Question

2.5 points Save Answer Given the following standard heats of formation: NH3 () --45.90 kJ/mol, O2(0) Okmol. H20 (3) --241.83 kJ/mol, and NO () -90.30 kJ/mol, the enthalpy of reaction would be Be careful of significant figures. 063.pdf Moving to another question will save this response

Answer 1

The enthalpy change for the formation of 1 mole of substance from its constituent in there standard state is called standard heat of formation . This is why heat of formation of Gaseous oxygen (O₂) is zero, because it is in standard state. The enthalpy change is related to standard heat of formation . Formula is heat of formation product side minus reactant side.

  

Enthalpy of reaction (AHrxn) 4 Hif (product) - OHI (reactant) Altran =

4 NO (g) + 6H₂O (9) 4 NH₃ (g) + 5 O₂(g) 소 reactant side product side. Given, H2O (9)=-241.83 K/ms) NH₃ = - 45.90 kJ/mol NO(g) = -90.30 K) mol

SH&xn = {{Axta *(-45.90) +0}}= {4x690.30) + 6x(+241.83) = -183.6 -(-1812-18) 1628.58 KJ/mol 2 (Reaction is Endothermic in nature)

The coefficients are must multiply with standard heat of formation. The enthalpy change is positive .Which means heat follows to the system and this reaction endothermic in nature.

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