Answer:
ΔH°c = - 1605.1 KJ
Step-by-step explanation:
- CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
∴ ΔHf CH4(g) = - 74.6 KJ/mol
∴ ΔHf CO2(g) = - 393.5bKJ/mol
∴ ΔHfH2O(g) = - 241.82 KJ/mol
standard enthalpy of combustion (ΔH°c):
⇒ ΔH°c = (2)(ΔHf H2O) + ΔHfCO2 - ΔHfCH4
⇒ ΔH°c = (2)(- 241.82) + ( - 393.5 ) - ( - 74.6 )
⇒ ΔH°c = - 802.54 KJ/mol
⇒ ΔH°c = ( - 802.54 KJ/mol )( 2 mol CH4 )
⇒ ΔH°c = - 1605.08 KJ