00:01
The enthalpy change of a reaction can be solved in terms of the entropy of formation of the product and the reactant as follows.
00:09
So the standard entropy of reaction is equal to the sum of the antelps of formation of all the products minus the sum of the entity of formation of all the reactants.
00:28
So in this problem, we're given with a combustion reaction of liquid c2h4.
00:36
That's acetyl the head.
00:38
So if it's a combustion reaction, then this is its reaction with oxygen gas.
00:44
And since the compound is made up of carbon and hydrogen, the products of which are carbon dioxide and water.
00:52
So this is the chemical equation for the, unbalanced chemical equation for the given combustion reaction.
00:59
The end of this formation are as follows.
01:02
For co2, you have negative 393 .5.
01:08
For h2o negative 285 .8 and the unit is kilojews per mole so this is these are the delta h standard antelps of formation what we want to find is the standard antelope of formation of acetyl deide if the entropy of the reaction specifically called as the entropy of combustion is equal to negative one 167 kilo juice per mole so we'll start by establishing the balance chemical equation.
01:46
So we have to make sure that both sides of the equation contain the same number of each element.
01:51
So let's write first year c2h4o plus o2, forming co2 plus h2o...