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Question
A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium, the partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium?
\[\ce{2HI (g) ⇌ H2 (g) + I2 (g)}\]
Solution
The initial concentration of HI is 0.2 atm. At equilibrium, it has a partial pressure of 0.04 atm. Therefore, a decrease in the pressure of HI is 0.2 - 0.04 = 0.16. The given reaction is:
| 2HI(g) | ↔ | H2(g) | I2(g) | |
| Initial conc. | 0.2 atm | 0 | 0 | |
| At equilibrium | 0.04 atm | `0.16/2` | `2.15/2` | |
| = 0.08 atm | = 0.08 atm |
Therefore
`"K"_"p" = ("p"_("H"_2) xx "p"_("I"_2))/("p"_("HI")^2)`
`= (0.08 xx 0.08)/((0.04)^2)`
`= 0.0064/0.0016`
= 4.0
Hence, the value of Kp for the given equilibrium is 4.0.
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