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

1 Answer
Ernest Z.
Apr 24, 2017

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"#

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