Question

1.38g of N₂O₄ was placed in a 1L reaction vessel at 400k and allowed to attain equilibrium N₂O₄​​_(g) ⇌ ​2NO_(g). The total pressure at equilibrium was found to be 9.15 bar. Calculate K_c, K_p, and partial pressure.

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{We}\mathrm{know}, \\ \mathrm{PV}=\mathrm{nRT} \\ \mathrm{Total}\mathrm{volume},\mathrm{V}=1\mathrm{L} \\ \mathrm{Molecular}\mathrm{mass}\mathrm{of}{\mathrm{N}}_2\mathrm{O}=92\mathrm{g} \\ \mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{N}}_2\mathrm{O}=\frac{13.8(\mathrm{it}\mathrm{should}\mathrm{be}13.8\mathrm{instead}\mathrm{of}1.38)}{92}=0.15 \\ \mathrm{R}=0.083{\mathrm{barLmol}}^{-1}{\mathrm{K}}^{-1} \\ \mathrm{T}=400\mathrm{K} \\ \mathrm{pV}=\mathrm{nRT} \\ \mathrm{p}\times 1\mathrm{L}=0.015\mathrm{mol}\times 0.083{\mathrm{barLmol}}^{-1}{\mathrm{K}}^{-1}\times 400\mathrm{K} \\ \mathrm{p}=4.98\mathrm{bar} \\ {\mathrm{N}}_2{\mathrm{O}}_4\to 2{\mathrm{NO}}_2 \\ \mathrm{Initial}\mathrm{pressure}4.98\mathrm{bar}0 \\ \mathrm{At}\mathrm{equilibrium}(4.98-\mathrm{x})\mathrm{bar}2\mathrm{x}\mathrm{bar} \\ {\mathrm{p}}_{\mathrm{total}}\mathrm{at}\mathrm{equilibrium}={\mathrm{p}}_{{\mathrm{N}}_2{\mathrm{O}}_4}+{\mathrm{p}}_{{\mathrm{NO}}_2} \\ 9.15=(4.98-\mathrm{x})+2\mathrm{x} \\ 9.15=4.98+\mathrm{x} \\ \mathrm{x}=9.15-4.98=4.17\mathrm{bar} \\ \mathrm{Partial}\mathrm{pressures}\mathrm{at}\mathrm{equilibrium},\mathrm{are}, \\ {\mathrm{p}}_{{\mathrm{N}}_2{\mathrm{O}}_4}=4.98-4.17=0.81\mathrm{bar} \\ {\mathrm{p}}_{{\mathrm{NO}}_2}=2\mathrm{x}=2\times 4.17=8.34\mathrm{bar} \\ {\mathrm{K}}_{\mathrm{p}}=\frac{({\mathrm{p}}_{{\mathrm{NO}}_2}{)}^2}{\left({\mathrm{p}}_{{\mathrm{N}}_2{\mathrm{O}}_4}\right)}=\frac{(8.34{)}^2}{(0.81)}=85.87 \\ {\mathrm{K}}_{\mathrm{p}}={\mathrm{K}}_{\mathrm{c}}(\mathrm{RT}{)}^{∆\mathrm{n}} \\ 85.87={\mathrm{K}}_{\mathrm{c}}(0.083\times 400{)}^1 \\ {\mathrm{K}}_{\mathrm{c}}=2.586=2.6 \end{gathered}$$