Given: N2 + 3H2 = 2NH3 + 92.5 KJ, what is the delta h of f for NH3?
The equation tells you that this is an exothermic reaction; therefore, dH = -92.5 kJ/2 mol = ? kJ/mol.
To find the enthalpy change of formation (ΔHf) for NH3, we need to use the equation you provided and the known enthalpy change of reaction (ΔH).
From the balanced equation: N2 + 3H2 = 2NH3, we can see that 2 moles of NH3 are produced from the reaction of N2 and 3 moles of H2.
The enthalpy change of reaction (ΔH) is given as 92.5 kJ. This means that 92.5 kJ of energy is released or absorbed when the reaction occurs.
To find the ΔHf for NH3, we need to express it in terms of the enthalpy change of reaction (ΔH).
Since 2 moles of NH3 are formed in the reaction, we can use stoichiometry to relate the enthalpy changes as follows:
(ΔHf)NH3 = (ΔH) / moles of NH3
Substituting the values, we have:
(ΔHf)NH3 = 92.5 kJ / 2 moles
Therefore, the enthalpy change of formation (ΔHf) for NH3 is 46.25 kJ per mole.