Question #e90b5

1 Answer

Answer:

You have to calculate #Δ_rG^°# and #K_text(c)#. Then you compare #Q_"c"# with #K_"c"#.

Explanation:

Your essential formulas are:

#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 #"NOCl"# or #Δ_rG^°# for the reaction.

I will use the tabulated value of #Δ_fH°# for #"NOCl"# (66.07 kJ/mol).

#color(white)(mmmmmmmm)"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_"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.

#color(white)(mmmmmll)"2NO(g)"color(white)(l) +color(white)(l) "Cl"_2("g")color(white)(l) ⇌ color(white)(l)"2NOCl(g)"#
#"I"//"mol·L"^"-1":color(white)(l) 0.050color(white)(mmm)0.045color(white)(mmmmll)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.