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What is using these equations below calculate the enthalpy for the reaction : 2H2 (g) +...

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sweetskittles24 | Student, Grade 11 | Salutatorian

Posted October 25, 2012 at 10:29 PM via web

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What is using these equations below calculate the enthalpy for the reaction : 2H2 (g) + O2 (g) -> 2H2O (g) ΔH = ?

1. 2O (g) -> O2 (g)                    ΔH= -294 kj/mol

2. H2O (l) -> H2O (g)                 ΔH= 44 kj/mol

3. 2H (g) + O (g) -> H2O (g)      ΔH= -803 kj/mol

4. C (gr.) + 2O (g) -> CO2 (g)    ΔH= -643 kj/mol

5. C (gr.) + O2(g) -> CO2 (g)     ΔH= -394 kj/mol

6. C (gr.) + 2H2 (g) -> CH4 (g)   ΔH= -75 kj/mol

7. 2H (g) -> H2 (g)                    ΔH= -436 kj/mol

8. H20 (l) -> H20 (g)                 ΔH= 10.25 kj/mol

(gr.) stands for graphite. I am wondering if you can multiply an equation by the value zero because there is no CH4 (g) in the final equation ?

Tagged with chemistry, hess's law, science

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jerichorayel | College Teacher | (Level 1) Senior Educator

Posted October 26, 2012 at 1:17 PM (Answer #1)

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Can you try to check the values? I see the items 2 and 8 have the same equation but have different delta H values. Another thing, if we try to add the combustion of methane (CH4), we can probably solve the problem.

CH4(g)  + 2O2 (g) --->2 H2O (g) + CO2 (g)    delta H = -891 kJ/mol

 
If ever the equation I mentioned will be included, we can find the reaction using Hess's law.
 
4O (g)    --> 2O2 (g)                   ΔH x 2
4H2O (l) --> 4H2O (g)                 ΔH x 4
2H2O (g) --> 4H(g) + 2O (g)       ΔH x(-2)
4H (g) --> 2H2 (g)                      ΔH x 2
CO2(g) --> C(gr) + 2O(g)           ΔH x (-1)
CO2(g) --> C(gr) + O2(g)           ΔH x (-1)
2C(gr) + 4H2(g) --> 2CH4 (g)     ΔH x 2
2CH4(g) + 4O2 (g) --> 4H2O(l)   ΔH x 2
                                 +2CO2(g) 
 
-----------------------------------------------
2H2 + O2 ---> 2H2O
 
 

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