Question

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

Answer 1

390/391

Answer 2

392/393

Answer 3

395/396/397

Answer 4

398

Answer 5

404/405/406

Answer 6

Ok this probably all I will post for today.

Answer 7

@Rushwr @taramgrant0543664 @woodward @sweetburger

Answer 8

Just pick a problem and go for it!

Answer 9

I'm going with 390 lol

Answer 10

390 A

Answer 11

sorry B

Answer 12

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

Answer 13

Yes B lol

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

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.

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

Yep ^^

Answer 23

I think my brain is all frozen !

Answer 24

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

Answer 25

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

Answer 26

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.

Answer 27

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

Answer 28

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

Answer 29

@Rushwr 396?

Answer 30

for 395 it's C right?

Answer 31

do you learn this formula?

\[DeltaG = -RTln(k) \]

Answer 32

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

Answer 33

=D correct

Answer 34

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 !

Answer 35

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

Answer 36

for 396

Answer 37

do you remember this formula? or learn it

Delta G = -RTlnK

Answer 38

they told us that delta G was positive

Answer 39

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.

Answer 40

so based off of that what would it be?

Answer 41

nop no I haven't learned that

Answer 42

let's say K is 0.5
and K is 2

Answer 43

\[\Delta G = -RTlnK\]

Answer 44

okai

Answer 45

IN this formula you notice that the sign is negative

Answer 46

-RTln(0.5) = ?

-RTln(2) = ?

Answer 47

what are those equal to?

Answer 48

a positive delta G or a negative delta G

Answer 49

negative delta G ?

Answer 50

for which?

Answer 51

the second one ?

Answer 52

the first one is positive !

Answer 53

yes

Answer 54

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

Answer 55

alright

Answer 56

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!