00:06
So this question wants us to calculate the dltg of evaporation of methanol, given that methanol evaporates at 5 degrees celsius.
00:21
So methanol is c3h whole h.
00:26
And now if we have the liquid state going to the gaseous stage, we have ch3 whole h gas.
00:35
Okay.
00:35
Okay, so we need to calculate the delta g of the reaction and we know the delta g standard of reaction would then be equal to summation n which is a number of mole of products multiplied by the delta g of formation of products, product as products, minus summation and number of mole of reactants multiplied by delta g's standard of formation of reactants.
01:11
Okay.
01:12
So in this case, our product is the methanol gas.
01:17
So we have one mole of it, put that one more.
01:23
One more of it multiplied by the delta g of formation of methanol gas.
01:30
And the delta g of formation of methanol gas is minus 162 .3 kilojillo.
01:39
Jules should remove the summation sign and minus the reactant in this case is methanol liquid and we have one more of it multiplied by the deltag of formation of methanol liquid is minus 166 .6 .6 .6m .m.
02:16
And this is also kilojoules per mole.
02:20
So this would give us what? this will give us 4 .3 kilojoules and that is the g for the reaction.
02:34
So we are also told to calculate the delta g had various non -standard conditions.
02:42
The first non -standard condition is when we have the partial pressure of methanol to be equals to 150 .0 millimeter of mercury.
03:00
So how do we go about this? we need to understand that the the delta g of reaction and the delta g standard for reaction are related by the expression delta g reaction is equal to delta g standard of reaction plus rt link q.
03:24
Okay, we are q is opposed to the partial pressure of methanol gas.
03:34
This is methanol gas in this case.
03:38
Now, knowing this, we can go ahead and perform the calculation on that different non -standard conditions...