a. You are going from 3 moles of gas to 2 moles of gas and 3 moles of solid. So, you are decreasing the number of moles of gas particles, so the entropy would decrease, giving a ΔS < 0.
b. Here, you're going from 2 moles of gas to 3, so ΔS would probably be positive.
c. Here the number of moles of gas is the same on both sides, but the total number of particles is increasing in the reaction so I would guess that ΔS would be slightly positive.
If this is not helpful then just tell me. I will try my best to explain it in another way. Also, are there any answer choices to go along with this question?
Can anyone help me with these three questions?
Predict the sign of delta S° and then calculate delta S° for each of the following reactions
a. 2H2S(g) + SO2(g) --> 3S(rhombic)(s) + 2H2O(g)
b. 2SO3(g) --> 2SO2(g) +O2(g)
c. Fe2O3(s) + 3H2(g) --> 2Fe(s) + 3H2O(g)
Thank you! I appreciate your help!
3 answers
I would just like to know how to calculate the delta S. And no, there are no answer choices with the question. However, thank you for taking the time to help me.
To calculate dS for each reaction, it is
dSrxn = (n*dSo products) - (n*dSo reactants)
You look up the dSo values in your text or notes.
dSrxn = (n*dSo products) - (n*dSo reactants)
You look up the dSo values in your text or notes.
1 answer