InterviewSolution
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InterviewSolution
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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)`. |
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Answer» `2CH_(4) +CO_(2) +1//2N_(2) rarr C_(3)H_(7)NO_(2), DeltaH^(Theta) = ?` First find `Delta_(f)H^(Theta) of CH_(4)` Given `CH_(4) +2O_(2) rarr CO_(2) +2H_(2)O, DeltaH^(Theta) =- 212.8` Using the definition of `DeltaH`, `DeltaH^(Theta) = [Delta_(f)H^(Theta)underset((CO_(2))).+2Delta_(f)H^(Theta)2(H_(2)O)]-Delta_(f)H^(Theta)(CH_(4))` (Note that `Delta_(f)H^(Theta)O_(2)=0)` `rArr -212.8 =[-94.05 +2)-68.32) -Delta_(f)H^(Theta)(CH_(4))]` ` rArr 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_(3)H_(7)NO_(2))-0] -[2xxDelta_(f)H^(Theta)(CH_(4))+Delta_(f)H^(Theta)(CO_(2)) +0]` `rArr DeltaH =(-133.57) -2(-17.89) -(-68.32)` `=- 374 kcal mol^(-1)` Now calculate `DeltaH` (combustion) of `C_(3)H_(7)NO_(2)`. `C_(3)H_(7)NO_(2)(s) +15//4O_(2) rarr 3CO_(2)(g) +1//2N_(2)(g) +7//2H_(2)O(g)` `Delta_("comb")H^(Theta) =3Delta_(f)H^(Theta) (CO_(2)) +0 +7//2Delta_(f)H^(Theta) (H_(2)O) -Delta_(f)H^(Theta) (C_(3)H_(7)NO_(2)) -0` `=3(-96.05) +7//2 (-68.32) -(-133.57)` `=- 387.70 kcal mol^(-1)` Find `DeltaU^(Theta)` using `DeltaU = DeltaH^(Theta) -DeltanRT` `DeltaU =- 387.70 -(1//4) xx2xx 10^(-3) (298)` `=- 387.72 kcal mol^(-1)` |
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