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"#.
