Answer to Question #308385 in General Chemistry for hyuka

Question

Calculate the standard enthalpy of formation of carbon disulfide (CS2) from

its elements, given that

C(graphite) + O2(g) ⟶ CO2(g) ∆H°rxn = - 393.5 kJ/mol

S(rhombic) + O2(g) ⟶ SO2(g) ∆H°rxn = - 296.4 kJ/mol

CS2 (l) + 3O2(g) ⟶ CO2(g) + 2SO2(g) ∆H°rxn = - 1073.6 kJ/mol

Answer 1

We need to manipulate the set of equations above such that we have C(graphite) and S(rhombic) on the reactant side of the net equation and Carbon disulfide (CS2).

We need to multiply equation (2) by 2 so that we now have 2SO2 on the product side.

Also, we need to multiply equation (3) by -1 so that the CS2 moves to the product side of the equation; so that we have;

ΔHorxn = (-395.5 - 592.8 + 1073.6) kJ/mol

ΔHorxn = 85.3kJ/mol.