For the equilibrium 2 ibr (g) i2 (g) + br2 (g) kp=8.5 ×10-3 at 150 oc. if 0.025 atm of ibr is placed in a 2.0-l container, what is the partial pressures of all substances after equilibrium is reached?

Answer : At equilibrium, the partial pressure of IBr, and is 0.0211 atm, 0.00195 atm and 0.00195 atm respectively.

Solution : Given,

Initial partial pressure = 0.025 atm

The given equilibrium reaction is,

initial pressure          0.025 atm          0         0

At equilibrium         (0.025 - 2x)          x          x

The expression of equilibrium constant for this reaction is,

Now put all the given values in this expression, we get value of 'x'.

By rearranging the terms, we get the value of 'x'.

x = 0.00195

Therefore,

The partial pressure of IBr = (0.025 - 2x) = (0.025 - 2(0.00195)) = 0.0211 atm

The partial pressure of = x = 0.00195 atm

The partial pressure of = x = 0.00195 atm

b is the correct

explanation:

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