posted by Bru .
calculate the standard molar entropy change for the combustion of methane gas using S naught values from standard thermodynamic tables. Assume that liquid water is one of the products.
CH4(g) + 2O2(g) ==> CO2(g) + H2O(l)
dSo rxn = (n*dSo poducts) - (dSo reactants)
I did this and the database said the answer was wrong:
(213.8 + 70.0) - (186.2 + 2*205.0)= -312.4 j/K*mol. Am I missing something?
I'm sorry. I didn't balance the equation. I should have written (and you should have checked)
CH4 + 2O2 ==> CO2 + 2H2O(l)
Try that and see if you have better luck. Thanks for letting me know. I almost always check the equations; this time I didn't and it caught me.