Question

A study of the system, 4NH3(g) + 7O2(g) <--> 2N2O4(g) + 6H2O(g), was carried out. A system was prepared with [NH3] = [O2] = 3.60 M as the only components initially. At equilibrium, [N2O4] is 0.600 M. Calculate the equilibrium concentration of NH3(g).

Answer 1

The equilbrium concentration of NH₃(g) is 2.4 M

Step-by-step explanation:

The balanced reaction is:

4 NH₃(g) + 7 O₂(g) ⇔ 2 N₂O₄(g) + 6 H₂O(g)

By stoichiometry of the reaction, 2 moles of N₂O₄ are formed from 4 moles of NH₃.

Considering that the concentration is
`concentration=(number of moles)/(volume)` and with a volume of 1 liter, it is possible to apply the following rule of three: if 2 M of N₂O₄ are formed from 4 M of NH₃, 0.6 M of N₂O₄ from what concentration of NH₃ are formed?

`concentration of NH_(3)=(0.6 M of N_(2)O_(3) *4MofNH_(3) )/(2 M of N_(2)O_(3) )`

concentration of NH₃= 1.2 M

By subtracting the moles of NH3 in equilibrium from the moles of NH₃ initially, you will see how many moles of NH₃ were converted and remain in equilibrium: 3.6 M - 1.2 M= 2.4 M

The equilbrium concentration of NH₃(g) is 2.4 M