1. Calculate the enthalpy change for the following chemical reaction:
CS2(1) + 3O2(g) ---> CO2(g) + 2SO2(g)
1.C(s) + O2(g) ---> CO2(g) ∆H° = - 393.5 kJ
2.S(s) + O2(g) ---> SO2(g) ∆H° = - 296.8 kJ
4.C(s) + 2S(s) ---> CS2(1) ∆H° = + 87.9 kJ
The enthalpy change of the given reaction can be deduced from the enthalpies given for reactions 1, 2 and 4. By combining these reactions together, the given one might be deduced:
reaction 1 + 2*reaction 2 +reverse reaction 4
The enthalpy of the given reaction is:
"\\Delta H = \\Delta H^o_1+2 \\times \\Delta H^o_2 - \\Delta H^o_4"
"\\Delta H = -393.5 + 2\\times (-296.8) - 87.9 = -1075.0 kJ"
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