How would you calculate the standard enthalpy change for the reaction 2A + B = 2C + 2D, if A = -269 B = -411 C = 189 D = -481?

1 Answer
Oct 26, 2015

Answer:

#DeltaH_"rxn" = +"365 kJ"#

Explanation:

Start by writing down the reaction

#2"A" + "B" -> 2"C" + 2"D"#

Now, I assume that the values given to you represent the respective standard enthalpies of formation for the species that take part in the reaction.

Standard enthalpies of formation are measured when one mole of each of these compounds is formed from its constituent elements in their pure state.

As a result, it is expressed in kilojoules per mole.

This is a very important thing to keep in mind - standard enthalpies of formation are always given in #"kJ/mol"#!

In your case, you would have

  • #"A" -> -"269 kJ/mol"#
  • #"B" -> -"411 kJ/mol"#
  • #"C" -> +"189 kJ/mol"#
  • #"D" -> -"481 kJ/mol"#

You can calculate the standard enthalpy change of reaction using standard enthalpies of formation by taking into account the fact that enthalpy changes are additive - this is known as Hess' Law.

In other words, you can express the enthlapy change of reaction by using each individual reaction that corresponds to the standard enthalpy change of formation for the products and for the reactants.

#color(blue)(DeltaH_"rxn"^@ = sum(n xx DeltaH_"f products"^@) - sum(m xx DeltaH_"f reactants"^@))#

Here #n#, #m# represent the stoichiometric coefficients of the compounds in the reaction.

So, considering that this reaction consumes 2 moles of #"A"# and one mole of #"B"#, and produces 2 moles of #"C"# and 2 moles of #"D"#, you will have

#DeltaH_"rxn"^@ = [2color(red)(cancel(color(black)("moles"))) xx (+189"kJ"/color(red)(cancel(color(black)("mole")))) + 2color(red)(cancel(color(black)("moles"))) xx (-481"kJ"/color(red)(cancel(color(black)("mole"))))] - [2color(red)(cancel(color(black)("moles"))) xx (-269"kJ"/color(red)(cancel(color(black)("mole")))) + 1color(red)(cancel(color(black)("mole"))) xx (-411"kJ"/color(red)(cancel(color(black)("mole"))))]#

#DeltaH_"rxn"^@ = -"584 kJ" - (-"949 kJ") = color(green)(+"365 kJ")#

Check out this video on how standard enthalpies of formation are used to find the enthalpy change of formation for a given reaction