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How to calculate standard enthalpy formation of Ca(OH)2 from the following data? 2 H2 (g) + 1/2O2> 2H20 (l) Delta H1 = 571.66 kJ mol ^1 CaO (s) + H2O (l) > Ca(OH)2 (s) Delta H2= 65.17 kJ mol ^1 2Ca(s) + O2 (g) > 2 CaO (s) Delta H3= 127...
We want the standard of formation for ##Ca(OH)_2##. Thus, our required equation is the equation where all the constituent 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)##
##=285.83635.1+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.