Calculate the standard enthalpy of formation of carbon disulfide (CS2) from it's elements, given that C (graphite) + O2 (g) → CO2 (g) ΔH o rxn = - 393.5 kJ/mol S (rhombic) + O2 (g) → SO2 (g) ΔH o rxn = - 296.4 kJ/mol CS2 + 3O2 (g) → CO2 (g) + 2SO2 (g) ΔH o rxn = - 1073.6 kJ/mol

Standard enthalpy of formation of Carbon disulfide CS2 = 87.3 KJ/mol

Explanation:

forming CS2 means that it should in the product side

C (graphite) + O2 → CO2 ΔH = - 393.5

2S (rhombic) + 2O2 → 2SO2 ΔH = - 296.4 x 2

CO2 + 2SO2 → CS2 + 3O2 ΔH = - 1073.6 x - 1

the second reaction is multiplied by 2 so that the SO2 and O2 can cancel out.

the third reaction is reversed (multiplied by - 1) so that CS2 will be on the product side.

after adding the reaction and cancelling out similarities, the final reaction is: C (graphite) + 2S (rhombic) → CS2

Add ΔH to find the enthalpy of formation of CS2

ΔHf = (-393.5) + (-296.4 x 2) + (-1073.6 x - 1) = 87.3 KJ/mol

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