Question

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Calculate ΔH
o
rxn
for the following:


CH4(g) + Cl2(g) → CCl4(l) + HCl(g)[unbalanced]

ΔH
o
f
[CH4(g)] = −74.87 kJ/mol
ΔH
o
f
[CCl4(g)] = −96.0 kJ/mol
ΔH
o
f
[CCl4(l)] = −139 kJ/mol
ΔH
o
f
[HCl(g)] = −92.31 kJ/mol
ΔH
o
f
[HCl(aq)] = −167.46 kJ/mol
ΔH
o
f
[Cl(g)] = 121.0 kJ/mol

Answer 1

CH4(g) + 4Cl2(g) → CCl4(l) + 4HCl(g)

ΔH= ((−139 kJ/mol) + 4 x (−92.31 kJ/mol)) - ((−74.87 kJ/mol) + (121.0 kJ/mol))= (-508.24 kJ/mol) - 46.13 kJ/mol = -554.37 kJ/mol.