00:01
The δ s of the surroundings is equal to negative δ h of the system divided by t.
00:14
So if δ h is negative, δ s surroundings is positive and vice versa.
00:20
So for the first reaction, we don't have to perform any calculations.
00:25
We just recognize that δ h is negative, so δ s surroundings is positive.
00:30
The reaction is 2 co gas reacts with o2 gas to produce 2 co2 gas and that's it.
00:45
Δ h is negative 566.
00:51
So, based upon the equation above, δ s surroundings is going to be positive.
01:05
And what about δ s system? well, δ s system is determined, its sign is determined, based upon the change in moles of gas.
01:14
We're going from 3 moles of gas to 2 moles of gas.
01:18
Decreasing moles of gas means we decrease the entropy of the system.
01:22
So it's negative.
01:25
Then for the second one, so i guess we would say, we'll use the notation they did, they have.
01:32
This is greater than zero and δ s system will be less than zero...