# 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 $P C {l}_{5} \left(g\right)$; 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 $P C {l}_{5}$ or $P C {l}_{3}$. Here's the information I got from the book... for $P C {l}_{5}$: -305.0 for $P C {l}_{3}$: -267.8

Apr 24, 2017

Warning! Long Answer. ${p}_{\textrm{\to t}} = \text{7.25 bar}$

#### 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.

$\textcolor{w h i t e}{m m m m m m m m m} {\text{PCl"_5 ⇌ "PCl"_3 + "Cl}}_{2}$
Δ_text(f)H^@"/kJ·mol"^"-1": color(white)(l)"-398.9"color(white)(m)"-306.4"color(white)(mm)0
${S}^{\circ} \text{/J·K"^"-1""mol"^"-1} : \textcolor{w h i t e}{m l} 353 \textcolor{w h i t e}{m m} 311.7 \textcolor{w h i t e}{m l l} 223$

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"

$K = {e}^{\text{-0.397}} = 0.672$

Calculate equilibrium concentrations

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

$\textcolor{w h i t e}{m m m m m m m m} {\text{PCl"_5 ⇌ "PCl"_3 + "Cl}}_{2}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m m} 0.0923 \textcolor{w h i t e}{m m m} 0 \textcolor{w h i t e}{m m l l} 0$
$\text{C/mol·L"^"-1":color(white)(mmm)"-"xcolor(white)(mmm)"+"xcolor(white)(mm)"+} x$
$\text{E/mol·L"^"-1} : \textcolor{w h i t e}{m} 0.0923 - x \textcolor{w h i t e}{m l l} x \textcolor{w h i t e}{m m l l} x$

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

${K}_{\textrm{c}} = \left(\left[{\text{PCl"_3]["Cl"_2])/(["PCl}}_{5}\right]\right) = {x}^{2} / \left(0.0923 - x\right) = 0.672$

Test for negligibility:

$\frac{0.0923}{0.672} = 0.14 < 400$. ∴ $x$ is not negligible. We must solve a quadratic equation.

${x}^{2} = 0.672 \left(0.0923 - x\right) = 0.0620 - 0.672 x$

${x}^{2} + 0.672 x - 0.0620 = 0$

$x = 0.0820$

Calculate total pressure

$\text{Total concentration" = (0.0923 - x + x + x") mol·L"^"-1" = "(0.0923 + 0.0820) mol/L" = "0.1743 mol/L}$

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"