question_answer 1)

A sample of HI (g) is placed in a flask at a pressure of 0.2aim. At equilibrium the partial pressure of HI (g) is 0.04atm. What is \[{{K}_{p}}\]for the given equilibrium? \[2HI(g)\rightleftharpoons {{H}_{2}}(g)+{{I}_{2}}(g)\]

Answer:

Information shadow: Initial pressure of \[HI\text{ }=\text{ }0.2\text{ }atm\] Equilibrium pressure of \[HI\text{ }=\text{ }0.04\text{ }atm\]Equilibrium constant \[{{K}_{p}}\] for the given reaction is required. Problem solving strategy: \[\underset{{{t}_{\begin{smallmatrix} 0 \\ {{t}_{eq}} \end{smallmatrix}}}}{\overset{{}}{\mathop{{}}}}\,\underset{\begin{smallmatrix} 0.2atm \\ (0.2-2x) \end{smallmatrix}}{\mathop{2HI(g)}}\,\,\,\,\,\,\rightleftharpoons \,\,\,\,\,\,\,\underset{\begin{smallmatrix} 0 \\ x \end{smallmatrix}}{\mathop{{{H}_{2}}(g)}}\,\,\,\,\,\,\,\,\,\,+\underset{\underset{x}{\mathop{0}}\,}{\mathop{{{I}_{2}}(g)}}\,\] Given \[0.2-2x=0.04\] \[2x=0.16\] \[x=0.08\] \[{{K}_{p}}=\frac{[{{p}_{{{H}_{2}}}}][{{p}_{{{I}_{2}}}}]}{{{[{{p}_{HI}}]}^{2}}}=\frac{0.08\times 0.08}{{{(0.04)}^{2}}}=4\]