Final answer:
The enthalpy change when 58.0 g of sulfur dioxide reacts with excess oxygen is calculated using the balanced thermochemical equation. By converting grams of SO2 to moles and using the stoichiometry of the reaction, the enthalpy change is found to be -89.6 kJ, revealing it's an exothermic reaction.
Step-by-step explanation:
The student has presented a chemical equation representing the reaction between sulfur dioxide and oxygen to form sulfur trioxide, and is asking how to calculate the enthalpy change when 58.0 g of sulfur dioxide is reacted with excess oxygen. The balanced thermochemical equation for the reaction is 2SO2(g) + O2(g) → 2SO3(g) and it releases 198 kJ of energy when 2 moles of SO2 are converted to 2 moles of SO3.
To calculate the enthalpy change for 58.0 g of SO2, we can use the molar mass of SO2 (64.07 g/mol) to find the number of moles of SO2, then use the thermochemical equation to find the energy change associated with this amount of reactant. The calculation is as follows:
Moles of SO2 = 58.0 g / 64.07 g/mol = 0.905 moles
The thermochemical equation shows that 2 moles of SO2 release 198 kJ, so 0.905 moles will release:
Enthalpy change = (198 kJ / 2 moles) × 0.905 moles
Enthalpy change = 99 kJ/mole × 0.905 moles
Enthalpy change = 89.6 kJ
Therefore, when 58.0 g of sulfur dioxide reacts with excess oxygen, the enthalpy change is -89.6 kJ (indicating an exothermic reaction).