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Author Topic: equilibrium&standard tanGibbs free energy; problems understanding relationship  (Read 356 times)

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scientific

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Cu2O (s) + 1/2 O2(g)  ::equil:: 2CuO (s)
ΔH° = -11.3 kJ

At 298 K and 1 atm, the closed system shown is at equilibrium. If the equilibrium is perturbed by isothermally decreasing the volume of the system, which of the following is NOT correct?
a. more product will be present after equilibrium is reestablished.
b. ΔG is less than zero for the process of reestablishing equilibrium
c. the equilibrium constant, Keq will decrease
d. the temperature remains constant
e. ΔG° remains unchanged

a. is because of le chatelier
b. is because as volume  :spindown:, temperature  :spinup: so for ΔG = ΔH - TΔS, a negative - larger positive becomes negative since originally ΔG = 0
d. is defined in question as isothermal

The correct is:
c. because the equilibrium constant should increase as more product is made




HOWEVER, I absolutely cannot figure out why e is false. Wouldn't e only be true at standard conditions?
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mjc123

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a. is true, but can you be more specific than just "Le Chatelier"?
b. is true, but your reason is wrong. Temperature does not rise - you are told it is isothermal. G is a minimum at equilibrium, so if you move from a non-equilibrium state to equilibrium, G must decrease.
c. is false, but your reason is again wrong. More product doesn't mean a higher equilibrium constant. (What is the definition of Keq in this case?) K is constant at constant T.
d. is true.
e. is false for the same reason as c. ΔG° = -RTlnK. If K is constant at consttant T, so is ΔG°.
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scientific

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a. As volume increases, the system prefers fewer moles of gas. In this case, that's the right side of the equlibrium because that side is all solid.
b. ΔG = ΔH - TΔS, and ΔH is negative. In order for ΔG to be negative, ΔH - TΔS must be (negative number - positive number) OR (negative number - smaller negative number), but since -11.3 seems small to me, I'll assume the first. For the term TΔS to be positive, ΔS should be positive. ΔS  (or Sfinal - Sinitial)  to be positive implies Sfinal > Sinitial
In the described system, the equilibrium gets shifted to the right (gas to solid) so Sfinal > Sinitial is true; ΔS is indeed positive,  ΔH - TΔS  is negative and ΔG is negative. Is that correct?
c. the Keq is actually unchanged, as opposed to  decrease-- it's dependent solely on temperature (in this case)
d. defined
e.


I'm still unclear on (e). My solutions guide says that (c) is the definite answer. Also, I used ΔG°=ΔH°−TΔS°, so wouldn't an isothermal system not be a problem?
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mjc123

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a. The volume decreases. The pressure increases, which favours consumption of gas.
b. Entropy decreases because you reduce the number of moles of gas. -11.3 may be small, but -11300 is not.
c. It's dependent solely on temperature always.
d. ok
e. is essentially the same as c. Sorry, I misread it the first time. K decreases is false. ΔG° is unchanged is true, for the same reason.
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