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

Answer:

heat rejected =-1427820 KJ/mol of $C_{2} H_{6}$

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

$h^{0} _{f} = enthalpy of formation at the standard reference state

From the enthalpy of formation tables  at 25 degrees  and 1 atm

Taking enathalpy of formation of [tex]CO_{2}$ = -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

$hc=(2\times-393520)+(3\times-241820)-(1\times-84680 )$

hc=-1427820 KJ/mol of $C_{2} H_{6}$

heat rejected = heat of enthalpy of formation

heat rejected =-1427820 KJ/mol of $C_{2} H_{6}$