chem

Consider the equation A <-> in a 1L container. At 7.57C, the molar concentrations of A and B are 7.228M and 1.976 (respectively). Suddenly, 1.831 moles of A are added to the system. What is the change in deltaG of the system (Answer in kJ)

At first, I thought I would use the following equation...
deltaG = deltaG^o + RTln(Q)

but I don't have deltaG^o or know which one is being reduced and oxidized. So which formula would I use?

Thanks

I assume your first sentence should read A <===> B in a 1 L container......
and I assume the first set of conditions is at equilibrium.
At equilibrium delta Go = 0 and Q = K. You know the numbers to calculate K and therefore, delta G may be calculated. THEN, use the second sent of conditions, still with delta Go = 0 (I assume the second set of numbers will eventually reach equilibrium), to calculate the new delta G. Using K as (products)/(reactants) means you don't need to worry about which is oxidized and which is reduced.
Check my thinking.

Ok, so the for the first one:

G = 0 + (8.314J)(305.57K)ln(1.976/7.228)
= -3294.7

2nd one:
G = 0 + (8.314J)(305.57K)ln(3.807/7.228)
= -1628.7

Difference is 1666J
In kJ 1.666kJ

Is this correct?

In rereadig my response, I notice I goofed (BIG TIME). At equilibrium, delta G = 0 and not delta Go. So you need to redo. I TOLD you to check my thinking! :-).
Probably you can take it from there.

What I'm trying to find is delta G though...

If delta G = 0, then how I could I find that?

I'm sorry, Im just confused.

Couldn't you set delta G = 0 for the first set of conditions (in which Q = K) and solve for delta Go ? (again I assume the first set of conditions are at equilibrium). Then redo, using the value of delta Go you found, put in the new values (this time for Q) and solve for delta G? Then do the difference? Wouldn't that work? Again, check my thinking!

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asked by Marisol

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