# Chemistry

The multiple Choice Question:

Determine delta G (free energy at standard condition s) for the reaction below, considering Kp = 8 at 25 degrees celcius.

2Ag -> B

A.-2.303 x (0.287) x (25) x log 8 [J]
B.-2.303 x (8.314) x (25) x log 8 [J]
C.-2.303 x (8.314) x (298) x log1/8 [J]
D.-2.303 R T log {[B]^2/[A]^2}
E.-2.303 x (8.314) x (298) x log 8 [J]

Do you have any thoughts on this? and why do you think that?

I was thinking C or E because its in Kelvins, but I am not to good with logs so I cant figure out if it should be 1/8 or 8. Then there is always the possibility its D but I don't think the B would be squared. So how would you figure out the log part?

I was thinking C or E because its in Kelvins, but I am not to good with logs so I cant figure out if it should be 1/8 or 8. Then there is always the possibility its D but I don't think the B would be squared. So how would you figure out the log part?

delta G = delta Go + RT ln Q
At equilibrium delta G = 0; so
0 = delta Go + RT ln K. The problem asks for delta G at standard conditions so that is delta Go. Solving for delta Go gives,
delta Go = -RT ln K where ln is the natural logarithm. If we want to use log base 10, it is
delta Go = -2.303*R*T*log K and since K = 8, this becomes
delta Go = -2.303*R*T*log(8)
R is 8.314 and T is 298.

NOW, neither A nor B can be right because T is not is Kelvin as you correctly point out. C is out because the answer has 1/8 and not 8 so they have substituted 1/K instead of K. D can't be right, as you correctly point out, because K = (B)/(A^2) and not (B^2) as it is in the answer. So guess what? Does what I have in bold print look anything like answer E?
I hope this helps.

It does indeed Dr. Bob. Thank you very much for your help!

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