# Calculate the change in entropy for the system and the surroundings

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1. May 2, 2017

### dancingdodo27

Calculate the change in entropy for the system, the surroundings and the Universe if 2 moles of ethane are completely combusted at 298 K. Standard entropies of C2H5(g), O2(g), H2O(l) and CO2(g) are 229.6, 205.1, 69.9, 213.7 J mol-1 K-1 , respectively. Standard enthalpy changes of formation of C2H5(g), H2O(l) and CO2(g) are -83.8, -285.8 and -393.5 kJ mol-1 , respectively.

I know that ΔS= Σproduct -Σreactants and that ΔS(uni)=ΔS(sys)+ΔS(surr)=0. But the question is worth 12.5 marks and I haven't even used the enthalpy values or moles.

Are those enthalpy values there to throw me off or am I missing something?

Any help is greatly appreciated and thanks in advance!

2. May 2, 2017

### Staff: Mentor

Let's see what you have done so far.

3. May 2, 2017

### dancingdodo27

2C2H6 +7O2→4CO2+6H2O

ΔS(sys)= 1274.2-1894.9=-620.7JK-1mol-1

ΔS(surr)= -ΔH/T

ΔH=-3288.8--167.6= -31121.2kJmol-1 = -3121200Jmol-1
∴ΔS(surr)=-3121200/298= -10473.8JK-1mol-1

ΔS(uni)=ΔS(sys)+ΔS(surr)
∴-620.7+-10473.8 = -11094.5JK-1mol-1

That's what I think it is but I'm not sure

4. May 2, 2017

### Staff: Mentor

What is your understanding of the initial and final thermodynamic equilibrium states of (a) the system and of (b) the surroundings in this question, in terms of temperature, pressure, and composition?

5. May 2, 2017

### dancingdodo27

All I know is that they're under standard conditions so I'm assuming its isothermal and isobaric

6. May 2, 2017

### Staff: Mentor

Are the reactants separate at 1 atm in the initial state and the products separate at 1 atm in the final state? That is, are the initial and final states truly the standard states?

7. May 2, 2017

### dancingdodo27

I don't really know

8. May 2, 2017

### Staff: Mentor

Thank you for being so forthcoming. When determining the change in entropy (or any other thermodynamic function of state) for a process, it is critical to precisely identify the initial- and final thermodynamic equilibrium states. The vagueness of the present problem statement prevents this. The initial and final states need to be established for both the system and for the surroundings (which, in a way, can be treated as just another system). Maybe you can get some clarification on the initial and final states from whomever posed this problem.

Chet

9. May 3, 2017

### I like Serena

As explained here (see example problem 19.2.3 which is similar), the heat transferred to the system is $Q=\Delta H$ and $\Delta S_{surr} = \frac{-\Delta H}{T}$.

So that:
$$\Delta S_{sys} = -620.7\text{ J/K} \\ \Delta H = -3121.2 \text{ kJ} \\ \Delta S_{surr} = \frac{--3121.2 \cdot 10^3}{298} = 10474 \text{ J/K} \\ \Delta S_{uni} = \Delta S_{sys} + \Delta S_{surr} = 9853 \text{ J/K}$$
As expected the reaction is exothermic ($\Delta S_{surr} > 0$).
And we see that the second law (total entropy increases) is observed ($\Delta S_{uni} > 0$).

As for the initial and final conditions, for the regular chemical process of combusting ethane I believe we would initially and finally assume a mixture of gasses at standard ambient temperature and pressure. The process would indeed be modelled as isobaric and isothermic.

10. May 3, 2017

### dancingdodo27

Thank you so much! That is huge help!

11. May 3, 2017

### Staff: Mentor

Your interpretation of the process and the initial and final states of the system makes a lot of sense. But, if that is the case, then the determination of the entropy change of the system using the standard entropies is incorrect. The standard entropies are based on the pure substances, each at 1 atm, and, in this problem, we are starting and ending with mixtures at 1 atm. So the entropy change has to be corrected for the entropy of mixing of the reactants (since liquid water is one of the products, the entropy of mixing of the products does not have to be included). As far as a correction to the heat of reaction is concerned, if we are assuming ideal gas behavior, no correction is needed, because the heat of mixing of ideal gases is zero.

12. May 3, 2017

### Staff: Mentor

Hi guys. I have a modified version of this problem that you might find interesting to consider.

The initial state of the system is 2 moles of ethane and 7 moles of O2 in separate containers at 298 K and 1 atm.

The final state is 4 moles of CO2 and 6 moles of H2O (l) in separate containers at 298 K and 1 atm.

I devise a reversible process path to transition from the initial state to the final state, and a corresponding reversible path is followed by the the surroundings. The gases are modeled as ideal gases.

What is the change in entropy of the system?
What is the change in entropy of the surroundings?
What is the change in entropy of the universe?
Devise a reversible process from the initial state to the final state.
If the initial and final state pressures are the same, why isn't the change in entropy of the surroundings equal to minus the standard change in enthalpy of the reaction divided by the temperature?
If the intial and final state pressures are the same, why isn't the standard change in entropy of the system equal to the change in enthalpy of the reaction divided by the temperature?

This is not a simple problem.

13. May 5, 2017

### I like Serena

Ok. This is as far as I got.

Process would be:
1. The reactant gasses mix, which is an irreversible isothermic expansion of ideal gasses with $\Delta S_i = -n R x_i \ln x_i$, where $n$ is the total number of moles, and $x_i$ is the fraction of moles of each reactant. No exchange of heat between the reactants or the surrounding. The entropy of the system and the universe increase. ($\Delta S_{sys} = \Delta S_{uni} = 16.9\text{ J/K}, \Delta S_{surr} = 0$).
2. Reversible chemical reaction with $\Delta S_{surr} = -\Delta H / T$, where $\Delta H$ is determined based on the other process steps.
3. Phase separation of $CO_2(g)$ and $H_2O(l)$. Still trying to figure out the entropy change. It should be spontaneous shouldn't it?
4. Separate the containers, which should be trivial since the liquid water should already be separated by gravity.

Since step 1 is irreversible, we can't devise a reversible process can we?

Edit: Oh wait. Suppose we don't let the gasses mix, but let them seep through semi permeable membranes into a small reaction chamber to combust. Then we wouldn't lose the entropy would we?
And then we can use more semi permeable membranes, draining off the water and the carbon dioxide separately. Would that work?

I guess we could also use a regular gas burner filled with ethane, carefully adding oxygen to it, catching the resulting carbon dioxide, and drain off the resulting water.

Last edited: May 5, 2017
14. May 5, 2017

### Staff: Mentor

Very nice. Your semipermeable membrane approach was the right idea. The inlets and outlets of a van't Hoff equilibrium box have semipermeable membranes that can be used to introduce the pure reactants (into a large mixture already at equilibrium) and to remove pure products. The pure reactants being introduced must be at the same partial pressures as in the mixture within the box, and the same goes for the pure products. The van't Hoff equilibrium box operates reversibly. So, the only other thing you need to do is to expand the reactants isothermally to their partial pressures in the box, and compress the products from their partial pressures in the box to 1 atm.

It was very perceptive of you to think of using semi permeable membranes.

PS, see my Physics Forums Challenge Problem