Question

Question #d6b18

Answer 1

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`

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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!

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