Question

The enthalpy of formation for \( \text{H}_2(g) \), \( \text{O}_2(g) \), and \( \text{H}_2O(l) \) are 0, 0, and -285.8 kJ/mol, respectively. What is the enthalpy change for the following reaction: \[ \text{2H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2O(l) \]

• A. \( -571.6 \, \text{kJ/mol} \)
• B. \( -285.8 \, \text{kJ/mol} \)
• C. \( 0 \, \text{kJ/mol} \)
• D. \( 571.6 \, \text{kJ/mol} \)

Hint:

The enthalpy change of a reaction can be calculated by subtracting the enthalpy of formation of the reactants from the enthalpy of formation of the products.

Answer 1

The Correct Option is A

Step 1: Understand the reaction. We are asked to calculate the enthalpy change for the reaction: \[ \text{2H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2O(l) \] Step 2: Use the enthalpy of formation values. The enthalpy change of a reaction can be calculated using the enthalpy of formation values: \[ \Delta H_{\text{reaction}} = \sum (\Delta H_f^{\circ} \text{products}) - \sum (\Delta H_f^{\circ} \text{reactants}) \] For this reaction: - The enthalpy of formation for \( \text{H}_2(g) \) and \( \text{O}_2(g) \) is 0. - The enthalpy of formation for \( \text{H}_2O(l) \) is -285.8 kJ/mol. Substitute into the equation: \[ \Delta H = \left( 2 \times (-285.8) \right) - \left( 2 \times 0 + 1 \times 0 \right) \] \[ \Delta H = -571.6 \, \text{kJ/mol} \] Answer: Therefore, the enthalpy change for the reaction is \( -571.6 \, \text{kJ/mol} \).