# Question e90b5

Mar 28, 2016

You have to calculate Δ_rG^° and ${K}_{\textrm{c}}$. Then you compare ${Q}_{\text{c}}$ with ${K}_{\text{c}}$.

#### Explanation:

color(blue)(|bar(ul(Δ_rH^° = ΣnΔ_fH^°("products") - ΣmΔ_fH^°("reactants"))|)

color(blue)(|bar(ul(Δ_rS^° = ΣnS^°("products") - ΣmS^°("reactants"))|)

color(blue)(|bar(ul(Δ_rG = Δ_rH – TΔ_rS)|)

color(blue)(|bar(ul(ΔG^@ = -RTlnK)|)

Step 1. Calculate the standard free energy of reaction.

We need either the enthalpy of formation of $\text{NOCl}$ or Δ_rG^° for the reaction.

I will use the tabulated value of Δ_fH° for $\text{NOCl}$ (66.07 kJ/mol).

$\textcolor{w h i t e}{m m m m m m m m} \text{2NO(g)"color(white)(l) +color(white)(l) "Cl"_2("g")color(white)(l) ⇌ color(white)(l)"2NOCl(g)}$
Δ_fH°//"kJ·mol"^"-1":color(white)(l) 90.29color(white)(mmmll)0color(white)(mmmmll)51.71
S°//"J·mol"^"-1""·K"^"-1":color(white)(l)210.65color(white)(mml)223.0color(white)(mmml)261.6
Δ_fG°//"kJ·mol"^"-1":color(white)(ll)86.60color(white)(mmmll)0color(white)(mmmml)66.07

Δ_rG^° = ΣnΔ_fG^°("products") - ΣmΔ_fG^°("reactants") = "2×66.07 kJ - 2×86.60 kJ" = "-41.06 kJ"

2. Calculate ${K}_{\text{eq}}$ for the reaction.

ΔG^° = -RTlnK

lnK = -(ΔG^°)/(RT) = ("41 060" color(red)(cancel(color(black)("J·mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1"))) × 298.15 color(red)(cancel(color(black)("K")))) = 16.56

K = e^16.56 = 1.56× 10^7

3. Calculate $Q$ for the reaction.

$\textcolor{w h i t e}{m m m m m l l} \text{2NO(g)"color(white)(l) +color(white)(l) "Cl"_2("g")color(white)(l) ⇌ color(white)(l)"2NOCl(g)}$
$\text{I"//"mol·L"^"-1} : \textcolor{w h i t e}{l} 0.050 \textcolor{w h i t e}{m m m} 0.045 \textcolor{w h i t e}{m m m m l l} 1.5$

Q =["NOCl"]^2/(["NO"]^2["Cl"_2]) = (1.5)^2/((0.050)^2× 0.045) = 2.0 × 10^"4"

4. Is the reaction at equilibrium?

Q ≠ K#, so the reaction is not at equilibrium.

5. In which direction will the reaction proceed?

$Q < K$, so we don't have enough products.

The reaction will proceed to the right.