion , deviation:
Ecomb.`(H_(2)O(g)) =sum_(i)n(i) Delta_(c)H-(i)=`
`=43.83 "mol" xx (-802.5 "kj mol"^(-1)) + 0.495 "mol" xx 0.5 xx(-2856.8 KJ "mol"^(-1))`
`=-35881 KJ`
`E_("comb")(H_(2)O(g))=-35881 KJ`
Deviation from PUC
EPUC `(H_(2)O(g)) =9.981 K Whm^(-3) xx 1 m^(3)xx 3600 KJ (kWh)^(-1) =35932 KJ`
Deviation: `DeltaE=(E_("comb")(H_(2)O(g))-E_(PUC)(H_(2)(g)) xx 100%xx [Ecomb.(H2)(g))]^(-1)`
`=(35881 KJ -35932 KJ ) xx 100% xx (35881 KJ )^(-1)=.^(-)C0.14%`
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