What is ΔHof of Al2S3(s)?
2Al2S3(s) + 9O2(g) → 2Al2O3(s) + 6SO2(g) ΔHo = -3684.2 kJ
ΔHof Al2O3(s) = -1675.7 kJ/mol
ΔHof SO2(g) = -296.8 kJ/mol
To find the standard enthalpy of formation (ΔHof) of Al2S3(s), we need to use the equation:
2Al2S3(s) + 9O2(g) → 2Al2O3(s) + 6SO2(g)
We have the values for ΔHo for the reaction (-3684.2 kJ) and the standard enthalpy of formation for Al2O3(s) (-1675.7 kJ/mol) and SO2(g) (-296.8 kJ/mol).
Using the equation:
ΔHo = ∑(ΔHof products) - ∑(ΔHof reactants)
We can rearrange it to solve for ΔHof Al2S3(s):
ΔHof Al2S3(s) = (∑(ΔHof products)) - (∑(ΔHof reactants))
ΔHof Al2S3(s) = (2mol * ΔHof Al2O3(s)) + (6mol * ΔHof SO2(g)) - (2mol * ΔHof Al2S3(s))
Substituting the given values:
ΔHof Al2S3(s) = (2mol * (-1675.7 kJ/mol)) + (6mol * (-296.8 kJ/mol)) - (-3684.2 kJ)
ΔHof Al2S3(s) = -3351.4 kJ - 1780.8 kJ + 3684.2 kJ
ΔHof Al2S3(s) = -1447.98 kJ
Therefore, the standard enthalpy of formation (ΔHof) of Al2S3(s) is -1447.98 kJ.