How to Determine Standard Enthalpy Change Using Hess's Law?

[nobb]

Hey. I am having trouble with this question:

What is the standard enthalpy change for the production of hydrogen from methane and steam: CH4(g) + H2O(g) -> CO(g) + 3H2(g)

2C(s) + O2(g) -> 2CO(g) Hc= -221.0 kJ
CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g) Hc= -802.7 kJ
CO(g) + H2O(g) -> CO2(g) + H2(g) Hr= -41.2 kJ
2H2(g) + O2(g) -> 2H2O(g) Hc= -483.6 kJ
C(s) + 2H2(g) -> CH4(g) Hf= -74.4 kJ
C(s) + H2O(g) -> CO(g) + H2(g) Hr= 131.3 kJ
2CO(g) + O2(g) -> 2CO2(g) Hc= -566.0 kJ
CO(g) + H2(g) + O2(g) -> CO2(g) + H2O(g) Hr= -524.8 kJ

This question is really complex and I do not know which equations to switch. I don't really know where to start. Could someone please offer some tips on how to do this question and others?

 

[chem_tr]

Add all reactions to one another to obtain the final (desired) one. Remember that switching a reaction from left to right also changes its sign; if it is negative, treat this as a positive one.

 

[nobb]

Yes I know that. I am confused because I do not know which ones to switch. I tried this question many times and I still can't get the answer

 

[chem_tr]

Switch #5 and add with #6. You'll find a value about 208 kJ.

 

[Gokul43201]

Actually, they have provided more reactions than you need. That's what makes it look complex, while in fact, the problem can be solved easily using only 3 of the above equations.

2C(s) + O2(g) -> 2CO(g) Hc= -221.0 kJ
CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g) Hc= -802.7 kJ
CO(g) + H2O(g) -> CO2(g) + H2(g) Hr= -41.2 kJ
2H2(g) + O2(g) -> 2H2O(g) Hc= -483.6 kJ
C(s) + 2H2(g) -> CH4(g) Hf= -74.4 kJ
C(s) + H2O(g) -> CO(g) + H2(g) Hr= 131.3 kJ
2CO(g) + O2(g) -> 2CO2(g) Hc= -566.0 kJ
CO(g) + H2(g) + O2(g) -> CO2(g) + H2O(g) Hr= -524.8 kJ

Remember, there are probably several ways to solve this using any number of the given equations. It may even be possible to use them all, but I'm certainly not going to try that when I've found a simpler way.

Look at the bolded equations and figure out how to manipulate them to get the desired result.

Having done this, try to come up with at least one other alternative route (there's one using only 2 equations). If you do these two things, you'll have little trouble with such problems in the future.

Edit : chem_tr has given away the simpler solution too, so try and look for a third one : I get 205.7 kJ, both ways.

 

[nobb]

Ok thanks. That makes much more sense now.