## Question 4J.7

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

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

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

VioletKo3F
Posts: 103
Joined: Sat Sep 07, 2019 12:18 am

### Question 4J.7

When calculating the enthalpy, why are only some reactants/products used to calculate it? For example, for part (a), you only use the enthalpy of formation of H2O and H2O2, and don't include O2.

Jordan Young 2J
Posts: 102
Joined: Thu Jul 25, 2019 12:17 am
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### Re: Question 4J.7

Enthalpy of formation of O2 is zero because it's in its most stable form of oxygen

VioletKo3F
Posts: 103
Joined: Sat Sep 07, 2019 12:18 am

### Re: Question 4J.7

Jordan Young 2J wrote:Enthalpy of formation of O2 is zero because it's in its most stable form of oxygen

How would you know that it's the most stable form? Would you have to memorize it?

Uisa_Manumaleuna_3E
Posts: 60
Joined: Wed Sep 21, 2016 2:56 pm

### Re: Question 4J.7

VioletKo3F wrote:
Jordan Young 2J wrote:Enthalpy of formation of O2 is zero because it's in its most stable form of oxygen

How would you know that it's the most stable form? Would you have to memorize it?

Yeah, it's kind of a memorization thing. It's a gas, bonded between two identical atoms. And O2 as a gas is literally one of the simplest elements to be found in nature, easily diffusing in blood cells, lungs, through cell membranes. Also, another tip, if you try looking up the enthalpy of formation of O2 on the constants sheet they give us on the tests, it doesn't exist. So you can just assume its zero and as such is in is most stable form.

Ruth Glauber 1C
Posts: 100
Joined: Wed Sep 18, 2019 12:20 am

### Re: Question 4J.7

Enthalpy of formation of O2 is 0 at its most stable form.

Eva Zhao 4I
Posts: 101
Joined: Sun Sep 29, 2019 12:16 am

### Re: Question 4J.7

Just to list a few more, H2, Cl2, Br2, and C(s, graphite) also have enthalpies of formation equal to 0.

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