Question

Ethane (C2H6) is burned at atmospheric pressure with a stoichiometric amount of air as the oxidizer. Determine the heat rejected, in kJ/kmol fuel, when the products and reactants are both at 25C, and the water vapor appears in the products as water vapor. (ANSWER: 1,427,820 kJ/kmol)

Answer 1

heat rejected =-1427820 KJ/mol of
`C_(2) H_(6)`

Step-by-step explanation:

Fuel ethane
`C_(2) H_(6)`

Burning Temperature = T =
`25^(0)`

Pressure = P = 1 atm

The stiochiometric equation for this reaction is

`C_(2) H_(6) +a_(th) (O_(2) +3.76N_(2)) >>>> 2CO_(2) + 3H_(2)O+3.76a_(th)N_(2)`

The enthalpy of the reaction is given as

`hc=H_(product) +H_(react)`

=
`\Sigma N_(p) h^(0) _(f.p) -\Sigma N_(r) h^(0) _(f.r)`

=
`\Sigma N_(p) h^(0) _(f.p)-\Sigma N_(r) h^(0) _(f.r)`

`=(Nh^(0) _(f) )_{CO_(2) }+(Nh^(0) _(f) )_{H_(2)O }-(Nh^(0) _(f) )_{C_(2) }H_(6)`

Where

N = number of poles

= -393520 KJ/mol

Taking enathalpy of formation of
`H_(2)O` = -241820 KJ/mol

Taking enathalpy of formation of
`C_(2) H_(6)` = -84680 KJ/mol

`=(Nh^(0) _(f) )_{CO_(2) }+(Nh^(0) _(f) )_{H_(2)O }-(Nh^(0) _(f) )_{C_(2) }H_(6)`

by putting values

`hc=(2*-393520)+(3*-241820)-(1*-84680 )\\`

hc=-1427820 KJ/mol of
`C_(2) H_(6)`

heat rejected = heat of enthalpy of formation

heat rejected =-1427820 KJ/mol of
`C_(2) H_(6)`