What is the enthalpy change for the following reaction #2SO_(2(g)) + O_(2(g))->2SO_(3(g))#?

1 Answer
Stefan V.
Jul 4, 2015

The enthalpy change of reaction is -197.7 kJ.

Explanation:

The easiest way to determine the enthalpy change of reaction is to use the standard enthalpy change of formation, #DeltaH_f^0#, of each compound that takes part in the reaction.

The equation that links the enthalpy change of reaction and the standard enthapy changes of formation looks like this

#DeltaH_"rxn" = sum(n * DeltaH_"f products"^0) - sum(m * DeltaH_"f reactants"^0)#, where

#n#, #m# - the stoichiometric coefficients (number of moles) of the products and of the reactants, respectively.

The standard enthaply change of formation for your compounds are

https://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_(data_table)

#DeltaH_("f "SO_2)^0 = "-296.84 kJ/mol"#

#DeltaH_("f "SO_3)^0 = "-395.7 kJ/mol"#

#DeltaH_("f "O_2)^0 = "0 kJ/mol"#

So, the enthalpy change of reaction will be

#DeltaH_"rxn" = [2cancel("moles") * (-395.7 "kJ"/cancel9"mol")] - [2cancel("moles") * -(296.84 "kJ"/cancel("mol")) + 1cancel("mole") * 0"kJ"/cancel("mol")]#

#DeltaH_"rxn" = -"791.4 kJ" + "593.68 kJ" = color(green)(-"197.7 kJ")#

Related questions
See all questions in Enthalpy
Impact of this question
35942 views around the world
You can reuse this answer
Creative Commons License