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I am having trouble with the following problem:

Dinitrogen pentoxide, $\ce{N2O5}$, is a solid with a high vapor pressure. Its vapor pressure at $\pu{7.5 °C}$ is $\pu{100 mmHg}$, and the solid sublimes at a pressure of $\pu{1.00 atm}$ at $\pu{32.4 °C}$. What is the standard Gibbs energy change for the process $\ce{N2O5(s) → N2O5(g)}$ at $\pu{25 °C}$?

Okay so my attempt is to first use the Clapeyron equation to find $\Delta H$ for usage in the reversible process entropy formula. The Clapeyron equation is $$\ln\left(\frac{P_2}{P_1}\right) = \frac{-\Delta H}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$ Plugging everything in I got $\pu{50807.79 J}$.

Next I found the change in entropy in a reversible process:

$$\Delta S = \frac{\Delta H_\mathrm{vap}}{T_\mathrm{vap}}.$$

Now here is where I have a problem. I calculated the enthalpy of vapor pressure. So I would expect to use the temperature that corresponds to the enthalpy of vaporization. Problem is, the book lists this as the wrong answer. I got within a few significant figures when I used $T_\mathrm{sublimation}$.

I calculated the final answer by using

$$\Delta G = \Delta H - T\Delta S$$

The corresponding values are plugged in. I am not sure if I even solved the problem correctly, but if I am, why is that temperature used?

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At the sublimation point ($\pu{1 atm}$), the vapor and solid have the same molar free energy, which allows you to write $\Delta G^\circ=0$, from which your second equation relating $\Delta S$ and $\Delta H$ at $T_\mathrm{sub}$ ($\pu{32.4^\circ C}$) follows (although you used vap instead of sub as subscript). From there you obtain $\Delta S$. However, the free energy difference is not non-zero for these two phases at $\pu{1 atm}$ and other temperatures. Since you are asked to compute the free energy change at $\pu{25^\circ C}$, you can use the values of $\Delta S$ and $\Delta H$ you just computed in the general expression $$\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ$$ but setting $T=\pu{298.15 K}$. You assume that the entropy and enthalpy changes are approximately constant over this T interval (that is, ignore changes due to the heat capacity differences).

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First, here is a way to solve it that avoids calculating entropy:

  1. Calculate the enthalpy from the given data.
  2. Calculate the vapor pressure at 25 $^\circ$C. This is equal to the equilibrium constant for the process.
  3. Calculate $\Delta G^\circ$ as -RT ln(K).

This way is efficient because you already set up the Clapeyron equation in the first step and you just use it again in the second. You could even solve 1. and 2. graphically, and then go from there.

I got within a few significant figures when I used Tsublimation.

You need to calculate $\Delta S^\circ$, i.e. the change in entropy when all species are at standard state. At the normal temperature of sublimation, $\ce{N2O5(g)}$ is at the standard state, so that is the temperature you need to use. The enthalpy is not concentration dependent, but the entropy and the Gibbs free energy are.

I am not sure if I even solved the problem correctly[...]

You solved it correctly in that you got the correct answer. You did not solve it as the problem author intended, I suspect, because you had to look up the normal sublimation point (unless you used the Clapeyron equation to estimate it).

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