Given the thermochemical equation #2SO_2+O_2 ->2SO_3#, #ΔH°_(rxn)# #= - 198 # #kJ#, what is the standard enthalpy change for the decomposition of one mole of #SO_3#?

1 Answer
Dec 13, 2015

Answer:

#DeltaH_"dec"^@ = +"99 kJ"#

Explanation:

Start by taking a look at the thermochemical equation given to you

#2"SO"_text(2(g]) + "O"_text(2(g]) -> 2"SO"_text(3(g])" " DeltaH_text(rxn)^@ = -"198 kJ/mol"#

Two important things to notice here

  • the balanced chemical equation stipulates the formation of two moles of sulfur trioxide when two moles of sulfur dioxide and one mole of oxygen react

  • the standard enthalpy change of reaction carries a negative sign

The fact that the standard enthalpy change of reaction carries a negative sign tells you that #"198 kJ"# heat is being given off when two moles of sulfur trioxide are formed by the reaction.

Simply put, the standard enthalpy change of reaction given to you corresponds to the formation of two moles of sulfur trioxide.

The first thing to do now is determine how much heat is given off when one mole of sulfur trioxide is formed by this reaction

#1 color(red)(cancel(color(black)("mole SO"_3))) * "198 kJ"/(2color(red)(cancel(color(black)("moles SO"_3)))) = "99 kJ"#

This tells you that the reaction

#"SO"_text(2(g]) + 1/2"O"_text(2(g]) -> "SO"_text(3(g])#

will have a standard enthalpy change of reaction equal to #DeltaH_"forw"^@ = -"99 kJ"#.

Since the formation of sulfur trioxide is exothermic process, i.e. it gives off heat, the decomposition of sulfur trioxide will be an endothermic process, i.e. it will require heat.

This means that you need to change the sign of the standard enthalpy change of reaction from negative to positive. Therefore, this reaction

#"SO"_text(3(g]) -> "SO"_text(2(g]) + 1/2"O"_text(2(g])#

will have a standard enthalpy change of reaction equal to

#DeltaH_"dec"^@ = - DeltaH_"forw"^@#

#DeltaH_"dec"^@ = color(green)(+"99 kJ")#