Question

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

Answer 1

Answer:

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

be aware of signs