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Photon336

  • one year ago

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

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  1. Photon336
    • one year ago
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    390/391

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  2. Photon336
    • one year ago
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    392/393

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  3. Photon336
    • one year ago
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    395/396/397

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  4. Photon336
    • one year ago
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    398

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  5. Photon336
    • one year ago
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    404/405/406

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  6. Photon336
    • one year ago
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    Ok this probably all I will post for today.

  7. Photon336
    • one year ago
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    @Rushwr @taramgrant0543664 @woodward @sweetburger

  8. Photon336
    • one year ago
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    Just pick a problem and go for it!

  9. taramgrant0543664
    • one year ago
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    I'm going with 390 lol

  10. Rushwr
    • one year ago
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    390 A

  11. Rushwr
    • one year ago
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    sorry B

  12. taramgrant0543664
    • one year ago
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    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

  13. taramgrant0543664
    • one year ago
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    Yes B lol

  14. Rushwr
    • one year ago
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    \[\Delta G= \Delta H- (T \Delta S)\]

  15. Rushwr
    • one year ago
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    So when Delta S positive -(TS) becomes negative . for a spontaneous reaction delta G should be negative.

  16. taramgrant0543664
    • one year ago
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    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

  17. Photon336
    • one year ago
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    \[\Delta G = \Delta H - T \Delta S\] 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.

  18. Photon336
    • one year ago
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    @taramgrant0543664 @Rushwr I agree. it's B.

  19. Photon336
    • one year ago
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    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

  20. taramgrant0543664
    • one year ago
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    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

  21. Photon336
    • one year ago
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    @taramgrant0543664 Yeah. me too because you keep Delta H constant, and you increase temperature. that means that -TDELTAS is going up. ^_^

  22. taramgrant0543664
    • one year ago
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    Yep ^^

  23. Rushwr
    • one year ago
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    I think my brain is all frozen !

  24. Photon336
    • one year ago
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    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

  25. taramgrant0543664
    • one year ago
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    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

  26. Photon336
    • one year ago
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    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.

  27. taramgrant0543664
    • one year ago
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    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

  28. Photon336
    • one year ago
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    yeah. I'm going to finish in about another 15 min lol

  29. Photon336
    • one year ago
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    @Rushwr 396?

  30. Rushwr
    • one year ago
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    for 395 it's C right?

  31. Photon336
    • one year ago
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    do you learn this formula? \[DeltaG = -RTln(k) \]

  32. Rushwr
    • one year ago
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    since it is vapourization I think the london dispersion forces are broken. Not the covalent bonds of Carbon and Chlorine >

  33. Photon336
    • one year ago
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    =D correct

  34. Rushwr
    • one year ago
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    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 !

  35. Photon336
    • one year ago
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    i guess it's the bonds between the molecules are london dispersion forces

  36. Photon336
    • one year ago
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    for 396

  37. Photon336
    • one year ago
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    do you remember this formula? or learn it Delta G = -RTlnK

  38. Photon336
    • one year ago
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    they told us that delta G was positive

  39. Photon336
    • one year ago
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    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.

  40. Photon336
    • one year ago
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    so based off of that what would it be?

  41. Rushwr
    • one year ago
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    nop no I haven't learned that

  42. Photon336
    • one year ago
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    let's say K is 0.5 and K is 2

  43. Photon336
    • one year ago
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    \[\Delta G = -RTlnK\]

  44. Rushwr
    • one year ago
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    okai

  45. Photon336
    • one year ago
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    IN this formula you notice that the sign is negative

  46. Photon336
    • one year ago
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    -RTln(0.5) = ? -RTln(2) = ?

  47. Photon336
    • one year ago
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    what are those equal to?

  48. Photon336
    • one year ago
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    a positive delta G or a negative delta G

  49. Rushwr
    • one year ago
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    negative delta G ?

  50. Photon336
    • one year ago
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    for which?

  51. Rushwr
    • one year ago
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    the second one ?

  52. Rushwr
    • one year ago
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    the first one is positive !

  53. Photon336
    • one year ago
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    yes

  54. Photon336
    • one year ago
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    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

  55. Rushwr
    • one year ago
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    alright

  56. anonymous
    • one year ago
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    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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