The standard enthalpy of formation of 3 NH is 1 kJmol.046 -- . If the enthalpy of formation of 2H from its atoms is 1 kJmol436 -- and that of 2 N is 1 kJmol712 -- , the average bond enthalpy of H N - bond in 3 NH ]

To find the average bond enthalpy of the H-N bond in ammonia (NH₃), we can use the given standard enthalpy of formation values and apply Hess's law. This approach allows us to relate the enthalpy changes of different reactions to find the desired bond enthalpy.

Understanding the Problem

We have the following information:

• The standard enthalpy of formation of NH₃ (3 NH) is +1 kJ/mol.
• The enthalpy of formation of H₂ from its atoms is +1 kJ/mol.
• The enthalpy of formation of N₂ from its atoms is +1 kJ/mol.

Setting Up the Reaction

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

1/2 N₂(g) + 3/2 H₂(g) → NH₃(g)

The enthalpy change for this reaction is the standard enthalpy of formation of NH₃, which is given as +1 kJ/mol.

Breaking Down the Components

Next, we need to consider the bond enthalpies involved in forming NH₃ from its elements. The formation involves breaking bonds in H₂ and N₂ and forming bonds in NH₃.

The enthalpy change for breaking the bonds can be expressed as:

• Breaking 1 mole of N≡N bond (from N₂): +1 kJ/mol
• Breaking 3 H-H bonds (from 3/2 H₂): +3/2 * 1 kJ/mol = +1.5 kJ/mol

Calculating Total Enthalpy Change

The total enthalpy change for breaking the bonds is:

ΔH(breaking) = ΔH(N₂) + ΔH(H₂) = 1 kJ/mol + 1.5 kJ/mol = 2.5 kJ/mol

Forming the Bonds in NH₃

When forming NH₃, we create three H-N bonds. Let’s denote the average bond enthalpy of the H-N bond as B. The enthalpy change for forming these bonds is:

ΔH(forming) = -3B

Applying Hess's Law

According to Hess's law, the total enthalpy change for the formation of NH₃ can be expressed as:

ΔH(reaction) = ΔH(breaking) + ΔH(forming)

Substituting the values we have:

1 kJ/mol = 2.5 kJ/mol - 3B

Solving for B

Now, we can rearrange the equation to solve for B:

3B = 2.5 kJ/mol - 1 kJ/mol

3B = 1.5 kJ/mol

B = 1.5 kJ/mol / 3 = 0.5 kJ/mol

Final Result

The average bond enthalpy of the H-N bond in ammonia (NH₃) is approximately 0.5 kJ/mol. This value reflects the energy required to break one mole of H-N bonds in the ammonia molecule, providing insight into the stability and reactivity of ammonia in various chemical contexts.