[LidaOliver]

A reaction A ----> B proceeds with H= -25 kJ/mol and S= -100 J/Kmol. Substance A was taken at the initial concentration 2mM. Determine the concentrations of A and B after the reaction reaches equilibrium. Assume T=298K.

I am a little lost and not sure exactly what to do, but here's what I have done so far:

A initial= 2mM = 0.02M
G=H - TS
G=(-25000 J/mol) - (298K*-100 J/Kmol)
G=4800 J/Kmol

ln(Keq)= -G/RT = (-4800)/(8.314*298)= -1.974
e^[ln(Keq)] = e^(-1.974)
Keq=0.1441

I am looking for equilibrium concentrations, and believe I found Keq, but I am not sure how to use the initial concentration of A...
(Thank you for your *delete me*!)

[Borek]

Once you know Keq it is just an equilibrium problem. From the reaction stoichiometry you know that [A]+[ B]=2 mM.

[Dan]

See also: http://en.wikipedia.org/wiki/ICE_table

[LidaOliver]

Once you know Keq it is just an equilibrium problem. From the reaction stoichiometry you know that [A]+[ B]=2 mM.

I don't understand what you mean. If A's initial is 0.02M, at equilibrium I can assume that it would be less than that as B increases. How do I find the exact concentrations at equilibrium?

[LidaOliver]

See also: http://en.wikipedia.org/wiki/ICE_table

How would I use ICE tables in this problem? I have two unknown equilibrium concentrations. I thought that you need atleast one of them? And I thought that was used for weak acids. Would I set it up

Keq=B²/0.02-A
Assume that A is negligible and say that B=sqrt(Keq×0.02)

That does not seem right to me....

[Borek]

Where did you got the square from?

Write formula for this reaction equilibrium constant.

[Dan]

Sorry, I glanced at this and massively overcomplicated the problem by suggesting an ICE table (you can use an ICE table, but it is not required). Sorry about that.

For A  B

Keq = [ B]/[ A]   (eq. 1)

You know, as Borek already said, that the total concentration [A] + [ B] = 2 mM   (eq. 2)

You have 2 unknowns and 2 equations, it's just a simple simultaneous equations matheatical problem. Rearrange one, substitute into the other, solve.

To do it with an ICE table, construct the table as normal:

     [A]   [ B]
I     2     0
C   -x     ?
E    ?      ?

I will let you complete it. Substitute the equilibrium concentrations for [A] and [ B] (i.e. row E) into eq. 1 above and solve.

[LidaOliver]

Where did you got the square from?

Write formula for this reaction equilibrium constant.

Sorry! I used the HH eq
pH= pKa + log(A/HA)
I should not have used that at all...

Keq=[product]/[reactant]
And I used ln(Keq)=-ΔG/RT to find a value for Keq.

[LidaOliver]

Sorry, I glanced at this and massively overcomplicated the problem by suggesting an ICE table (you can use an ICE table, but it is not required). Sorry about that.

For A  B

Keq = [A]/[ B]   (eq. 1)

You know, as Borek already said, that the total concentration [A] + [ B] = 2 mM   (eq. 2)

You have 2 unknowns and 2 equations, it's just a simple simultaneous equations matheatical problem. Rearrange one, substitute into the other, solve.

To do it with an ICE table, construct the table as normal:

     [A]   [ B]
I     2     0
C   -x     ?
E    ?      ?

I will let you complete it. Substitute the equilibrium concentrations for [A] and [ B] (i.e. row E) into eq. 1 above and solve.

Okay 

        [A]           [ B]
I.     0.02           0
C.      -x.          +x
E.     0.02-x      +x


Keq=[ B]/[A]
0.144=x/(0.02-x)
Solving for x I get x=0.00252

Then [A]= 0.02-0.00252=0.0175
And [ B]=0.00252

Is this correct?! 

[Borek]

Looks OK to me (although I have not checked numbers).

[Dan]

Is this correct?! 

Units are required. Also bear in mind that 2 mM is not 0.02 M.

[Borek]

Oh my. Shouldn't answer before morning coffee.

Not that I drink coffee in the morning, it was just a figure of speech

[LidaOliver]

Is this correct?! 

Units are required. Also bear in mind that 2 mM is not 0.02 M.

Right, it's 0.002M. Simply enough to correct from there...