How would you calculate the standard enthalpy change for the following reaction at 25 °C: H2O (g) + C (graphite)(s) --> H2 (g) + CO (g)?

1 Answer
Oct 26, 2015

Answer:

#-"131.3 kJ"#

Explanation:

You can calculate the standard enthalpy change of reaction by using the standard enthalpies of formation of the species that take part in the reaction, i.e. the reactants and the products.

You can find the standard enthalpies of formation, #DeltaH_"f"^@#, here:

https://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_%28data_table%29

In your case, you would have

  • #"For H"_2"O"_text((g]):" " -"241.82 kJ/mol"#
  • #"For C"_text((s]):" " "0 kJ/mol"#
  • #"For H"_2:" " "0 kJ/mol"#
  • #"For CO: " -"110.53 kJ/mol"#

To find the standard enthalpy change of reaction, #DeltaH_"rxn"^@#, using the standard enthalpies of formation, you need to take into account the fact that, for a given chemical reaction, the change in enthalpy is independent of the pathway taken.

In other words, you can go from these two reactants, water vapor and graphite, to those products, hydrogen gas and carbon monoxide, by using the reactions that describe the formation of each of these compounds.

SInce the enthalpy changes for those respective reactions are the standard enthalpies of formation, it follows that you can say

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

Here #n# and #m# represent the stoichiometric coefficients of the products and of the reactants, respectively.

As you can see, standard enthalpies of formation are given per mole, so you need to take into account how many moles of each compounds you have.

In this case, the reaction consumes one mole of water and one mole of graphite, and produce one mole of hydrogen gas and one mole of carbon monoxide.

Therefore, you have

#DeltaH_"rxn"^@ = [1color(red)(cancel(color(black)("mole"))) * (-241.82"kJ"/color(red)(cancel(color(black)("mole")))) + 1color(red)(cancel(color(black)("mole"))) * 0"kJ"/color(red)(cancel(color(black)("mole"))))] - [1color(red)(cancel(color(black)("mole"))) * 0"kJ"/color(red)(cancel(color(black)("mole"))) + 1color(red)(cancel(color(black)("mole"))) * (-110.53"kJ"/color(red)(cancel(color(black)("mole"))))]#

#DeltaH_"rxn"^@ = -"241.82 kJ" - (-"110.53 kJ") = color(green)(-"131.3 kJ")#