Question

What is the enthalpy change for the following reaction: C3H8(g) + 5O2 (g) --> 3CO2 (g) + 4H2O (l)ΔHof C3H8 (g) = -103.8 kJΔHof CO2 (g) = -393.5 kJΔHof H20 (l) = -285.8 kJ

Answer 1

`\Delta\text{H}_{\text{ rxn}}\text{ = 3}\Delta\text{H}_{\text{CO}_2}\text{ + 4}\Delta\text{H}_{\text{H}_2\text{O}}\text{ }-\text{ }\Delta\text{H}_{\text{ C}_3\text{H}_8}\text{ }-5\Delta\text{H}_{\text{O}_2}`
`\Delta\text{H}_{\text{ rxn}}\text{ = 3\lparen-393.5 kJ/mol\rparen + 4\lparen-285.8 kJ/mol\rparen }-\text{ \lparen-103.8 kJ/mol\rparen }-5(0\text{ kJ/mol})`
`\Delta\text{H}_{\text{ rxn}}\text{ = -2219.9 kJ/mol}`

The answer is -2219.9 kJ/mol