AP Chemistry Gas Laws

Question

 Gases N2 (g) and H2 (g) are added to a previously evacuated container and react at a constant temperature according to the following chemical equation:

N2 (g)  + H2 (g) →  NH3 (g)

     If the initial pressure of N2 (g) was 1.2 atm, and that of H2 (g) was 3.8 atm, what is the partial pressure of NH3 (g)   when the partial pressure of N2 (g) has decreased to 0.900 atm? (*** BALANCE 1st!!!)



a. 0.30 atm b. 0.60 atm c. 0.90 atm d. 1.80 atm e. 3.80 atm

Answer 1

The previous answer provided by Anthony T. is perfectly correct, but you might be better able to understand it (see it) if you set up an ICE table. Hopefully, you are familiar with ICE tables.

N₂ + 3H₂ ==> 2NH₃ ... balanced equation

1.2......3.8.........0........Initial

-0.3...-0.9........+0.6....Change

0.9....2.9..........0.6.....Equilibrium

So answer is 0.6 atm



So, there is no need to use 3.8 atm when solving. You simply just fill in the ICE table with the information provided. You are told the initial pressure (I) of N2 is 1.2 and the final (E) is 0.9 so the change (C) must be 0.3. Once you know that, you can fill in the rest of the ICE table because the change (C) for H2 will be 3x that of N2 (3:1 in balanced equation), etc.

Answer 2

The partial pressures of the gases are proportional to the number of moles at constant temperature and volume. The balanced equation for the reaction is N₂ + 3H₂ → 2NH₃. This means that for every mole of N₂ that reacts 2 moles of NH₃ are formed. As the partial pressure of N₂ decreased to 0.900 atm, this means that 0.30 atm of N₂ reacted. As the ratio of moles NH₃ produced is twice the number of N₂ moles reacted, the partial pressure of NH₃ would be 0.60 atm.