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Question: S(s)+ O2(g) --> SO2(g) Delta H=-296.8 kJ/mol S(s) + 3/2(O2)(g) --> SO3(g) Delta H= -395.6kJ/mol Determine the enthalpy change for the decomposition reaction. 2SO3(g)--> 2SO2(g) + O2(g)
S(s)+ O2(g) --> SO2(g) Delta H=-296.8 kJ/mol
S(s) + 3/2(O2)(g) --> SO3(g) Delta H= -395.6kJ/mol

Determine the enthalpy change for the decomposition reaction.

2SO3(g)--> 2SO2(g) + O2(g)

Question Ss O2g gt SO2g Delta H2968 kJmol Ss 32O2g gt SO3g Delta H 3956kJmol Determine the enthalpy change for the decomposition reaction 2SO3ggt 2SO2g O2g Ss O class=


Answer :

Answer:So, the enthalpy change for the decomposition reaction is −989.2 kJ/mol−989.2kJ/mol.

Explanation:To find the enthalpy change for the decomposition reaction 2SO3(g)→2SO2(g)+O2(g)2SO3​(g)→2SO2​(g)+O2​(g), we apply Hess's Law.Given reactions:S(s)+O2(g)→SO2(g)S(s)+O2​(g)→SO2​(g) with ΔH=−296.8 kJ/molΔH=−296.8kJ/molS(s)+32O2(g)→SO3(g)S(s)+23​O2​(g)→SO3​(g) with ΔH=−395.6 kJ/molΔH=−395.6kJ/molDouble the second reaction and reverse it: 2SO3(g)→2S(s)+3O2(g)2SO3​(g)→2S(s)+3O2​(g)Reverse the sign of its enthalpy change: ΔH=395.6 kJ/molΔH=395.6kJ/molMultiply the first reaction by 2: 2S(s)+2O2(g)→2SO2(g)2S(s)+2O2​(g)→2SO2​(g)Now, sum up the modified reactions:ΔHdecomposition=(2×−296.8)+395.6ΔHdecomposition​=(2×−296.8)+395.6ΔHdecomposition=−593.6+395.6=−989.2 kJ/molΔHdecomposition​=−593.6+395.6=−989.2kJ/molSo, the enthalpy change for the decomposition reaction is −989.2 kJ/mol−989.2kJ/mol.