Calculate the standard enthalpy of formation of carbon disulfide (CS2) from its elements, given that.

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

This problem requires some knowledge of Hess' Law (review if necessary).

C(graphite) + 2S(rhombic) --> CS2(l)

Given:

eq.1: C(graphite) + O2(g)-->CO2(g) ... ∆H°rxn= -393.5 kJ/mol

eq.2: S(rhombic) + O2(g)-->SO2(g) ... ∆H°rxn= -296.4 kJ/mol

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

Procedure:

copy eq.1: C(graphite) + O2(g)-->CO2(g) ... ∆H°rxn= -393.5 kJ/mol

multiply eq.2 by 2: 2S(rhombic) + 2O2(g)--> 2SO2(g) ... ∆H°rxn= -592.8 kJ/mol (multiply by 2)

reverse eq.3: CO2(g)+ 2SO2(g) --> CS2(l) +3O2(g) ... ∆Hºrxn = +1073.6 (change sign)

Add all three and combine/cancel like terms:

C(graphite) + O2(g) + 2S(rhombic) + 2O2(g) + CO2(g)+ 2SO2(g) --> CO2(g) + 2SO2(g) + CS2(l) +3O2(g)

C(graphite) + 2S(rhombic) --> CS2(l) TARGET EQUATION

∆Hºrxn = -393.5 + -592.8 + 1073.6 = 87.3 kJ/mol

Answer 2

Multiply the 2^nd equation by 2 (which doubles its ∆H°_rxn) and

reverse the direction of the 3rd equation (which changes the sign of its ∆H°_rxn from negative to positive).

Then add the resulting equations together:

C(graphite) + O₂(g) --> CO₂(g) ∆H°_rxn= –393.5 kJ/mol

2S(rhombic) + 2O₂(g) --> 2SO₂(g) ∆H°_rxn = 2(–296.4) kJ/mol = –592.8 kJ/mol

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

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C(graphite) + 3O₂(g) + 2S(rhombic) + CO₂(g) + 2SO₂(g) --> CO₂(g) + 2SO₂(g) + CS₂(l) + 3O₂(g)

Now cancel out the molecules that appear with equal coefficients on left and right to obtain

C(graphite) + 2S(rhombic) --> CS₂(l)

which is the required equation for formation of CS₂(l) from its component elements.

We get ∆H°_rxn by summing the three enthalpies:

∆H°_rxn = –393.5 – 592.8 +1073.6 = 87.3 kJ/mol

So we calculate the enthalpy of formation of CS₂ to be 87.3 kJ/mol