Standard enthalpy of formation of `C_(3)H_(7)NO_(2)(s),CO_(2)(g)` and `H_(2)O(l)` are `133.57,-94.05` and `-68.32kcal mo1^(-1)` respectively Standard enthalpy combustion of `CH_(4)` at `25^(@)C` is `-212.8 kcal mo1^(-1)` Calculate `DeltaH^(Theta)` for the reaction:
`2CH_(4)+CO_(2)+1//2N_(2)rarrC_(3)H_(7)NO_(2)(s)+1//2H_(2)`
Calculate `DeltaU` for combustion of `C_(3)H_(7)NO_(2)(s)`.

Explanation

`2CH_4 + CO_2 + 1//2 N_2 to C_3 H_7NO_2 , DeltaH^(Theta) =?`
First find `Delta_(f)H^(Theta) "of" CH_(4)`
Given `CH_(4) + 2O_(2) to CO_2 + 2H_2O , DeltaH^(Theta) = -212.8`
Using the definition of `Delta H`,
`DeltaH^(Theta) = [Delta_(f)H^(Theta) (CO_2) + 2Delta_(f)H^(Theta)(H_2O)]-Delta_(f)H^(Theta)(CH_4) ("Note that" Delta_(f)H^(Theta)O_2 = 0)`
`implies -212.8 = [-94.05 + 2(-68.32) - Delta_(f)H^(Theta)(CH_4)]`
`implies Delta_(f)H^(Theta)(CH_4) = -17.89 kcal//mol`
Now find the `DeltaH` of the required equation using `Delta_(f)H^(Theta)(CH_4)`.
`[Delta_(f)H^(Theta)(C_3H_7NO_2)-0] - [2 xx Delta_(f)H^(Theta)(CH_4)+Delta_(f)H^(Theta)(CO_2)+0]`
` implies Delta H = (-133.57) - 2(-17.89) - (-68 . 32) = -374 kcal mol^(-1)`
Now calculate `DeltaH` (combustion) of `C_3H_7NO_2`
`C_3H_7NO_2(s_ + 15//4O_2 to 3CO_2(g) + 1//2 N_2(g)+ 7//2H_2O(g)`
`Delta_("Comb") H^(Theta) = 3Delta_fH^(Theta) (CO_2) + 0 + 7//2Delta_(f)H^(Theta)(H_2O) - Delta_fH^(Theta)(C_3H_7NO_2)-0`
`=3 (-94.05) + 7//2(-68.32) - (-133.57) = -387.70 kcal mol^(-1)`
Find `DeltaU^(Theta)` using `DeltaU = Delta_fH^(Theta) - Delta nRT`
`DeltaU = -387.70 - (-1//4) xx 2 xx 10^(-3) (298) = -387.72 kcal mol^(-1)`.