Final answer:
To calculate the enthalpy change for the reaction of SO2 with oxygen, the moles of SO2 that 58.0 g represents are first calculated using its molar mass, and this amount is used to proportionally calculate the enthalpy change, resulting in approximately -90 kJ.
Step-by-step explanation:
To calculate the enthalpy change (ΔH) for the reaction of sulfur dioxide (SO2) with oxygen (O2), we first need to know the balanced chemical equation and the enthalpy change associated with the reaction:
2SO2(g) + O2(g) → 2SO3(g) + 198 kJ
Since the enthalpy change for the formation of 2 moles of SO3 is +198 kJ, for 1 mole of SO3 produced it would be +99 kJ. We need to determine the number of moles of SO2 that corresponds to 58.0 grams, and then apply this proportion to find the enthalpy change for that amount.
The molar mass of SO2 is 64.07 g/mol (32.07 g/mol for S + 2 × 16.00 g/mol for O). Therefore, 58.0 g of SO2 corresponds to 58.0 g / 64.07 g/mol = 0.905 mol SO2.
The reaction shows that 2 moles of SO2 produce 2 moles of SO3 and release 198 kJ, thus 0.905 moles of SO2 will produce 0.905 moles of SO3 and release 0.905 / 2 × 198 kJ. After calculating, the ΔH for 58.0 g of SO2 reacted is approximately -90 kJ (since the reaction is exothermic).