Answer
-800KJ/mol
Work Step by Step
There are two ways to solve this problem.
Way 1--
Step 1: Compute $\Delta$$H_{R}$$^{\circ}$(298.15K) for the reaction C$H_{4}$(g)+2$O_{2}$(g)--> C$O_{2}$(g)+2$H_{2}$O(l)
**Notice the water product is in liquid phase**
$\Delta$$H_{R}$$^{\circ}$(298.15K)= $\Delta$$H_{f}$$^{\circ}$(C$O_{2}$(g))+2$\Delta$$H_{f}$$^{\circ}$($H_{2}$O(l))-$\Delta$$H_{f}$$^{\circ}$(C$H_{4}$(g))-2$\Delta$$H_{f}$$^{\circ}$($O_{2}$(g))
= -393.51KJ/mol+2$\times$(-285.83KJ/mol)-(-74.81KJ/mol)-0
= -890.36 KJ/mol
Step 2: Compute the individual changes of enthanpy for 1 mol of all components ($φ_{i}$) and the net change of enthanpy from 298.15K to 500K:
φ(C$O_{2}$(g))=$\int^{500K}_{298.15K}$$C_{P,m (CO_{2},g)}$dT=$\int^{500K}_{298.15K}$(44.22+8.79$\times$$10^{-3}$T-$\frac{8.62\times10^{5}}{T^{2}}$)(J/mol-K)dT
=8.47 KJ/mol
φ($H_{2}O$)=$\int^{373.15K}_{298.15K}$$C_{P,m (H_{2}O,l)}$dT+$\Delta$$H^{^{\circ}}_{vap}$(373.15K)+$\int^{500K}_{373.15K}$$C_{P,m (H_{2}O,g)}$dT
=$\int^{373.15K}_{298.15K}$(75.29J/mol-K)dT+40.7KJ/mol+$\int^{500K}_{373.15K}$(33.58J/mol-K)dT
= 50.61 KJ/mol
φ($O_{2}$(g))=$\int^{500K}_{298.15K}$$C_{P,m (O_{2},g)}$dT=$\int^{500K}_{298.15K}$(29.96+4.18$\times$$10^{-3}$T-$\frac{1.67\times10^{5}}{T^{2}}$)(J/mol-K)dT
=6.16 KJ/mol
φ(C$H_{4}$(g))=$\int^{500K}_{298.15K}$$C_{P,m (CH_{4}, g)}$dT=$\int^{500K}_{298.15K}$(35.31 J/mol-K)dT
=7.13 KJ/mol
The net change =φ(C$O_{2}$(g))+2φ($H_{2}O$)-φ(C$H_{4}$(g))-2φ($O_{2}$(g))
=(8.47+2$\times$50.61-7.13-2$\times$6.16) KJ/mol=90.24 KJ/mol
Step 3: $\Delta$$H_{R}$$^{\circ}$(500K)=$\Delta$$H_{R}$$^{\circ}$(298.15K)+(Net change of enthalpy from step 2)
= (-890.36+90.24)KJ/mol$\approx$ -800KJ/mol
Way 2--
Step 1: Compute $\Delta$$H_{R}$$^{\circ}$(298.15K) for the reaction C$H_{4}$(g)+2$O_{2}$(g)--> C$O_{2}$(g)+2$H_{2}$O(g)
**Notice the water product is in gaseous phase**
$\Delta$$H_{R}$$^{\circ}$(298.15K)= $\Delta$$H_{f}$$^{\circ}$(C$O_{2}$(g))+2$\Delta$$H_{f}$$^{\circ}$($H_{2}$O(g))-$\Delta$$H_{f}$$^{\circ}$(C$H_{4}$(g))-2$\Delta$$H_{f}$$^{\circ}$($O_{2}$(g))
= -393.51KJ/mol+2$\times$(-241.82KJ/mol)-(-74.81KJ/mol)-0
= -802.34 KJ/mol
Step 2: Compute the individual changes of enthanpy for 1 mol of all components ($φ_{i}$) and the net change of enthanpy from 298.15K to 500K:
Values of φ(C$O_{2}$(g)), φ($O_{2}$(g)) and φ(C$H_{4}$(g)) are the same as way 1, and the only difference is φ($H_{2}O$):
φ($H_{2}O$)=$\int^{500K}_{298.15K}$$C_{P,m (H_{2}O,g)}$dT=$\int^{500K}_{298.15K}$(33.58J/mol-K)dT= 6.78 KJ/mol
The net change= φ(C$O_{2}$(g))+2φ($H_{2}O$)-φ(C$H_{4}$(g))-2φ($O_{2}$(g))
=(8.47+2$\times$6.78-7.13-2$\times$6.16) KJ/mol=2.58 KJ/mol
Step 3: $\Delta$$H_{R}$$^{\circ}$(500K)=$\Delta$$H_{R}$$^{\circ}$(298.15K)+(Net change of enthalpy from step 2)
= (-802.34+2.58)KJ/mol$\approx$ -800KJ/mol
