Question

A nitrogen-hydrogen mixture initially in the molar ratio of 1 : 3 reached equilibrium to form (d) 20 ammonia when 25% of the n, and h, had reacted. If the total pressure of the system was atm, the partial pressure of ammonia at the equilibrium was

Answer 1

Given :

Molar ratio of Nitrogen-Hydrogen = 1:3

Percentage of Nitrogen and Hydrogen reacted = 25%

Total pressure of the system = 21 atm

To Find :

Partial pressure of Ammonia at equillibrium

Solution :

• The reaction between nitrogen and hydrogen is

                               $N_{2} + 3H_{2} \Longleftrightarrow 2NH_{3}$

          At t=0            a       3a             0

         At t=t₁           a-x    3a-3x         2x

• Total number of moles = $(a-x) + (3a-3x) + 2x$

                          $= 4a -2x$    

                          $=4a - \frac{a}{2}$      

                          $=\frac{7a}{2}$

• Partial pressure of NH₃ = $\frac{Noles\:of\:NH_{3} }{Total\:no\:molee} \times Total\:pressure$

                         $=\frac{\frac{a}{2} }{\frac{7a}{2} } \times 21$

                         $=3atm$

       The partial pressure of NH₃ is 3atm