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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

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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

adioabiola

The standard enthalpy of formation of carbon disulfide (CS2) from it's elements is; ΔHorxn = 85.3kJ/mol

To calculate the standard enthalpy of formation of Carbon disulfide (CS2) from it's elements, given the following;

  • C(graphite) + O2(g) → CO2(g); .............eqn(1)

C(graphite) + O2(g) → CO2(g); .............eqn(1)ΔHorxn = −393.5 kJ/mol

  • C(graphite) + O2(g) → CO2(g); .............eqn(1)ΔHorxn = −393.5 kJ/molS(rhombic) + O2(g) → SO2(g) ..............eqn(2)

C(graphite) + O2(g) → CO2(g); .............eqn(1)ΔHorxn = −393.5 kJ/molS(rhombic) + O2(g) → SO2(g) ..............eqn(2)ΔHorxn = −296.4 kJ/mol

  • C(graphite) + O2(g) → CO2(g); .............eqn(1)ΔHorxn = −393.5 kJ/molS(rhombic) + O2(g) → SO2(g) ..............eqn(2)ΔHorxn = −296.4 kJ/molCS2 + 3O2(g) → CO2(g) + 2SO2(g)

C(graphite) + O2(g) → CO2(g); .............eqn(1)ΔHorxn = −393.5 kJ/molS(rhombic) + O2(g) → SO2(g) ..............eqn(2)ΔHorxn = −296.4 kJ/molCS2 + 3O2(g) → CO2(g) + 2SO2(g) ΔHorxn = −1073.6 kJ/mol

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;

  • C(graphite) + O2(g) → CO2(g); .............eqn(1)

C(graphite) + O2(g) → CO2(g); .............eqn(1)ΔHorxn = −393.5 kJ/mol

  • C(graphite) + O2(g) → CO2(g); .............eqn(1)ΔHorxn = −393.5 kJ/mol2S(rhombic) + 2O2(g) → 2SO2(g) ........eqn(2)

C(graphite) + O2(g) → CO2(g); .............eqn(1)ΔHorxn = −393.5 kJ/mol2S(rhombic) + 2O2(g) → 2SO2(g) ........eqn(2)ΔHorxn = −296.4 kJ/mol × 2 = -592.8kJ/mol

  • C(graphite) + O2(g) → CO2(g); .............eqn(1)ΔHorxn = −393.5 kJ/mol2S(rhombic) + 2O2(g) → 2SO2(g) ........eqn(2)ΔHorxn = −296.4 kJ/mol × 2 = -592.8kJ/molCO2(g) + 2SO2(g) → CS2 + 3O2(g)

C(graphite) + O2(g) → CO2(g); .............eqn(1)ΔHorxn = −393.5 kJ/mol2S(rhombic) + 2O2(g) → 2SO2(g) ........eqn(2)ΔHorxn = −296.4 kJ/mol × 2 = -592.8kJ/molCO2(g) + 2SO2(g) → CS2 + 3O2(g)ΔHorxn = −1073.6 kJ/mol × -1 = 1073.6kJ/mol

Therefore, the net equation is the algebraic sum of all the equations; so that we have;

C(graphite) + 2S(rhombic) → CS2

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

ΔHorxn = 85.3kJ/mol.

In essence, the standard enthalpy of formation of carbon disulfide (CS2) from it's elements is;

ΔHorxn = 85.3kJ/mol

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