Question

The enthalpy change for the reaction
 \(2H_2(g) + O_2(g) → 2H_2O(l)\)
is -572 kJ/mol. What is the enthalpy change for the formation of 1 mole of \(H_2O(l)\)}?

• A. -572 kJ/mol
• B. -286 kJ/mol
• C. 286 kJ/mol
• D. 572 kJ/mol

Hint:

For enthalpy changes in reactions, scale the \(\Delta H\) by the stoichiometric coefficient of the desired species.

Answer 1

The Correct Option is B

The given reaction is: \[ \ch{2H2(g) + O2(g) → 2H2O(l)}, \quad \Delta H = -572 \, \text{kJ} \] This enthalpy change corresponds to the formation of 2 moles of \(H_2O(l)\). To find the enthalpy change for 1 mole of \(H_2O(l)\), divide by 2: \[ \Delta H_{\text{per mole}} = \frac{-572}{2} = -286 \, \text{kJ/mol} \] Thus, the enthalpy change for the formation of 1 mole of \(H_2O(l)\) is: \[ {-286} \]