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 \[0.2-2x=0.04\]
\[2x=0.16\]
\[x=0.08\]
\[{{K}_{p}}=\frac{[{{p}_{{{H}_{2}}}}][{{p}_{{{I}_{2}}}}]}{{{[{{p}_{HI}}]}^{2}}}=\frac{0.08\times
0.08}{{{(0.04)}^{2}}}=4\]
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