00:03
All right, here it is given the thermochemical equations, right? so let is just write that it is given 2fe solid plus 6h2o in liquid form that gives us 2 feoh goldquoise.
00:23
Gold thrice you can say in solid state, right, plus 3h2 gas, right? and now delh equals to plus 32 kilojoules right and secondly it is given f a 203 plus this is in solid form right plus 3h2o that gives 2 f e o h whole thrice in solid form right so now here delh equals to plus 289 kilojones right now thirdly it is given 2 h2 plus o2 gas right so this gives us 2 h2 in liquid form so now the enthalpy for this equation must be equals to delh equals to plus 572 kilojoules right now what you could do is you need to find the enthalpy for the oxidation of ion, right? we need to find from these three equations, we want fe to fe solid, right, plus 3 by 2 o2.
02:14
This is in gaseous form.
02:16
This gives us fe2 o3.
02:19
Right.
02:20
So now we need to find the enthalpy for this.
02:23
Now what you could do is you, you could just observe it right first of all the equation must be we want two fe here right so you can get fe from here right so simply write this equation number one right so we will add with two fe something right what we want to add 3 by 2 o2 right 3 by 2 o2 from where we will be getting 3 by 2 o2 because out of these two three equations we are getting only o2 here right so if we are getting o2 here that means this equation must be multiplied by 3 by 2 right and enthalpy will also be changed according to this multiplication or division form right like if it has 572 kilojoules now it becomes 3 by 2 of 572 after multiplying this equation by 3 by 2.
03:28
Now, what it gives is this is equation number 2 now and this is equation number 3.
03:34
Right? this is what we are doing...