Question

Calculate heat of formation of HCl gas from following reaction.
$\text{H}_{2(g)} + \text{Cl}_{2(g)} \rightarrow 2\text{HCl}_{(g)}$ ; $\Delta\text{H} = -194\ \text{kJ}$

• A. $-143\ \text{kJ}\ \text{mol}^{-1}$
• B. $-286\ \text{kJ}\ \text{mol}^{-1}$
• C. $-92\ \text{kJ}\ \text{mol}^{-1}$
• D. $-97\ \text{kJ}\ \text{mol}^{-1}$

Hint:

Always pay close attention to the definition of "enthalpy of formation"—it is explicitly normalized for exactly 1 mole of product. Simply dividing the total given reaction energy by the stoichiometric coefficient of the product ($2$) leads directly to the correct option.

Answer 1

The Correct Option is D

Step 1: Understand the question.
The reaction $\text{H}_2 + \text{Cl}_2 \rightarrow 2\text{HCl}$ has $\Delta H = -194$ kJ. We must find the heat of formation of HCl gas.

Step 2: Recall the meaning of heat of formation.
Heat of formation is the energy change when exactly one mole of a substance is made from its elements. So it is always per one mole.

Step 3: Note how many moles the equation makes.
The given equation makes 2 moles of HCl, and releases $-194$ kJ for those 2 moles.

Step 4: Set up the per mole formula.
\[ \Delta H_f = \frac{\Delta H_{\text{reaction}}}{\text{moles of HCl}} \]

Step 5: Divide to get per mole.
\[ \Delta H_f = \frac{-194}{2} = -97 \;\text{kJ mol}^{-1} \]

Step 6: Pick the answer.
The heat of formation of HCl is $-97$ kJ mol$^{-1}$, which is option 4. The trap is forgetting to divide by 2.
\[ \boxed{-97\ \text{kJ mol}^{-1}} \]