In the haber process for ammonia synthesis, k " 0.036 for n 2 (g) ! 3 h 2 (g) ∆ 2 nh 3 (g) at 500. k. if a 2.0-l reactor is charged with 1.42 bar of n 2 and 2.87 bar of h 2 , what will the equilibrium partial pressures in the mixture be?

Answer : The partial pressure of at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution :  Given,

Initial pressure of = 1.42 bar

Initial pressure of = 2.87 bar

= 0.036

The given equilibrium reaction is,

Initially                   1.42      2.87             0

At equilibrium    (1.42-x)  (2.87-3x)     2x

The expression of will be,

Now put all the values of partial pressure, we get

By solving the term x, we get

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of + Partial pressure of + Partial pressure of

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

well in order to answer this question you have to provide more information < 3

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"