Answer: The equilibrium partial pressure of [tex]Cl_{2}[/tex] is 0.964 atm.
Explanation:
Given: [tex]K_{p} = 0.636[/tex]
[tex]P_{COCl_{2}} = 0.836 atm[/tex]
[tex]P_{CO} = 0.551 atm[/tex]
The given reaction equation is as follows.
[tex]COCl_{2}(g) \rightleftharpoons CO(g) + Cl_{2}(g)[/tex]
Formula used to calculate the partial pressure of [tex]Cl_{2}[/tex] is as follows.
[tex]K_{p} = \frac{P_{CO} \times P_{Cl_{2}}}{P_{COCl_{2}}}[/tex]
Substitute the values into above formula as follows.
[tex]K_{p} = \frac{P_{CO} \times P_{Cl_{2}}}{P_{COCl_{2}}}\\0.636 = \frac{0.551 atm \times P_{Cl_{2}}}{0.836 atm}\\P_{Cl_{2}} = 0.964 atm[/tex]
Thus, we can conclude that the equilibrium partial pressure of [tex]Cl_{2}[/tex] is 0.964 atm.
