How do I solve this enthalpy of formation problem?

Question

The heats of combustion of C, H2, and CH4 at 298K and 1 atm are respectively -393 kJ/mol, -286 kJ/mol, and -892 kJ/lol. How do I calculate the enthalpy of formation for CH4 using the given information?

Answer 1

This is an example of using Hess's Law.  Since you are given ∆H combustion, you should write the combustion reactions to get started.

C + O₂ ==> CO₂   ∆H = -393 kJ

H₂ + 1/2 O₂ ==> H₂O   ∆H = -286 kJ

CH₄ + 2O₂ ==> CO₂ + 2H₂O   ∆H = -892 kJ

 

Now, to use Hess's Law, you need to rearrange the various equations above so that you cancel what's not wanted and keep what is wanted.  You essentially want to get it in the form C + 2H₂ == CH_4.  So, here's what to do.

Re-write the first equation as is.  Then multiply the 2nd equation by 2.  Finally, reverse the 3rd equation.  You'll have this:

 

C + O2 ==> CO2   ∆H = -393 kJ

2H2 + O2 ==> 2H2O   ∆zh = 2 x -286 kJ = -572 kJ

CO2 + 2H2O ==> CH4 + 2O2   ∆H = +890 kJ (NOTE THAT YOU MUST CHANGE THE SIGN)

________________________________________  add the 3 equations

C + 2H₂ ==> CH₄  after cancelling the H₂O's and the O₂'s on each side.

Then add up the corresponding ∆H values which would be -393 kJ + (-572 kJ) + 890 kJ = -75 kJ This is the answer.

Note that it is negative, meaning an exothermic reaction.