Question

If the standard enthalpy change (\(\Delta H^o\)) for reaction \[ \text{H}_2(g) + \text{Br}_2(l) \rightarrow 2 \text{HBr}(g) \] is \(-72.8 \, \text{kJ}\), the standard enthalpy of formation (\(\Delta H_f^o\)) of \(\text{HBr}(g)\) in \(\text{kJ mol}^{-1}\) is:

• A. \(-36.4 \, \text{kJ/mol}\)
• B. \(+36.4 \, \text{kJ/mol}\)
• C. \(-18.2 \, \text{kJ/mol}\)
• D. \(+18.2 \, \text{kJ/mol}\)

Hint:

To calculate the enthalpy of formation, divide the given enthalpy change by the number of moles of the substance involved in the reaction.

Answer 1

The Correct Option is A

The enthalpy change for the reaction is given as: \[ \Delta H^\circ = -72.8 \, \text{kJ} \] For the reaction where 2 moles of \(\text{HBr}(g)\) are formed, we divide the total enthalpy change by 2 to find the enthalpy of formation for one mole of \(\text{HBr}(g)\): \[ \Delta H_f^\circ (\text{HBr}(g)) = \frac{-72.8 \, \text{kJ}}{2} = -36.4 \, \text{kJ/mol} \] Thus, the standard enthalpy of formation of \(\text{HBr}(g)\) is \(-36.4 \, \text{kJ/mol}\).

Answer 2

Step 1: Understand the given reaction and data.
The reaction is:
\[ \text{H}_2(g) + \text{Br}_2(l) \rightarrow 2 \, \text{HBr}(g), \] with a standard enthalpy change \(\Delta H^o = -72.8 \, \text{kJ}\).

Step 2: Relation to standard enthalpy of formation.
The standard enthalpy of formation \(\Delta H_f^o\) of a compound is defined as the enthalpy change when 1 mole of the compound is formed from its elements in their standard states.

Here, the formation of 2 moles of HBr from elemental hydrogen and bromine corresponds to \(\Delta H^o = -72.8 \, \text{kJ}\).

Step 3: Calculate \(\Delta H_f^o\) for 1 mole of HBr.
Since the reaction forms 2 moles of HBr, the enthalpy change per mole of HBr is:
\[ \Delta H_f^o = \frac{-72.8 \, \text{kJ}}{2} = -36.4 \, \text{kJ/mol}. \]

Step 4: Conclusion.
The standard enthalpy of formation of \(\text{HBr}(g)\) is \(\boxed{-36.4 \, \text{kJ/mol}}\).