When 2 moles of #H_2(g)# and 1 mole of #O_2(g)# react to give liquid water, 572 kJ of heat evolve. #2H_2(g)# + #O_2(g)# #rarr# #2H_2O(l)# ; #DeltaH# = -572 kJ a) What is the themochemical eq. for 1 mole of liquid water?

You know that when #2# moles of hydrogen gas react with #1# mole of oxygen gas, you get #2# moles of water and #"572 kJ"# of heat are evolved.

#2"H"_ (2(g)) + "O"_ (2(g)) -> 2"H"_ 2"O"_ ((l))" "DeltaH = -"572 kJ"#

Don't forget that the enthalpy change of reaction must be negative here to illustrate the fact that heat if being given off by the reaction.

Now, in order for this reaction to produce #1# mole of water, all the coefficients of the chemical equation must be halved.

#(1/2 * 2)"H"_ (2(g)) + 1/2"O"_( 2(g)) -> (1/2 * 2)"H"_ 2"O"_ ((l))#

This will get you

#"H"_ (2(g)) + 1/2"O"_ (2(g)) -> "H"_ 2"O"_ ((l))#

Now, the enthalpy change for this reaction will be half the value of the enthalpy change for the reaction that produced #2# moles of water.

#DeltaH_ ("1 mole H"_ 2"O") = 1/2 * DeltaH_ ("2 moles H"_ 2"O")#

#DeltaH_ ("1 mole H"- 2"O") = (-"572 kJ")/2 = -"286 kJ"#

This means that the thermochemical equation that describes the formation of #1# mole of water looks like this

#"H"_ (2(g)) + 1/2"O"_ (2(g)) -> "H"_ 2"O"_ ((l))" " DeltaH = -"276 kJ"#

To write the thermochemical equation that describes the decomposition of #1# mole of water into hydrogen gas and oxygen gas, you need to reverse the chemical equation

#"H"_ 2"O"_ ((l)) -> "H"_ (2(g)) + 1/2"O"_ (2(g))#

and change the sign of the enthalpy change of reaction.

#DeltaH_"reverse" = -DeltaH_"forward"#

This means that the thermochemical equation will look like this

#"H"_ 2"O"_ ((l)) -> "H"_ (2(g)) + 1/2"O"_ (2(g))" " DeltaH = +"276 kJ"#

This means that when #1# mole of water undergoes decomposition, #"276 kJ"# of heat are being absorbed!