Calculate the enthalpy change for the reaction SO3(g) + H2O(l) - H2SO4(l) using standard heats of formation.

Question

Calculate the enthalpy change for the reaction SO3(g) + H2O(l) -> H2SO4(l) using standard heats of formation.

ScanSolve AI Tutor · Accepted Answer

To calculate the enthalpy change for the reaction, we use the formula:

\Delta H_{\text{rxn}} = \sum \Delta H_f^\circ (\text{products}) - \sum \Delta H_f^\circ (\text{reactants})

The reaction described is the formation of sulfuric acid from sulfur trioxide and water:

\text{SO}_3(\text{g}) + \text{H}_2\text{O}(\text{l}) \rightarrow \text{H}_2\text{SO}_4(\text{l})

List the given standard heats of formation ( $\Delta H_f^\circ$ ).
- $\Delta H_f^\circ (\text{SO}_3(\text{g})) = -395 \text{ kJ/mol}$
- $\Delta H_f^\circ (\text{H}_2\text{O}(\text{l})) = -286 \text{ kJ/mol}$
- $\Delta H_f^\circ (\text{H}_2\text{SO}_4(\text{l})) = -811 \text{ kJ/mol}$
Identify the products and reactants from the balanced equation.
Reactants: $\text{SO}_3(\text{g})$ and $\text{H}_2\text{O}(\text{l})$
Product: $\text{H}_2\text{SO}_4(\text{l})$
Substitute the values into the enthalpy change formula.
$\Delta H_{\text{rxn}} = [\Delta H_f^\circ (\text{H}_2\text{SO}_4(\text{l}))] - [\Delta H_f^\circ (\text{SO}_3(\text{g})) + \Delta H_f^\circ (\text{H}_2\text{O}(\text{l}))]$ $\Delta H_{\text{rxn}} = [-811 \text{ kJ/mol}] - [-395 \text{ kJ/mol} + (-286 \text{ kJ/mol})]$
Calculate the sum of the heats of formation for the reactants.
$\text{Sum of reactants} = -395 \text{ kJ/mol} - 286 \text{ kJ/mol} = -681 \text{ kJ/mol}$
Calculate the enthalpy change of the reaction.
$\Delta H_{\text{rxn}} = -811 \text{ kJ/mol} - (-681 \text{ kJ/mol})$ $\Delta H_{\text{rxn}} = -811 \text{ kJ/mol} + 681 \text{ kJ/mol}$ $\Delta H_{\text{rxn}} = -130 \text{ kJ/mol}$
The enthalpy change accompanying this reaction is:
$\text{-130 kJ/mol}$

Calculate the enthalpy change for the reaction SO3(g) + H2O(l) -> H2SO4(l) using standard heats of formation.