The standard enthalpy of formation of NH3 is -46.0kj/mol . if the enthalpy of formation of H2 from its atoms is -436kj/ mol and that of N2 is -712 kj/mol , the average bond enthalpy of N - H bond in NH3 is ?

To find the average bond enthalpy of the N-H bond in ammonia (NH3), we can use the given enthalpy values and apply Hess's law. This law states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction. Let's break down the process step by step.

Understanding the Formation of Ammonia

The formation of ammonia from its elements can be represented by the following reaction:

• N_2(g) + 3H_2(g) → 2NH_3(g)

We need to consider the enthalpy changes involved in this reaction. The standard enthalpy of formation of NH3 is given as -46.0 kJ/mol. This means that when 1 mole of NH3 is formed from its elements, 46.0 kJ of energy is released.

Breaking Down the Enthalpy Changes

Next, we need to look at the enthalpy changes for the formation of the reactants:

• The enthalpy of formation of H2 from its atoms is -436 kJ/mol.
• The enthalpy of formation of N2 from its atoms is -712 kJ/mol.

Since we are forming NH3 from N2 and H2, we can express the enthalpy change for the reaction in terms of the bond enthalpies. The overall reaction can be thought of as breaking the bonds in N2 and H2 and then forming the N-H bonds in NH3.

Calculating the Total Enthalpy Change

To find the average bond enthalpy of the N-H bond, we can set up the equation based on the enthalpy changes:

• ΔH (reaction) = ΔH (bonds broken) + ΔH (bonds formed)

For the reaction:

• ΔH (reaction) = -46.0 kJ (for 2 moles of NH3)
• ΔH (bonds broken) = ΔH (N≡N) + 3 × ΔH (H-H)
• ΔH (bonds formed) = 6 × ΔH (N-H) (since there are 6 N-H bonds in 2 moles of NH3)

Plugging in the Values

Now, we can substitute the known values into the equation:

• ΔH (N≡N) = 712 kJ/mol (since breaking bonds requires energy, we take the positive value)
• ΔH (H-H) = 436 kJ/mol

Now, substituting these into the equation:

• -46.0 kJ = (712 kJ + 3 × 436 kJ) - 6 × ΔH (N-H)

Calculating the left side:

• 3 × 436 kJ = 1308 kJ
• 712 kJ + 1308 kJ = 2020 kJ

Now, substituting back into the equation:

• -46.0 kJ = 2020 kJ - 6 × ΔH (N-H)

Solving for the Average Bond Enthalpy

Rearranging the equation gives:

• 6 × ΔH (N-H) = 2020 kJ + 46.0 kJ
• 6 × ΔH (N-H) = 2066 kJ
• ΔH (N-H) = 2066 kJ / 6
• ΔH (N-H) = 344.33 kJ/mol

Thus, the average bond enthalpy of the N-H bond in ammonia is approximately 344.33 kJ/mol. This value reflects the energy required to break one mole of N-H bonds in ammonia, providing insight into the stability of the ammonia molecule.