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.
