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 ?
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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
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