[dathrilla]

Hi!

I have a problem with the 7/2(O2) in this equation...

2NH3(g)   +   7/2(O2)(g)  ----------->   2NO2(g)   +   3H2O(g)           HfNO2= +8.60(kcal/mole)   HfH2O= -57.8

127(kcal) are being released. We have to determine the ammoniac in the problem, my problem is that i don't  know what the 02 is equal to in terms of (kcal)

thanx

[Albert]

Oxygen gas is the stable allotropic form of oxygen. So, its enthalpy of formation is dHf = 0.

[dathrilla]

Thanks a lot.

But the proposed answer is -14.6 (kcal/mole)

I find -141.6, could you tell me why?

[Albert]

If 127 kcal are released, you deltaH of reaction is negative.

[T_N_T]

Hi!

I have a problem with the 7/2(O2) in this equation...

2NH3(g)   +   7/2(O2)(g)  ----------->   2NO2(g)   +   3H2O(g)           HfNO2= +8.60(kcal/mole)   HfH2O= -57.8

127(kcal) are being released. We have to determine the ammoniac in the problem, my problem is that i don't  know what the 02 is equal to in terms of (kcal)

thanx

Hf(O2g)=zero

DetalH(reaction) = [2*Hf(NO2) + 3*Hf(H2O)] – [2*Hf(NH3) + Hf(O2)]

<=> -127 = [2*8.60 + 2*(-98.4)] – 2*Hf(NH3)

è Hf(NH3) = -14,6(kcal/mol)

Goodluck.