Question #d6b18

1 Answer
Tanish J.
Nov 29, 2014

We want the standard enthalpy of formation for #Ca(OH)_2#. Thus, our required equation is the equation where all the constituent elements combine to form the compound, i.e.:

#Ca +H_2+O_2->Ca(OH)_2#

Let us now write down the given equations:

[The first equation mentioned is incorrect, and so I have revised it.]

#(1)# # 2H_2 (g) + O_2(g)->2H_2O (l)# and #DeltaH_1=-571.66 kJmol^-1#

#(2)# #CaO (s) + H_2O (l) -> Ca(OH)_2 (s)# and #DeltaH_2=-65.17 kJmol^-1#

#(3)# #2Ca(s)+O_2(g)->2CaO(s)# and #DeltaH_3=-1270.2 kJmol^-1#

Now, our aim is to use the 4 operators of mathematics (multiplication, division, addition and subtraction) to get the required equation. An easy way to do this is:

Step 1. Look for elements, other than #O_2#, (that are present in the required equation) in the given equations.

We find that #Ca# in is #(3)#, #H_2# is in #(1)# and #Ca(OH)_2# is in #(2)#.

Step 2. Multiply or divide given equations to make the amounts of the elements the same as those in the required equation. That is,

  • In the required equation, there is one atom of #Ca# but in equation #(3)#, there are two atoms of #Ca#. Thus, we must divide the equation #(3)# by 2. This will affect the enthalpy of reaction as well, which will also be divided by 2.

  • In the required equation, there is one molecule of #H_2# but in equation #(1)#, there are two molecules of #H_2#. Thus, we must divide the equation #(1)# by 2. The enthalpy of reaction will also be divided by 2.

  • In the required equation as well as #(2)#, there is an equal number of #Ca(OH)_2# molecules. However, in the required equation, the molecule lies on the products side, while in #(2)#, it lies on the reactants side. So, we must reverse equation #(2)#. This would mean that the sign of enthalpy of reaction of #(2)# will be changed.
    (#+ -> -# [or] #- -> +#)

We can now simply add the three new values of enthalpies that we have calculated!

So, enthalpy of formation of #Ca(OH)_2# is

#DeltaH_f##=(DeltaH_1)/2+(DeltaH_3)/2+(-DeltaH_2)#

#=(-571.66)/2+(-1270.2)/2-(-65.17)#

#=-285.83-635.1+65.17#

#=-855.76 kJmol^-1#

What I have essentially done is that I have manipulated and added the equations to get the required equation.
Try adding the manipulated equations of Step 2 and see what you get!

I know that this is rather long and complex, so I hope this will help you understand better.

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