## midterm 6 b

$\Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}$

$\Delta G^{\circ}= -RT\ln K$

$\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$

WGaines_2E
Posts: 47
Joined: Mon Nov 18, 2019 12:18 am

### midterm 6 b

For which process will dH and dG be expected to be the most similar?

a)2Al(s) + Fe2O3(s) -> 2Fe(s) +2Al2O3(s)
b) 2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2(g)
c) 2NO2(g) -> N2O4(g)
d) 2H2(g) + O2(g) -> 2H2O(g)

Kylie Lim 4G
Posts: 110
Joined: Sat Aug 17, 2019 12:15 am

### Re: midterm 6 b

A, because the reaction has the smallest deltaS values, meaning in the equation deltaG = deltaH - TdeltaS, change in gibbs and change in enthalpy will be the closest to each other

anjali41
Posts: 109
Joined: Fri Aug 09, 2019 12:15 am

### Re: midterm 6 b

The states of matter hint at the answer, as all the products and reactants in answer choice a are solids.

CynthiaLy4F
Posts: 103
Joined: Sat Jul 20, 2019 12:16 am

### Re: midterm 6 b

For this problem, the answer would be the reaction that has the smallest change in entropy, which in this case is A.

John Liang 2I
Posts: 102
Joined: Fri Aug 30, 2019 12:18 am

### Re: midterm 6 b

since dG=dH-TdS, you want to find the smallest dS value. we see this in the first choice because the solid states remain most solid. hope this helps

AniP_2D
Posts: 95
Joined: Sat Aug 17, 2019 12:17 am

### Re: midterm 6 b

For deltaH and deltaG to be very similar, the change in entropy has to be minimal, so you would look to see which reaction has the least change in entropy, which would be A since the phase remains a solid throughout the reaction.

Rida Ismail 2E
Posts: 139
Joined: Sat Sep 07, 2019 12:16 am

### Re: midterm 6 b

In order to solve this problem, you need to see which choice gives you an answer that has the smallest delta S. A would be the answer because that is the only one in which the state stays solid throughout, and solids have the lowest entropy values.