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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 K_p for the given equilibrium?
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 atm00At equilibrium0.04 atm0.1622.152=0.08 atm =0.08 atm
Therefore,
Kp=PH2×PI2PHI=0.08×0.08(0.04)2=0.00640.0016=4.0
Hence, the value of Kp for the given equilibrium is 4.0.


Chemistry
NCERT Textbook
Standard XI

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