Question

http://oi68.tinypic.com/33muayw.jpg

Answer 1

@Photon336

Answer 2

I only need the delta G part. I've been trying to work it and getting myself confused. The equation for it is:
I2 + H2 <--> 2HI

Kc = [HI]^2/[I2][H2]

Answer 3

@jebonna

Answer 4

Yeah, interesting thing here is that the reaction is already at equilibrium

Answer 5

\[kc =\frac{ [HI]^2}{ [H_2][ I_2] }\]

Answer 6

We know that it's at equilibrium already so those concentrations will be constant.

So, I think we can calculate the number of moles of HI easily.

\[0.017mol~H_{2}*(\frac{ 2HI }{ H_2 }) = 0.034~moles~HI\]

Answer 7

since we know that the volume of the container is 1L to find the concentration is easy. it's just moL/L

i.e. 0.017mol/L, for both HI and I2

now next step would be to put all the information into the formula for Kc

Answer 8

@Photon336 So HI= .017M and I2= .017M. H2 = .017

Answer 9

Then do I put it in the prod/reactant equation? [.017]^2/[.017][.017]

Answer 10

@Photon336 and if I do that I get 1

Answer 11

@Photon336 so is that the delta G then or do we need to do more to get it?

Answer 12

ok wait lol sorry we need to solve for delta G in terms of Kp

Answer 13

Alright. I was confused on the delta G part.

Answer 14

\[\Delta G = RTln(Kp)\]

Answer 15

733K

Answer 16

@staldk3 R = 8.314 J mol K and T = 733

Answer 17

delta G = 8.314(733)ln(1)

Answer 18

Do you still raise it to e or keep it as ln?

Answer 19

\[-(8.314 J~mol^-1~K^-1)(733K)(\ln(1)) = \Delta G \]

Answer 20

delta G = 0

Answer 21

why does this make sense @staldk3 ?

Answer 22

Because it's at equilibrium

Answer 23

:in the problem you're told that the reaction is at equilibrium already, and you're then given the equilibrium concentrations of your products and reactants.

Answer 24

So Q=K

Answer 25

which means G=0

Answer 26

@Photon336 or should it rather be Kp=Kc

Answer 27

exactly, are you familiar with what Q is?

Answer 28

Vaguely. I know the formula for it is similar to K

Answer 29

well Q is basically the same as K

Q is basically the ratio of the concentration of products over reactants at any time in the reaction other than equilibrium.

\[\frac{ [PRODUCTS] }{ [REACTANTS] } = \frac{ [C][D] }{ [A][B] } = Q\]

Q is the ratio of the concentration of the reactants and products at equilibrium

\[\frac{ [PRODUCTS] }{ [REACTANTS] } = \frac{ [C][D] }{ [A][B] } = k\]

Answer 30

That seems simple enough

Answer 31

Also it's not wanting to take delta Go = 0

Answer 32

But that's right...right?

Answer 33

Since it's at equilibrium

Answer 34

what?

Answer 35

The delta G deg = 0, right? Since it's at equilibrium.

Answer 36

yep

Answer 37

So the answer for c should be 0

Answer 38

when Q = K equilibrium was established.

Answer 39

Does it matter if it's delta G rxn?

Answer 40

yeah delta G means the delta G for the reaction

Answer 41

Bleh. It keeps counting it wrong for some reason. Did we miss something?

Answer 42

@staldk3 what did it say for Kc?

Answer 43

As in the a and b part?

Answer 44

first two parts

Answer 45

a) Kc>1
b) Kp = Kc

Answer 46

Kp=Kc(RT)Δn

Answer 47

@aaronq @Cuanchi

Answer 48

@Cuanchi We are stuck on the part c

Answer 49

not sure why it's wrong?

Answer 50

there is a small detail
deltaG = deltaGo + RT ln Q

because your reaction is not at standard condition the delta G standard of the reaction

deltaGo = - RT ln K

Answer 51

-(8.314)(733)ln(1)

Answer 52

@Cuanchi like this?

Answer 53

K is not 1

Answer 54

Then what's the K?

Answer 55

interesting.. that's the mistake I made. I assume the reaction was at equilibrium

Answer 56

it is at equilibrium but not a standard condition the temperature and the pressures of teh reactantas and products are not 273 and 1 atm or 1 molar

Answer 57

So how do I need to set this up?

Answer 58

completely missed that!

Answer 59

I think the K is 16

Answer 60

How did you get 16?

Answer 61

the problem said each molecule is 0.017 mole, that is not the concentration of each gas in the mix

Answer 62

did you count haw many molecules of HI, I2, H2 are in the figure?

Answer 63

Yes. 9HI 13 I2 and 12 H2

Answer 64

no

Answer 65

you have 9 HI, 2 of I2 and 2H2

it is hard to see in the still picture but in your animation you should be able to differentiate the Hi from I2

Answer 66

Yeah. Kinda. It moves a little fast so it's still hard to see.

Answer 67

[9]^2/[2][2]

Answer 68

yea but now you have to multiply each of the molecules by 0.017 mole to calculate the molarity (I think is not 16 either)

Answer 69

20.25

Answer 70

Wait why do you multiply by .017?

Answer 71

yes it is the same if you dont do the multiplication

Answer 72

81/4=20.25

Answer 73

Mk that's what I did and I got that answer.

Answer 74

good!

Answer 75

now use that value of K in the formula for delta g standard of reaction with the value of temperature of the reaction 733
deltaGo = - RT ln K

Answer 76

-8.314(733)ln(20.25)

Answer 77

getting closer?

Answer 78

=-18332.18263

Answer 79

joules?

Answer 80

J/mol

Answer 81

Sorry

Answer 82

no?

Answer 83

the answer was wrong?

Answer 84

Maybe I did it wrong.

Answer 85

-8.314(733)=-6094.162ln(20.25)

Answer 86

for the ln do I need to set it to e?

Answer 87

I got the same

Answer 88

The -18332.18263?

Answer 89

like e ^-6094

Answer 90

@Cuanchi I'm confused what to do with the ln

Answer 91

you are fine I got the same answer than you 18332 J/mol

Answer 92

Should it be positive or negative?

Answer 93

negative

Answer 94

Alright and for the sig figs is it okay to keep all of them>

Answer 95

looks like than 2-3 SF will be enough, the data in the problem has 2 or 3 SF. Just doble check the number of molecules in the figure if it is correct