7.11 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 ?

2HI (g) H2 (g) + I2 (g)

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:    

                            2HIg          H2g     +      I2gInitial conc.      0.2 atm                0                     0At equilibrium   0.04 atm           0.162              0.162                                                =0.08 atm      =0.08 atmTherefore,                               Kp=pH2×pI2p2HI                                    =0.08×0.080.042                                    =0.00640.0016                                    =4.0 

Hence, the value of Kp for the given equilibrium is 4.0.