# Question 199e5

Jan 22, 2018

The answer is (C) ${45}^{\circ} \text{C}$.

#### Explanation:

Your goal here is to figure out the temperature at which this reaction

${\text{N"_ 2"O"_ (5(s)) -> "N"_ 2"O}}_{5 \left(g\right)}$

becomes spontaneous, which happens when the change in Gibbs free energy for this reaction, $\Delta G$, becomes $\le 0$. Keep in mind that in order for the process to be spontaneous, you need to have $\Delta G < 0$, but you can use the equilibrium condition $\Delta G = 0$ to find the temperature at which the process becomes spontaneous.

Your tool of choice here will be the equation

color(blue)(ul(color(black)(DeltaG = DeltaH - T * DeltaS)))" " " "color(darkorange)("(*)")

Here

• $\Delta H$ is the change in enthalpy of the process
• $T$ is the absolute temperature at which the process takes place
• $\Delta S$ is the change in entropy of the process

Now, the change in enthalpy and the change in entropy can be calculated using Hess' Law.

For the change in enthalpy, you have

$\Delta H = \sum n \cdot \Delta {H}_{\text{f product"^@ - sum m * DeltaH_ "f reactant}}^{\circ}$

Here $n$ and $m$ represent the number of moles of each chemical species. Plug in your values to find

DeltaH = overbrace(1 color(red)(cancel(color(black)("mole"))) * "13.3 kJ"/(1color(red)(cancel(color(black)("mole")))))^(color(blue)("for N"_ 2"O"_(5(g)))) - overbrace( 1 color(red)(cancel(color(black)("mole"))) * (-"43.1 kJ"/(1color(red)(cancel(color(black)("mole"))))))^(color(blue)("for N"_ 2"O"_ (5(s))

$\Delta H = + \text{56.4 kJ}$

Notice that the change in enthalpy is positive, which is consistent with the fact that sublimation is an endothermic process, i.e. energy is required in order to get the molecules to perform a solid $\to$ gas phase change.

Next, calculate the change in entropy by using

$\Delta S = \sum n \cdot \Delta {S}_{\text{f product"^@ - sum m * DeltaS_ "f reactant}}^{\circ}$

Plug in your values to find--I'll use the values in kilojoules per mole Kelvin!

DeltaS = overbrace(1 color(red)(cancel(color(black)("mole"))) * "0.3557 kJ"/(1color(red)(cancel(color(black)("mole")))"K"))^(color(blue)("for N"_ 2"O"_(5(g)))) - overbrace( 1 color(red)(cancel(color(black)("mole"))) * "0.1782 kJ"/(1color(red)(cancel(color(black)("mole")))"K"))^(color(blue)("for N"_ 2"O"_ (5(s))

$\Delta S = + {\text{0.1775 kJ K}}^{- 1}$

The fact that you have $\Delta S > 0$ is consistent with the fact that the entropy of the system is increasing as a result of the solid $\to$ gas phase change--the molecules are no longer held in place in a crystal structure--and it tells you that this process is entropy driven.

Now, you want to find the temperature at which the sublimation of dinitrogen pentoxide becomes spontaneous, so set the equilibrium condition using equation $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{(*)}}$

$\Delta G = 0 \implies \Delta H = T \cdot \Delta S$

and rearrange the equation to solve for $T$

$T = \frac{\Delta H}{\Delta S}$

You know the change in enthalpy and the change in entropy associated with this reaction, so plug the values into the equation to find

T = (+56.4 color(red)(cancel(color(black)("kJ"))))/(+0.1775 color(red)(cancel(color(black)("kJ"))) "K"^(-1)) = "317.75 K"

Finally, convert the temperature to degrees Celsius to get

t[""^@"C"] = "317.75 K" - 273.15 = 44.60^@"C" ~~ 45^@"C"#

Notice that according to equation $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{(*)}}$, in order to have

$\Delta G \le 0$

you need to have

$T \ge \text{317.75 K}$

which means that at temperatures that are lower than this value, the sublimation of dinitrogen pentoxide is not a spontaneous process.