# 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.