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Consider the reduction of gaseous sulfur trioxide, SO₃(g), to form sulfur dioxide:

\[ \text{SO}_3(g) \rightleftharpoons \text{SO}_2(g) + \frac{1}{2} \text{O}_2(g) \]

ΔH = 98.3 kJ

(a) What is the enthalpy change for the reverse reaction?
ΔH_reverse = ______ kJ/mol.

Answer :

Final answer:

The enthalpy change for the reverse reaction of SO3 to SO2 is -98.3 kJ/mol. To calculate the enthalpy change for reacting 58.0 g of SO2 with excess oxygen, convert the mass to moles and use the known enthalpy change per mole to find the total energy released.

Explanation:

The enthalpy change for the reverse reaction of the reduction of sulfur trioxide (SO3) to sulfur dioxide (SO2) can be determined by considering the forward reaction's enthalpy change. Since the forward reaction (the reduction of SO3 to SO2) has an enthalpy change (H) of +98.3 kJ, the reverse reaction, which is the oxidation of SO2 to form SO3, will have an enthalpy change of -98.3 kJ. This is because enthalpy changes are effectively reversed in sign when a reaction is reversed.

Regarding the calculation of the enthalpy change when a specific mass of SO2 is reacted with excess oxygen, first, we need to convert the mass of SO2 to moles using its molar mass (64.07 g/mol for SO2). For every 2 moles of SO2 that react according to the given balanced equation (2SO2 + O2 -> 2SO3), 198 kJ of energy is released. Since 198 kJ is released for every 2 moles, we can simply calculate the energy released for the actual amount of moles present in 58.0 g of SO2.

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