Question #275b6

1 Answer
Apr 3, 2016

Answer:

Here's what I got.

Explanation:

The idea here is that you can calculate the standard enthalpy change of reaction, #DeltaH^@#, for the decomposition of potassium chlorate by using the standard enthlapies of formation*, #DeltaH_f^@#, of the products and of the reactant.

More specifically, you can use the fact that standard enthalpy of reaction is equal to the difference between the sum of the standard enthalpies of formation of the products and the sum of the enthalpies of formation of the reactant - think Hess' Law.

#color(blue)(|bar(ul(color(white)(a/a)DeltaH^@ = sum (n xx DeltaH_"f prod"^@) - sum (m xx DeltaH_"f react"^@)color(white)(a/a)|)))#

Here #n#, #m# represent the stoichiometric coefficients of the products and of the reactant, respectively, in the balanced chemical equation.

The standard enthlapies of formation of the chemical species involved in your reaction can be found here:

https://en.wikipedia.org/wiki/Standard_enthalpy_of_formation

So, you know that

#"For KClO"_3: " " DeltaH_f^@ = -"391.4 kJ mol"^(-1)#

#"For KCl: " color(white)(aaa)DeltaH_f^@ = -"436.68 kJ mol"^(-1)#

#"For O"_2: color(white)(aaaaa)DeltaH_f^@ = "0 kJ mol"^(-1)#

Now, it's important to keep in mind that standard enthalpies of formation represent the enthalpy change of reaction when one mole of a given compound is formed from its constituent elements in their most stable form under standard conditions.

The balanced chemical equation for your reaction looks like this

#color(purple)(2)"KClO"_ (3(s)) -> color(blue)(2)"KCl"_ ((2)) + 3"O"_(2(g))#

Notice that the reaction consumes #color(purple)(2)# moles of potassium chlorate, #"KClO"_3#, and forms #color(blue)(2)# moles of potassium chloride, #"KCl"#, and #3# moles of oxygen gas, #"O"_2#.

This means that the standard enthalpy change of reaction for this decomposition will be

#DeltaH^@ = (color(blue)(2) xx DeltaH_ ("f KCl")^@ + color(blue)(2) xx DeltaH_ ("f O"_ 2)^@) - (color(purple)(2) xx DeltaH_ ("f KClO"_ 3)^@)#

Plug in the values to find

#DeltaH^@ = [color(blue)(2)color(red)(cancel(color(black)("moles"))) * (-436.68"kJ"/color(red)(cancel(color(black)("mol")))) + 3color(red)(cancel(color(black)("moles"))) * 0"kJ"/color(red)(cancel(color(black)("mol")))] - [color(purple)(2)color(red)(cancel(color(black)("moles"))) * (-391.4"kJ"/color(red)(cancel(color(black)("mol"))))]#

which will get you

#DeltaH^@ = color(green)(|bar(ul(color(white)(a/a)-"90.56 kJ"color(white)(a/a)|)))#