Have a question on predicting entropy signs, in the reaction for combustion of ethane,
2C2H6 (g) + 7O2(g) --> 4CO2 (g) + 6H2O(l)
To deduce the whether entropy is increasing or decreasing, should we say
1) entropy is decreasing as , entropy of gases >> liquids > solids, and we have less molecules of gases (9 --> 4) in the end.
2) entropy is increasing as we have more molecules of substances ( 9--> 10) in the end?
Originally posted by atomos:Have a question on predicting entropy signs, in the reaction for combustion of ethane,
2C2H6 (g) + 7O2(g) --> 4CO2 (g) + 6H2O(l)
To deduce the whether entropy is increasing or decreasing, should we say
1) entropy is decreasing as , entropy of gases >> liquids > solids, and we have less molecules of gases (9 --> 4) in the end.
2) entropy is increasing as we have more molecules of substances ( 9--> 10) in the end?
Yes, that's correct. Gases have much greater entropy compared to liquids or solids. Hence, for most questions, it'll suffice to compare number of moles of gases on LHS vs RHS. In your qn above, moles of gas has been reduced from 9 (on LHS) to 4 (on RHS), and therefore it can be deduced that entropy change is negative.
This example problem demonstrates how to examine the reactants and products to predict the sign of the change in entropy of a reaction. Knowing if the change in entropy should be positive or negative is a useful tool to check your work on problems involving changes in entropy. It is easy to lose a sign during thermochemistry homework problems.
http://chemistry.about.com/od/workedchemistryproblems/a/Entropy-Change-Example-Problem.htm
Thanks all for the reply. This was actually one part of a larger question.
The combustion is feasible at 25 degrees celcius.
The next part would be where I had difficulty in:
Question: State and explain how the sign and magnitude would be affected by increase in temp.
Approach 1
If delta S was deduced to be negative, an increase in temp will cause delta G to be less negative (or even positive at some stage)?
Approach 2
Whereas if S was deduced to be positive, an increase in temperature will simply cause delta G to be always negative.
Approach 1 or approach 2?
Intuitively, the 2nd option allows a "neater conclusion", but it meant I had to assume delta S was positive instead of negative.
Originally posted by bycai:This example problem demonstrates how to examine the reactants and products to predict the sign of the change in entropy of a reaction. Knowing if the change in entropy should be positive or negative is a useful tool to check your work on problems involving changes in entropy. It is easy to lose a sign during thermochemistry homework problems.
http://chemistry.about.com/od/workedchemistryproblems/a/Entropy-Change-Example-Problem.htm
Thanks for the link. I tried looking around, but couldn't find questions that invoved
a decrease in gas molecules but an overall increase in all molecules. Without acutual S values to calculate, that posed a dilemma to me.
PAP : "What to do? It's already happened. Let's move on."
When you say that RHS has more moles of gases, are you viewing the H2O as gaseous states?
Cause if I based it on the equation, the state symbol of H2O was reflected as liquid. So RHS moles of gas is less than LHS moles of gases in this case?
Originally posted by atomos:When you say that RHS has more moles of gases, are you viewing the H2O as gaseous states?
Cause if I based it on the equation, the state symbol of H2O was reflected as liquid. So RHS moles of gas is less than LHS moles of gases in this case?
Yes, you're right, my error. Based on the reaction occuring under standard conditions, the gaseous water vapour would condense back into liquid state. So under standard conditions, there is indeed an overall negative entropy change.
Use your "Approach 1" answer. Tell your student to write this answer in the exams.
For your "Approach 2" answer, you said :
Whereas if S was deduced to be positive, an increase in temperature will simply cause delta G to be always negative.
Based on the negative entropy change for the reaction under standard conditions, there is no reason to deduce entropy change is positive, and therefore no reason to expect the reaction to be feasible at higher temperatures. The higher the temperature, the greater the effect of entropy on feasibility, whether it's a positive or negative effect.
Ethene is reacted with aqueous HBr. What would be the observation? I think it will be reddish brown to colorless as that case of Br2..
Originally posted by hoay:Ethene is reacted with aqueous HBr. What would be the observation? I think it will be reddish brown to colorless as that case of Br2..
Ethene, hydrobromic acid, and ethyl bromide are all colorless species. Hence, there would not be a color change. Only visible reaction, would be the ethene gas dissolving into the aqueous HBr, forming two immiscible layers : water and ethyl bromide.