Question

$N_2\left(g\right)+{3H}_2\left(g\right)\rightleftharpoons 2{NH}_3\left(g\right)+heat$
The reaction above is carried out in a sealed vessel.
Which changes would both shift the equilibrium position to favor in the formation of products?

• A. heat the vessel, Increase the pressure
• B. cool the vessel, Increase the pressure
• C. heat the vessel, decrease the pressure
• D. cool the vessel, decrease the pressure

Answer 1

The correct option is B cool the vessel, Increase the pressure

- The pressure
In the reaction, $N_2\left(g\right)+3H_2\left(g\right)<-->2N{H}_3\left(g\right)$ notice that there are 4 molecules on the left-hand side of the equation, but only 2 on the right. According to Le Chatelier's Principle, the system will respond by favouring the reaction which produces fewer molecules. That will cause the pressure to fall again. Hence, the reaction will be more product favoured.
- The temperature
In order to produce the maximum possible amount of ammonia in the equilibrium mixture. The forward reaction $N_2\left(g\right)+3H_2\left(g\right)<-->2N{H}_3\left(g\right)$ (the production of ammonia) is exothermic. According to Le Chatelier's Principle, this will be favoured if you lower the temperature. , a very low temperature will cause a reaction to occur very slowly and hence, not efficient. Therefore, 400 - 450°C is a compromise temperature producing a high proportion of ammonia in the equilibrium mixture