What is the reaction for the standard enthalpy of formation for CdSO₄(s)? (Cd(s) is the stable form of cadmium.)

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The standard enthalpy of formation for CdSO₄ can be determined by using the following reaction: Cd(s) + 1/2 S2(g) + 1/2 O2(g) → CdSO4(s). The standard enthalpy of formation for CdSO4 is -435.1 kJ/mol. So, we can calculate the standard enthalpy of formation for CdSO₄ using Hess's law: Cd(s) + 1/2 S2(g) + 1/2 O2(g) → CdSO4(s) ΔH° = -435.1 kJ/mol CdSO4(s) + 1/2 O2(g) → CdSO₄(s) + 1/2 SO2(g) ΔH° = x Cd(s) + 1/2 S2(g) + O2(g) → CdSO₄(s) ΔH°f = x - 435.1 kJ/mol We need to find x, which is the standard enthalpy of formation for CdSO₄. The correct answer is C, which shows a balanced chemical equation that corresponds to the Hess's law calculation above.

What Is the Reaction for the Standard Enthalpy of Formation