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)##

##=-285.83-635.1+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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