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### Midterm Q6B

Posted: Wed Feb 19, 2020 8:15 pm
For which process will ∆H^o and ∆G^o be expected to be most similar to 2H2(g) + 2Cl2(g) → 4 HCl(g), with ∆H^o = –92.3 kJ?

(A) 2Al(s) + Fe2O3(s) → 2Fe(s) + Al2O3(s)
(B) 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
(C) 2NO2(g) → N2O4(g)
(D) 2H2(g) + O2(g) → 2H2O(g)

Why is the answer A? How do we approach this problem without using ∆Hf or ∆Gf since they weren't in our constants sheet?

### Re: Midterm Q6B

Posted: Wed Feb 19, 2020 8:24 pm
For this question, you have to look at the equation ∆G= ∆H-T∆S. ∆G= ∆H when ∆S is approximately 0. In answer a., the change in entropy is approximately zero because there are no phase changes. The reactants and the products are both all solids.

### Re: Midterm Q6B

Posted: Wed Feb 19, 2020 8:30 pm
Remember that delta G = delta H - T*delta S, so the scenario where delta H and delta G are most similar would be the case where delta S is the smallest, so the reaction where the change in entropy is the smallest. The reason it is A is because this reaction has no phase change, hence a very low reaction change in entropy. Question B has the production of a gas, and questions C and D both have a decrease in the moles of gas, meaning a negative change in entropy. Thus, the one that has the least change in entropy is reaction A.

### Re: Midterm Q6B

Posted: Wed Feb 19, 2020 10:58 pm
From the equation ∆G= ∆H-T∆S. ∆G= ∆H when the second term is close to 0. Therefore, you should find an equation with very little entropy change, which is often when the reactants and products are in the same state with the same amount of moles (best when all solid).

### Re: Midterm Q6B

Posted: Thu Feb 20, 2020 12:04 am
Since DeltaG=DeltaH-T*DeltaS, in order to have a nearly equivalent DeltaG and DeltaH means an insignificant DeltaS (disorder). Thus, A has 3 solids from reactant to 3 solids from the product, indicates DeltaS of the reaction is pretty much zero. Assuming constant temperature for the four reactions, DeltaH and DeltaG of A are the most similar.

### Re: Midterm Q6B

Posted: Sun Mar 15, 2020 7:08 am
The answer is A because we first realize that chnage in gibbs free energy is equal to the change in enthalpy subracted by the product of temperature and the change in entropy or, ∆G= ∆H-T∆S. We also know that ∆G= ∆H when ∆S is approximately 0. Because answer choice has reactants and products all in the same phase (solid) it will yield the right answer.