A sample of {Hl}_{{g})} is placed in a flask at a pressure of 0.2 \mathrm{~atm}. At equilibrium, the partial pressure of \mathrm{Hl}_{(\mathrm{g})} is 0.04 \mathrm{~atm}.What is \mathrm{K}_{\mathrm{o}} for the given equilibrium? 2 \mathrm{Hl}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{l}_{2}(\mathrm{~g})
A sample of {Hl}_{{g})} is placed in a flask at a pressure of 0.2 \mathrm{~atm}. At equilibrium, the partial pressure of \mathrm{Hl}_{(\mathrm{g})} is 0.04 \mathrm{~atm}.What is \mathrm{K}_{\mathrm{o}} for the given equilibrium? 2 \mathrm{Hl}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{l}_{2}(\mathrm{~g})

Answer:

The initial concentration of HI is 0.2 atm. It has a partial pressure of 0.04 atm when it is in equilibrium with the surrounding environment.

The pressure of HI drops by 0.2-0.04=0.16 as a result of the decrease in HI. The following is the suggested reaction:

2 \mathrm{Hl}(\mathrm{g}) \quad \rightleftharpoons \quad \mathrm{H}_{2}(\mathrm{~g})+\mathrm{l}_{2}(\mathrm{~g})

Evaluating the value of equilibrium constant at constant pressure,
K_{p=} \frac{p_{H_{2}} \times p_{l_{2}}}{p_{H I}^{2}}
=\frac{0.08 \times 0.08}{(0.04)^{2}}
=\frac{0.0064}{0.0016}
=4.0