Question

What is the total pressure of the gases in the flask at this point?

Introduced into a 1.30 −L flask is 0.120 mol of `PCl_5(g)`; the flask is held at a temperature of 227 °C until equilibrium is established.
`PCl_5(g)⇌PCl_3(g)+Cl_2(g)`

What is the total pressure of the gases in the flask at this point? [Hint: Use data from Appendix D in the textbook and appropriate relationships from this chapter.]

From what I know, I need to apply `\DeltaG°`, but I don't know whether I need that for `PCl_5` or `PCl_3`.
Here's the information I got from the book...
for `PCl_5`: -305.0
for `PCl_3`: -267.8

Answer 1

Warning! Long Answer. `p_text(tot) = "7.25 bar"`

Explanation:

Yes, you need `ΔG^@`, but at 500 K, not 298 K.

Calculate `ΔG^@` at 500 K

You can calculate it from the values tabulated at 298 K.

`color(white)(mmmmmmmmm)"PCl"_5 ⇌ "PCl"_3 + "Cl"_2`
`Δ_text(f)H^@"/kJ·mol"^"-1": color(white)(l)"-398.9"color(white)(m)"-306.4"color(white)(mm)0`
`S^@"/J·K"^"-1""mol"^"-1":color(white)(ml)353color(white)(mm)311.7color(white)(mll)223`

`Δ_text(r)H^@ = sumΔ_text(f)H^@("products") - sumΔ_text(f)H^@("reactants") = "(-306.4 + 398.9) kJ/mol" = "92.5 kJ/mol"`

`Δ_text(r)S^@ = sumS^@("products") - sumS^@("reactants") = (311.7 + 223 - 353) color(white)(l)"kJ/mol" = "181.7 kJ/mol"`

`ΔG^@ = ΔH^@ -TΔS^@`

∴ At 227 °C (500 K),

`ΔG^@ = "92 500 J·mol"^"-1" -500 "K" × "181.7 J"·"K"^"-1""mol"^"-1" = "(92 500 - 90 850) J·mol"^"-1" = "1650 J·mol"^"-1"`

Calculate `K`

`ΔG^@ = "-"RTlnK`

`lnK =(-ΔG^@)/(RT)= ("-1650 J·mol"^"-1")/(8.314 "J·K"^"-1""mol"^"-1" ×500 "K") = "-0.397"`

`K = e^"-0.397" = 0.672`

Calculate equilibrium concentrations

Finally, we can set up an ICE table to calculate the equilibrium concentrations.

`color(white)(mmmmmmmm)"PCl"_5 ⇌ "PCl"_3 + "Cl"_2`
`"I/mol·L"^"-1": color(white)(mm)0.0923color(white)(mmm)0color(white)(mmll)0`
`"C/mol·L"^"-1":color(white)(mmm)"-"xcolor(white)(mmm)"+"xcolor(white)(mm)"+"x`
`"E/mol·L"^"-1":color(white)(m)0.0923 -xcolor(white)(mll)xcolor(white)(mmll)x`

`["PCl"_5]_0 = "0.120 mol"/"1.30 L" = "0.0923 mol/L"`

`K_text(c) = (["PCl"_3]["Cl"_2])/(["PCl"_5]) = x^2/(0.0923-x)= 0.672`

Test for negligibility:

`0.0923/0.672 = 0.14 < 400`. ∴ `x` is not negligible. We must solve a quadratic equation.

`x^2 = 0.672(0.0923-x) = 0.0620 - 0.672x`

`x^2 + 0.672x - 0.0620 = 0`

`x = 0.0820`

Calculate total pressure

`"Total concentration" = (0.0923 - x + x + x") mol·L"^"-1" = "(0.0923 + 0.0820) mol/L" = "0.1743 mol/L"`

`p = (nRT)/V = cRT = (0.1743 color(red)(cancel(color(black)("mol·L"^"-1"))) × "0.083 14 bar"·color(red)(cancel(color(black)("L·K"^"-1""mol"^"-1"))) × 500 color(red)(cancel(color(black)("K")))) = "7.25 bar"`