N2(g) + O2(g) >< N2O(g)

Question

At a very high temperature, nitrogen at an initial concentration of 1.20 mol/L is mixed with oxygen at an initial concentration of 1.00 mol/L. This reaction mixture reaches equilibrium when the concentration of the produced dinitrogen monoxide, commonly called nitrous oxide, us 1.00 mol/L

For the reaction described above, calculate the equilibrium concentration of O2(g).

Cant solve this question. Help much appreciated.

Answer 1

Hi Alex!

So the first thing you'd want to do for these types of problems is make sure that the equation is balanced. In this case, it isn't so we can go ahead and do that first. This makes the new equation 2N2(g)+O2(g)↔ 2N2O(g). For this problem, it's easiest to set up an ICE table (also called RICE table).

We are given the initial concentrations of both of the reactants which are 1.20 mol/L and 1.00 mol/L respectively. We can also assume that before the reaction took place, no products existed in solution, so the initial concentration of N₂O_(g) is 0.00 mol/L. We are also given the final concentration of the N₂O_(g) which is 1.00 mol/L so we can insert that into the table as well.

Now, we can take the final concentration of the products (1.00 mol/L) minus the initial concentration of the products (0.00 mol/L). 1.00-0.00= 1.00 mol/L so the change in concentration of the products is +1.00mol/L.

When we look back to out equation, we see that the coefficients on the N₂O_(g) and the N_2(g) are both 2. This means that they have the same mole ratio, so if the change in N₂O_(g) concentration was +1.00 mol/L, then the change in N_2(g) was -1.00 mol/L. To get the equilibrium concentration of N_2(g) we take 1.20 mol/L - 1.00 mol/L to give us 0.20 mol/L.

On the other hand, the coefficient on the oxygen gas is only a 1, therefore in order to get the final concentration we must take our 1.00mol/L change in product concentration and divide it by 2. This makes the change in concentration of O_2(g) -0.50 mol/L. Then we can take 1.00 mol/L - 0.50 mol/L = 0.50 mol/L to get the equilibrium concentration for O_2(g).

2N2(g)+O2(g) ↔ 2N2O(g)

I (initial) 1.20 mol/L 1.00 mol/L 0.00 mol/L

C (change) -1.00mol/L -0.50 mol/L +1.00 mol/L

E (equilibrium) 0.20 mol/L 0.50 mol/L 1.00 mol/L

I hope this helps!!