Question

Consider the following reaction, which is carried out at 300 K:
CH3CH2OH + 3O2 → 2CO2 + 3H2O ∆H = –1,235 kJ/mol and ∆S = +0.22kJ/mol•K
• What is the value of ∆G? Show your work.
• Classify the reaction as spontaneous or nonspontaneous.
• Suggest why the value for ∆S is positive in this reaction.

Answer 1

1)
\[\Large \Delta G = \Delta H - T \Delta S\]
2)
Is the \(\Delta G\) negative or positive?

3) Any good suggestion what might be the answer? try look at how much is being created from how much is being used. [Entropy being a measure for order and disorder OR the distribution of stuff]

Answer 2

We can go over each one if you like, but I want you to solve the problem.

Answer 3

yes please

Answer 4

Okay. Lets look at the first one.
The most known equation for Gibbs free energy at constant pressure at temperature \(T\) is\[\Large \Delta G = \Delta H- T \Delta S\]
Here are, \(\Delta G\) change in Gibbs free energy, \(\Delta H\) the change in enthalpy and \(\Delta S\) the change in entropy.

As the information we have been give is all at the same temperature we can calculate the change in Gibbs free energy at 300 K.


Try plug in the numbers given in the equation above and calculate the Gibbs free energy. (I check your answer when you post it)

Answer 5

so so i plug in 1.235 and .22 to the formual ?

Answer 6

Yes and the 300 at T

Answer 7

I got 205.7

Answer 8

Hmmm I think you have done something wrong.
Did you calculate the following?
\[\Large \Delta G=-1235 \frac{ kJ }{ mol }-300 ~ K \times 0.22 \frac{ kJ }{ mol \times K }\]

Answer 9

As you can see the overall result is going to be negative

Answer 10

Oh I did know it was negative. But this time I got -357.5

Answer 11

I don't see how you get that number. Lets break it up by parts:
We got from above that:
\[\Large \Delta G=-1235 \frac{ kJ }{ mol }-300 ~ K \times 0.22 \frac{ kJ }{ mol \times K }\]
And notice that
\[\Large \Delta H =-1235 \frac{ kJ }{ mol }\]
\[\Large T \Delta S=300 K \times 0.22 \frac{ kJ }{ K \times mol }= 66\frac{ kJ }{ mol }\]
\[\Large \Delta G=-1235 \frac{ kJ }{ mol }-66 \frac{ kJ }{ mol }=1301 \frac{ kJ }{ mol }\]
@aaronq please check... I'm getting tired and want to be sure I don't make a mistake.

Answer 12

Correction!
\[\Large \Delta G=-1301 \frac{ kJ }{ mol }\]

Answer 13

Oh okay i was mutliplying wrong

Answer 14

It happens :)
So the answer is (Just want you to write it)

Answer 15

-1301kJ/mol

Answer 16

Exactly. :) that is the answer to 1)

Answer 17

For number 2 we are going to use some knowledge about the Gibbs free energy. \(\Delta G\) is a among others a indicator for spontaneous change. The ability for the reaction to happen spontaneously is given weather \(\Delta G\) is positive or negative.

Do you know, or can you guess, if \(\Delta G\) should be positive or negative for a spontaneous reaction to occur?

Answer 18

I think it should be negative

Answer 19

You are thinking correctly! For a spontaneous reaction \(\Delta G<0\), and for a non-spontaneous reaction \(\Delta G>0\).


So is the reaction CH3CH2OH + 3O2 → 2CO2 + 3H2O spontaneous at 300 K?

Answer 20

its spontaneous right?

Answer 21

It sure is. Well done so far! :)

Now to the last one, you need to know what entropy (\(\Delta S\)) is. Do you have an idea what entropy is about?

Answer 22

No

Answer 23

I get it

Answer 24

Right sorry I deleted the answer, but I try make a general conclusion you should just remember (it will work until you study thermodynamics more closely):

"For a change from a state we call X into a state we call Y (donated \(X \to Y\)), the state with the most accessible energy levels / most accessible states, will be favorer by an increase in entropy."

Answer 25

So for your example it would be:

CH3CH2OH + 3O2 → 2CO2 + 3H2O

Note we got 4 particles on the reactant side, but 5 particles on the product side.
Using the general conclusion we can write \(4 X \to 5Y\)

Meaning we got a positive entropy.

Answer 26

So we are trying to change it from a negative to a positive?

Answer 27

Not really. Entropy is all about how many accessible states particles can take (vibrational, translational, rotational ect.). But for simplicity we ONLY focus on how many particles there are on each side of the reaction and assume that all particles contributes with exactly one state.