For the reaction #4"XY"_3 + 7"Z"_2 -> 6"Y"_2"Z" + 4"XZ"_2#, what is the enthalpy change? The steps are shown below.
#"X"_2 + 3"Y"_2 -> 2"XY"_3# , #DeltaH_1 = -"320 kJ"#
#"X"_2 + 2"Z"_2 -> 2"XZ"_2# , #DeltaH_2 = -"140 kJ"#
#2"Y"_2 + "Z"_2 -> 2"Y"_2"Z"# , #DeltaH_3 = -"230 kJ"#
1 Answer
#DeltaH_(rxn) = DeltaH_1' + DeltaH_2' + DeltaH_3' = ???# (
#DeltaH_i' ne DeltaH_i# .)
This is asking you to apply Hess's law in a VERY general setting. The basic ideas are:
- Multiply reaction step
#i# by a constant#-># multiply#DeltaH_i# by that constant. - Reverse reaction step
#i# #-># switch the sign of#DeltaH_i# . - Anything that matches on the reactants and products side of any reaction steps in the mechanism cancels out.
- You must achieve the overall reaction when adding up the individual steps.
In general, this will guarantee you get the correct answer:
- Look for all intermediates and catalysts, and plan your actions out so that they will cancel out.
- Use the overall reaction as a guide, as you must get the right number of products and reactants on each side.
- Apply the above methods to flip reactions, scale them, etc., so that all intermediates and catalysts cancel out.
So, for this reaction, I will demonstrate the above by blindly looking to produce the correct product and reactant coefficients:
- Multiply step
#1# by#2# and reverse it. That brings#4XY_3# to the left side. - Multiply step
#2# by#2# . That allows#4XZ_2# to be produced. - Multiply step
#3# by#3# . That allows#6Y_2Z# to be produced.
#-2(X_2 + 3Y_2 -> 2XY_3)# ,#-2(DeltaH_1 = -"320 kJ")#
#2(X_2 + 2Z_2 -> 2XZ_2)# ,#2(DeltaH_2 = -"140 kJ")#
#3(2Y_2 + Z_2 -> 2Y_2Z)# ,#3(DeltaH_3 = -"230 kJ")#
Now apply the coefficients. A negative coefficient takes into account reversing the reaction.
Note that I foolishly did not look for what were intermediates or catalysts... and yet... magic. If this doesn't work, then there is a typo in the reaction steps!
#4XY_3 -> cancel(2X_2) + cancel(6Y_2)# ,#DeltaH_1' = +"640 kJ"#
#cancel(2X_2) + 4Z_2 -> 4XZ_2# ,#DeltaH_2' = -"280 kJ"#
#cancel(6Y_2) + 3Z_2 -> 6Y_2Z# ,#DeltaH_3' = -"690 kJ"#
#"----------------------------------------------------"#
#4XY_3 + 7Z_2 -> 6Y_2Z + 4XZ_2# ,
#color(blue)(DeltaH_(rxn) = DeltaH_1' + DeltaH_2' + DeltaH_3' = ???)#
And just to see if you're paying attention, I'll let you add this up. Rather than give you the straight-up answer, I want you to go through the process, because this is best learned by doing.
