the standard enthalpy of formation of ammonia is -46 kj/mol. if the enthalpy of formation of hydrogen molecule is -436 kj/mol and that of nitrogen molecule is -712 kj/mol find the average bond enthalpy of N-H bond in ammonia
Gueh
To find the average bond enthalpy of the N-H bond in ammonia (NH3), we need to use the concept of enthalpy of formation and the bond enthalpies of the hydrogen (H2) and nitrogen (N2) molecules.
The enthalpy of formation of a compound is the energy change accompanying the formation of one mole of that compound from its constituent elements in their standard states.
Given data:
Enthalpy of formation of ammonia (NH3) = -46 kJ/mol
Enthalpy of formation of hydrogen molecule (H2) = -436 kJ/mol
Enthalpy of formation of nitrogen molecule (N2) = -712 kJ/mol
We can write the balanced reaction for the formation of ammonia as follows:
N2(g) + 3H2(g) → 2NH3(g)
From the given data, we know the enthalpies of formation for N2, H2, and NH3.
Using Hess's Law, we can calculate the enthalpy change for the formation of ammonia (NH3) from its elements N2 and H2:
∆H = [2ΔHf(NH3)] - [ΔHf(H2) + 3ΔHf(N2)]
Plugging in the given values:
∆H = [2(-46 kJ/mol)] - [(-436 kJ/mol) + 3(-712 kJ/mol)]
Simplifying:
∆H = -92 kJ/mol + 436 kJ/mol + 2136 kJ/mol
∆H = 2480 kJ/mol
The enthalpy change (∆H) for the formation of ammonia is equal to 2480 kJ/mol.
Given that we have 3 N-H bonds in one molecule of ammonia, we can divide the enthalpy change by the number of bonds to get the average bond enthalpy.
Average bond enthalpy = ∆H / Number of bonds
Average bond enthalpy = 2480 kJ/mol / 3
Average bond enthalpy = 826.6 kJ/mol
Therefore, the average bond enthalpy of the N-H bond in ammonia (NH3) is 826.6 kJ/mol.