Discussion/ (a lot of questions attached) please pick an answer and justify why it's right along with why the others are wrong.

- Photon336

- jamiebookeater

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- Photon336

390/391

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- Photon336

392/393

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- Photon336

395/396/397

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- Photon336

398

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- Photon336

404/405/406

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- Photon336

Ok this probably all I will post for today.

- Photon336

- Photon336

Just pick a problem and go for it!

- taramgrant0543664

I'm going with 390 lol

- Rushwr

390 A

- Rushwr

sorry B

- taramgrant0543664

In order for a reaction to be spontaneous in the forward direction delta G has to be negative
G=H-TS (I can't draw deltas on here)
H has to be negative and S has to be positive to make sure G is negative so A

- taramgrant0543664

Yes B lol

- Rushwr

\[\Delta G= \Delta H- (T \Delta S)\]

- Rushwr

So when Delta S positive -(TS) becomes negative . for a spontaneous reaction delta G should be negative.

- taramgrant0543664

It is definitely B I looked up and saw A and I typed it lol I'm getting to tired for this ill be leaving soon

- Photon336

\[\Delta G = \Delta H - T \Delta S\]
We have to consider the effect on T delta S. because Tdelta S is negative. so if we have delta S that's positive t delta S will be < 0. and if Enthalpy dELTA s IS <0 THEN we can clearly see that (negative)-(negative) = - negative number. both signs are negative so that means our delta G < 0.
Delta H < 0 Delta S > 0
-T delta S < 0
Delta H > 0
this would be negative every time.

- Photon336

@taramgrant0543664 @Rushwr I agree. it's B.

- Photon336

the other's can have a negative delta G but it depends on how big delta S or delta H were. so it's not clear cut. the last one D. if delta H is greater than 0 AND DELTA S < 0 both of them would be positive hence non spontaneous. delta g > 0

- taramgrant0543664

For 391 I was thinking D if G needs to be negative to be spontaneous then if heating was involved then S would be positive so -TS can stay negative and H would be positive so it would be spontaneous at high temp and non spontaneous in the forward direction at low temps

- Photon336

@taramgrant0543664 Yeah. me too because you keep Delta H constant, and you increase temperature. that means that -TDELTAS is going up. ^_^

- taramgrant0543664

Yep ^^

- Rushwr

I think my brain is all frozen !

- Photon336

392. change in internal energy \[U = Q + W \] i'm not too good at these.
work is being done on the gas so work would be positive +1475J
Heat is being lost so that's -375
\[-375+1475j = 1,100 j \]
answer C

- taramgrant0543664

For 392 that makes sense but I didn't even remember how to start that one I haven't done this for like 9 months so it's been awhile lol

- Photon336

yeah
work being done on the system is positive
work being done by the system is negative because the system is doing work on surroundings.
heat flowing into the system is positive.
heat flowing out of the system is negative.

- taramgrant0543664

Ya it makes sense! I'm thinking I'm calling it a night for me I'll look at the rest tomorrow but it's almost 1 am here and stuff is starting to blend together is seems lol

- Photon336

yeah. I'm going to finish in about another 15 min lol

- Photon336

@Rushwr 396?

- Rushwr

for 395 it's C right?

- Photon336

do you learn this formula?
\[DeltaG = -RTln(k) \]

- Rushwr

since it is vapourization I think the london dispersion forces are broken. Not the covalent bonds of Carbon and Chlorine >

- Photon336

=D correct

- Rushwr

for 396 I think it is D cuz they seem to be unrelated/ G talks about spontanity while Equilibrium constant talks about the something else !

- Photon336

i guess it's the bonds between the molecules are london dispersion forces

- Photon336

for 396

- Photon336

do you remember this formula? or learn it
Delta G = -RTlnK

- Photon336

they told us that delta G was positive

- Photon336

the natural log (ln) of a number less than one will give you a negative number
the natural log (ln) of a number greater than one will give you a positive number.

- Photon336

so based off of that what would it be?

- Rushwr

nop no I haven't learned that

- Photon336

let's say K is 0.5
and K is 2

- Photon336

\[\Delta G = -RTlnK\]

- Rushwr

okai

- Photon336

IN this formula you notice that the sign is negative

- Photon336

-RTln(0.5) = ?
-RTln(2) = ?

- Photon336

what are those equal to?

- Photon336

a positive delta G or a negative delta G

- Rushwr

negative delta G ?

- Photon336

for which?

- Rushwr

the second one ?

- Rushwr

the first one is positive !

- Photon336

yes

- Photon336

so.. for the first one K < 1 so delta G would be positive because -RTln(k) > 0
the second one K >1 so Ln(K>1) is also greater than one. so we have Delta G < 0

- Rushwr

alright

- anonymous

Sure you can use \(\Delta G = \Delta H - T \Delta S\) for 390 and 391 but it should be intuitive.
\(\Delta H\) represents the change of internal heat energy. Losing heat is something that happens naturally, gaining heat requires something else of higher temperature to be nearby. I think they jokingly call this the "zeroth law of thermodynamics" haha.
\(\Delta S\) Is entropy, disorder! It's always increasing! The only way you can decrease entropy is by increasing something else's entropy more than the decrease you caused.
I just wanted to like say this cause it just sounded like everyone was like talking based on algebra but like there's stuff goin on there!

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