Question

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Calculate the enthalpy of the reaction:

N204(g) + 3CO(g) ———->N20(g) + 3CO2(g)

Given that;

∆fH–CO(g) = – 110 kj mot-1; ∆fHC02(g) = – 393 kj mol-1
∆fHN20(g) = 81 kj mot-1; ∆fN2O4(g) = 9.7 kj mol-1

Answer 1

$\mathfrak{Answer :-}$

$\texbf{}$

$\implies Enthalpy\: of \:reaction \:$(delta ∆r,H) =

$\implies$[81 + 3 (- 393)] - [9.7 + 3 (- 110)]

$\implies$ [81 - 1179] - [9.7 – 330]

$\implies$ - 778 kj mol-1

Answer 2

Given :

∆fHC02(g) = – 393 kj mol -1

∆fCO(g) = – 110 kj

∆f N20(g) = 81 kj

∆fN2O4(g) = 9.7 kj mol-1

N204(g) + 3CO(g) ———->N20(g) + 3CO2(g)

ENTHALPHY OF REACTION .

81 + 3 (- 393) - ( 9.7 + 3 (- 110) )

= > 81 - 1179 - 9.7 + 330

= > - 777.77 KJ mol ^ ( - 1 )

ANSWER :

- 777.77 KJ mol ^ ( - 1 )