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aleksley [76]
4 years ago
10

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)
Engineering
1 answer:
Amanda [17]4 years ago
7 0

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 stateFrom the enthalpy of formation tables  at 25 degrees  and 1 atmTaking 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}

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