Part 1. Balance the equation.
"PbS" + "O"_2 → "SO"_2 + "PbO"
Start with the most complicated formula. Put a 1 in front of "SO"_2.
"PbS" + "O"_2 → color(red)(1)"SO"_2 + "PbO"
Balance "S". Put a 1 in front of "PbS.
color(blue)(1)"PbS" + "O"_2 → color(red)(1)"SO"_2 + "PbO"
Balance "Pb". Put a 1 in front of "PbO".
color(blue)(1)"PbS" + "O"_2 → color(red)(1)"SO"_2 + color(green)(1)"PbO"
Balance "O". Put 1.5 in front of "O"_2.
color(blue)(1)"PbS" + color(orange)(1.5)"O"_2 → color(red)(1)"SO"_2 + color(green)(1)"PbO"
Remove fractions. Multiply by 2.
"2PbS" + "3O"_2 → "2SO"_2 + 2PbO"
Part 2. Calculate ΔH_"rxn".
color(white)(mmmmmmmmm)"2PbS(s)" + "3O"_2("s") → "2SO"_2("g")color(white)(l) + "2PbO(s)"
ΔH_f^°"/kJ·mol⁻¹":color(white)(m)"-100.4"color(white)(mmm) "0"color(white)(mmmm) "-296.83"color(white)(mm) "-217.32"
For most chemistry problems involving ΔH_f^°, you need the equation:
ΔH_(rxn)^° = ΣΔH_f^°("p") - ΣΔH_f^°("r"),
where "p" = products and "r" = reactants.
ΣΔH_f^°("p") = "[2(-296.83) + 2(-217.32)] kJ = -1028.30 kJ"
ΣΔH_f^°("r") = "2(-100.4 kJ)" = "-200.8 kJ"
ΔH_("rxn")^° = ΣΔH_f^°("p") - ΣΔH_f^°("r") = "-1028.30 kJ + 200.8 kJ" = "-827.5 kJ"
The standard enthalpy change is "-827.5 kJ".