Question

What is enthalpy of formation of $\text{NH}_3$ if bond enthalpies are as $(\text{N}\equiv\text{N}) = 941\ \text{kJ}$, $(\text{H}-\text{H}) = 436\ \text{kJ}$, $(\text{N}-\text{H}) = 389\ \text{kJ}$?

• A. $-84.5\ \text{kJ}$
• B. $-21.25\ \text{kJ}$
• C. $-42.5\ \text{kJ}$
• D. $-63.45\ \text{kJ}$

Hint:

Always remember that the definition of the enthalpy of formation ($\Delta_f H^\circ$) explicitly requires the formation of exactly one mole of product. A common error is selecting $-85\ \text{kJ}$ because you forget to divide by 2!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given individual chemical bond enthalpies and need to compute the standard enthalpy of formation ($\Delta_f H^\circ$) for a single mole of ammonia gas ($\text{NH}_3$).

Step 2: Key Formula or Approach:
The standard chemical equation representing the synthesis of 2 moles of $\text{NH}_3$ gas from its elemental states is: $$\text{N}_{2(g)} + 3\text{H}_{2(g)} \rightarrow 2\text{NH}_{3(g)}$$ The total reaction enthalpy can be computed using the individual gaseous bond enthalpies via the relation: $$\Delta_r H^\circ = \sum \Delta H^\circ_{\text{Bonds Broken (Reactants)}} - \sum \Delta H^\circ_{\text{Bonds Formed (Products)}}$$ Since standard enthalpy of formation is defined per mole of substance produced, we must divide the total reaction enthalpy by 2: $$\Delta_f H^\circ(\text{NH}_3) = \frac{\Delta_r H^\circ}{2}$$

Step 3: Detailed Explanation:
Let's count the chemical bonds broken and formed in this reaction:

Bonds broken: 1 mole of $\text{N}\equiv\text{N}$ bonds and 3 moles of $\text{H}-\text{H}$ bonds.

Bonds formed: 2 moles of $\text{NH}_3$ molecules, where each molecule contains 3 individual $\text{N}-\text{H}$ bonds, giving a total of $2 \times 3 = 6$ moles of $\text{N}-\text{H}$ bonds. Substitute the provided numerical values into our equation: $$\Delta_r H^\circ = [1 \times \Delta H^\circ(\text{N}\equiv\text{N}) + 3 \times \Delta H^\circ(\text{H}-\text{H})] - [6 \times \Delta H^\circ(\text{N}-\text{H})]$$ $$\Delta_r H^\circ = [941 + 3(436)] - [6(389)]$$ $$\Delta_r H^\circ = [941 + 1308] - [2334]$$ $$\Delta_r H^\circ = 2249 - 2334 = -85\ \text{kJ}$$ Now, find the enthalpy of formation for one single mole of $\text{NH}_3$: $$\Delta_f H^\circ = \frac{-85\ \text{kJ}}{2} = -42.5\ \text{kJ/mol}$$

Step 4: Final Answer:
The molar enthalpy of formation of ammonia is $-42.5\ \text{kJ}$, corresponding to option (C).