Question #275b6
1 Answer
Here's what I got.
Explanation:
The idea here is that you can calculate the standard enthalpy change of reaction,
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
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
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)|)))#