# Question 050be

Aug 17, 2016

I get K = 1.93 × 10^11 for $\text{BrCl}$ and 2.03 × 10^4 for ${\text{N"_2"O}}_{4}$.

#### Explanation:

For $\text{BrCl}$

$\text{Br"_2"(g)" + "Cl"_2"(g)" ⇌ "2BrCl(g)}$
Δ_fH^° = "14.6 kJ·mol"^"-1"
ΔS^° = "240.0 J·K"^"-1""mol"^"-1"

Δ_fG^° = Δ_fH^° - TΔS^° = "14.6 kJ·mol"^"-1" - 610 color(red)(cancel(color(black)("K"))) × "0.2400 kJ"·color(red)(cancel(color(black)("K"^"-1")))"mol"^"-1"= "14.6 kJ·mol"^"-1" - "146.4 kJ·mol"^"-1" = "-131.8 kJ·mol"^"-1"

ΔG = "-"RTlnK

lnK = ("-"ΔG)/(RT) = ("131 800" color(red)(cancel(color(black)("J·mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1"))) × 610 color(red)(cancel(color(black)("K")))) = 25.99

K = e^25.99 = 1.93 × 10^11

For $\text{N"_2"O"_4}$

$\textcolor{w h i t e}{m m m m m m m m m} 2 {\text{NO"_2 ⇌ "N"_2"O}}_{4}$
Δ_fH^°"/kJ·mol"^"-1":color(white)(ll)33.10color(white)(mmm)9.08
S^°"/J·K"^"-1""mol"^"-1": color(white)(ll)240.04color(white)(mm)304.38

The formula for enthalpy of reaction is

color(blue)(|bar(ul(color(white)(a/a) Δ_rH = sumΔ_fH_text(products) - sumΔ_fH_text(reactants)color(white)(a/a)|)))" "

 Δ_rH = "[1(9.08) - 2(33.01)] kJ" = "-56.94 kJ"

The formula for the entropy of reaction is

color(blue)(|bar(ul(color(white)(a/a)Δ_rS = sumS_text(products) - sumS_text(reactants)color(white)(a/a)|)))" "

Δ_rS = "(1×304.38 - 2×240.04) J/K" = "-175.80 J/K"

color(blue)(|bar(ul(color(white)(a/a)ΔG = ΔH - TΔScolor(white)(a/a)|)))" "

Δ_rG = "-56.94 kJ" - 610 color(red)(cancel(color(black)("K"))) × ("-0.175 80 kJ"·color(red)(cancel(color(black)("K"^"-1")))) = "-56.94 kJ" + "107.238 kJ" = "-50.30 kJ"

The formula for $K$ is

color(blue)(|bar(ul(color(white)(a/a)ΔG = -RTlnKcolor(white)(a/a)|)))" "

lnK = ("-"ΔG)/(RT) = ("50 300" color(red)(cancel(color(black)("J·mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1"))) × 610 color(red)(cancel(color(black)("K")))) = 9.918

K = e^"9.918" = 2.03 × 10^4#