Photon336
  • Photon336
Discussion/ (a lot of questions attached) please pick an answer and justify why it's right along with why the others are wrong.
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Photon336
  • Photon336
390/391
1 Attachment
Photon336
  • Photon336
392/393
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Photon336
  • Photon336
395/396/397
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Photon336
  • Photon336
398
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Photon336
  • Photon336
404/405/406
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Photon336
  • Photon336
Ok this probably all I will post for today.
Photon336
  • Photon336
@Rushwr @taramgrant0543664 @woodward @sweetburger
Photon336
  • Photon336
Just pick a problem and go for it!
taramgrant0543664
  • taramgrant0543664
I'm going with 390 lol
Rushwr
  • Rushwr
390 A
Rushwr
  • Rushwr
sorry B
taramgrant0543664
  • 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
  • taramgrant0543664
Yes B lol
Rushwr
  • Rushwr
\[\Delta G= \Delta H- (T \Delta S)\]
Rushwr
  • Rushwr
So when Delta S positive -(TS) becomes negative . for a spontaneous reaction delta G should be negative.
taramgrant0543664
  • 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
  • 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
  • Photon336
@taramgrant0543664 @Rushwr I agree. it's B.
Photon336
  • 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
  • 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
  • Photon336
@taramgrant0543664 Yeah. me too because you keep Delta H constant, and you increase temperature. that means that -TDELTAS is going up. ^_^
taramgrant0543664
  • taramgrant0543664
Yep ^^
Rushwr
  • Rushwr
I think my brain is all frozen !
Photon336
  • 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
  • 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
  • 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
  • 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
  • Photon336
yeah. I'm going to finish in about another 15 min lol
Photon336
  • Photon336
@Rushwr 396?
Rushwr
  • Rushwr
for 395 it's C right?
Photon336
  • Photon336
do you learn this formula? \[DeltaG = -RTln(k) \]
Rushwr
  • Rushwr
since it is vapourization I think the london dispersion forces are broken. Not the covalent bonds of Carbon and Chlorine >
Photon336
  • Photon336
=D correct
Rushwr
  • 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
  • Photon336
i guess it's the bonds between the molecules are london dispersion forces
Photon336
  • Photon336
for 396
Photon336
  • Photon336
do you remember this formula? or learn it Delta G = -RTlnK
Photon336
  • Photon336
they told us that delta G was positive
Photon336
  • 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
  • Photon336
so based off of that what would it be?
Rushwr
  • Rushwr
nop no I haven't learned that
Photon336
  • Photon336
let's say K is 0.5 and K is 2
Photon336
  • Photon336
\[\Delta G = -RTlnK\]
Rushwr
  • Rushwr
okai
Photon336
  • Photon336
IN this formula you notice that the sign is negative
Photon336
  • Photon336
-RTln(0.5) = ? -RTln(2) = ?
Photon336
  • Photon336
what are those equal to?
Photon336
  • Photon336
a positive delta G or a negative delta G
Rushwr
  • Rushwr
negative delta G ?
Photon336
  • Photon336
for which?
Rushwr
  • Rushwr
the second one ?
Rushwr
  • Rushwr
the first one is positive !
Photon336
  • Photon336
yes
Photon336
  • 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
  • Rushwr
alright
anonymous
  • 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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