1
$\begingroup$

In this answer, I don't understand how this step worked.

Using $\mu_i = \mu^\circ_i + RT\ln \frac{P_i}{\pu{1 bar}}$ \begin{align} \Delta G &= (c\mu^\circ_\ce{C} + d\mu^\circ_\ce{D} - a\mu^\circ_\ce{A} - b\mu^\circ_\ce{B}) + RT \ln\frac{P_\ce{C}^c P_\ce{D}^d}{P_\ce{A}^a P_\ce{B}^b}\\ \Delta G &= \Delta G^\circ + RT\ln Q \end{align}

How does $P_i=\dfrac{P_\ce{C}^c P_\ce{D}^d}{P_\ce{A}^a P_\ce{B}^b}$? And how does that equal the quotient $Q$?

If this is wrong, then please provide me the correct way to proceed from there.

$\endgroup$
1
  • $\begingroup$ Refer to the balanced equation and combine. The $i$ will refer to the exact species of interest (i.e., A, B, C, or D). $\endgroup$
    – Zhe
    Commented Jan 12, 2021 at 15:36

1 Answer 1

2
$\begingroup$

You have, for each compound, $\mu_i = \mu^\circ_i + RT\ln(P_i/P^\circ)$.

So, replacing each potential in $\Delta G$ by its expression, assuming $P^\circ = 1$ bar, we get : $$\Delta G = c(\mu^\circ_C + RT\ln(P_C)) + d(\mu^\circ_D + RT\ln(P_D)) - a((\mu^\circ_A + RT\ln(P_A)) - b(\mu^\circ_B + RT\ln(P_B))$$

Then we develop and group standard potentials and logs:

$$\Delta G = (c\mu^\circ_C + d\mu^\circ_D - a\mu^\circ_A - b\mu^\circ_B) + RT(c\ln(P_C) + d\ln(P_D) - a\ln(P_A)-b\ln(P_B))$$

Then, using calculation rules on logarithms:

$$\Delta G = (c\mu^\circ_C + d\mu^\circ_D - a\mu^\circ_A - b\mu^\circ_B) + RT\left(\ln({P_C}^c)+\ln({P_D}^d)-\ln({P_A}^a)-\ln({P_B}^b)\right)$$

And finally, since $-\ln(x) = \ln(1/x)$ and $\ln(x)+\ln(y)=\ln(xy)$, $$\boxed{\Delta G = (c\mu^\circ_C + d\mu^\circ_D - a\mu^\circ_A - b\mu^\circ_B) + RT\ln\left(\frac{{P_C}^c{P_D}^d}{{P_A}^a{P_B}^b}\right)}$$

Then, try expressing the quotient of the reaction in terms of partial pressures, assuming standard pressure is 1 bar and you'll find out it gives the inside of the log!

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.