# Question #199e5

##### 1 Answer

The answer is **(C)**

#### Explanation:

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

#"N"_ 2"O"_ (5(s)) -> "N"_ 2"O"_ (5(g))#

**becomes spontaneous**, which happens when the change in Gibbs free energy for this reaction, *equilibrium condition* **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

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

For the change in enthalpy, you have

#DeltaH = sum n * DeltaH_"f product"^@ - sum m * DeltaH_ "f reactant"^@#

Here

#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))#

#DeltaH = +"56.4 kJ"# Notice that the change in enthalpy is

positive, which is consistent with the fact that sublimation is anendothermic process, i.e. energy isrequiredin order to get the molecules to perform a solid#-># gasphase change.

Next, calculate the change in entropy by using

#DeltaS = sum n * DeltaS_"f product"^@ - sum m * DeltaS_ "f reactant"^@#

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))#

#DeltaS = +"0.1775 kJ K"^(-1)# The fact that you have

#DeltaS > 0# is consistent with the fact that theentropy of the systemisincreasingas a result of the solid#-># gas phase change--the molecules are no longer held in place in a crystal structure--and it tells you that this process isentropy driven.

Now, you want to find the temperature at which the sublimation of dinitrogen pentoxide becomes spontaneous, so set the equilibrium condition using equation

#DeltaG = 0 implies DeltaH = T * DeltaS#

and rearrange the equation to solve for

#T = (DeltaH)/(DeltaS)#

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

#DeltaG <=0#

you need to have

#T >= "317.75 K"#

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