Given the following reactions where X represents a generic metal or metalloid,
1) H2(g)+1/2O2 –> H2O…………..DeltaH1 = 241.8 kJ
2) X(s)+2Cl2(g) –>XCl4(s)……….DeltaH2 = 207.7 kJ
3) 1/2H2(g)+1/2Cl2(g) –>HCl(g)..DeltaH3 = 92.3 kJ
4) X(s)+O2(g) –> XO2(s)…………..DeltaH3 = 810.1 kJ
5) H2O(g) –>H2O(l)…………………DeltaH5= 44.0 kJ
What is the enthalpy (DeltaH) for this reaction:
XCl(s) + 2H2O(l) –> XO2(s) + 4HCl(g)
Can I get a step by step?
1 Answer

We know that XCl4 is a reactant. That means equation two will have to be reversed.
We know that 4HCl is a product. That means equation three will have to multiplied by 4.
XO2, in equation four, is in the right place and the right amount. Leave that equation alone.
We need 2H2O(l) as a reactant. Reverse equation five and multiply by 2.
2H2O(g), from our changed equation five, needs to be removed. We do that by reversing equation one and multiplying it by 2.
Now, modify the equations according to the above:
1) 2H2O(g) —> 2H2(g)+O2……….DeltaH1 = +483.6 kJ
2) XCl4(s) —> X(s)+2Cl2(g)……….DeltaH2 = 207.7 kJ
3) 2H2(g)+2Cl2(g) –>4HCl(g)…….DeltaH3 = 369.2 kJ
4) X(s)+O2(g) –> XO2(s)……………DeltaH4 = 810.1 kJ
5) 2H2O(l) –>2H2O(g)………………DeltaH5= +88.0 kJ
If we did everything correctly, adding the five modified equations should wind up in the target equation. H2O(g) will cancel, as will H2, O2, Cl2 and X(s).
Add up the five modified enthalpies for the final answer. Make sure to check that I did all the sign changes and multiplications correctly.
I have a bunch more Hess’ Law examples here:
http://chemteam.info/Thermochem/HessLawIntro1b.htm…
They are mostly four data equationtype problems but there are some with five (like yours).