here is an interesting question about the chemical equilibrium and Le Chatelier's principle.

the following pressure show the partial that produced in the equilibrium reaction at time (t1 and t2)

$$\ce{N2 + 3H2 <=> 2NH3 \Delta H =-92 KJ}$$

at point t1 hydrogen is added to the previous equilibrium system at this point on the curve and after a period a new equilibrium state took place at point t2 so A = ... B = ... C = ...

Partial pressure graph

what i say is the answer is A=H2 ,B=N2 and C=NH3 becuase by adding more hydtogen thus increasing the conc. the reaction will move forward so the temp will increase increasing the pressure according to The Pressure Temperature Law so the the pressure of N2 increases but again also the pressure of NH3 should increase because the no of moles will increase as the reaction will shift forward so I am lost here


1 Answer 1


Since, H2 is added at the equilibrium state, it's initial pressure will be more and starts decreasing as it goes into the reaction with N2. This is clearly depicted by graph A. While the pressure of N2 starts decreasing from the level it was in equilibrium state, represented by graph C. Also, the reaction shifts forwards in order to gain equilibrium, pressure of NH3 will increase. So, the correct answer would be,

A = H2, B = NH3, C = N2


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