How would you use a standard enthalpies of formation table to determine the change in enthalpy for these reactions?

#2CO(g) + O_2(g) -> 2CO_2(g)#

#CH_4(g) + 2O_2(g) -> CO_2(g) + 2H_2O(l)#

#2H_2S(g) + 3O_2(g) -> 2H_2O(l) + 2SO_2(g)#

1 Answer
Al E.
Jun 28, 2017

#-566kJ#
#-999kJ#
#-1125kJ#

Explanation:

Our equation of interest to apply for this problem is:
#DeltaH_(rxn)^° = sum(n*p roducts)-sum(m* react ants)#

Where n and m are moles of substances. We'd subsequently look the values up, do the math, and voila, so in my text, here's what I get. The reason some reactants/products have 0 formation energy is because when the molecule is the most stable form, no energy is needed to form.

#2CO(g)+O_2(g) to 2CO_2(g)#

#DeltaH_(rxn)^° = (2*-393.5)-(0+2(-110.5)) = -566kJ#

#CH_4(g)+2O_2(g) to CO_2(g)+ 2H_2O(l)#

#DeltaH_(rxn)^° = (-393.5+2*-285.8)-(-74.1+2*0) = -999kJ#

#2H_2S(g) + 3O_2(g) to 2H_2O(l) + 2SO_2(g)#

#DeltaH_(rxn)^° = (2*-285.8+2*-296.9)-(2*-20.17+3*0) = -1125kJ#

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