Question

NH₄NO₃(s) <---> N₂O (g) + 2H₂O(g) K = 0.665 at 525K

If a 2L reaction vessel contains 0.65 moles of N₂O (g), 1.2 moles of 2H₂O (g) and 1.3 moles of NH₄NO₃ (s), which of the following statements are correct?

a. The amount of N₂O (g) will increase and K will not change

b. The amount of N₂O (g) will increase and K will get larger

Please explain!!!!!!!!!!!!!!!

If the volume of the container is decreased, where will the equilibrium shift?

Answer 1

NH₄NO₃(s) <---> N₂O (g) + 2H₂O(g) K = 0.665 at 525K

If a 2L reaction vessel contains 0.65 moles of N₂O (g), 1.2 moles of 2H₂O (g) and 1.3 moles of NH₄NO₃ (s), which of the following statements are correct?

a. The amount of N₂O (g) will increase and K will not change

b. The amount of N₂O (g) will increase and K will get larger

Please explain!!!!!!!!!!!!!!!

If the volume of the container is decreased, where will the equilibrium shift?

First calculate the moalrities as follows:

N₂O (g) = 0.65 / 2.0 = 0.325 M

2H₂O (g) = 1.2 /2 = 0.6 M

NH₄NO₃(s) = 1.3/2 = 0.65 M

Now Q= [N₂O] [H₂O]^2/[ NH₄NO₃]

But for solid NH₄NO₃ = 1

Q= 0.325* (0.6)^2

= 0.117

And K = 0.665

Here Q < Keq thus the system will achieve equilibrium by
shifting to the right. The amount of N2O will increase and the K will not change

a. The amount of N₂O (g) will increase and K will not change

When there is a decrease in volume, the equilibrium will shift towards the side of the reaction with fewer moles.

NH₄NO₃(s) <---> N₂O (g) + 2H₂O(g) K = 0.665 at 525K

Thus this reaction shifts backwards side