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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 ⇌ H2(g) + I2(g). |
Answer» pHI = 0.04 atm pH2 = 0.08 atm pI2 = 0.08 atm Kp = \(\frac{pH_2 \times pI_2}{p^2HI}\) = \(\frac{(0.08\, atm)\times (0.08\, atm)}{(0.04\,atm)\times (0.04\,atm)}\) = 4 |
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