At a certain temperature, the following reaction is at equilibrium: N2 ( 𝑔 ) + 3 H2 ( 𝑔 ) ⇌ 2 NH3 ( 𝑔 ) + heat Predict the direction the equilibrium will shift in response to each change.

Question

a) Predict the direction the equilibrium will shift in response to each of the following changes. Explain your reasoning using Le Châtelier’s Principle.

1. Additional H₂ gas is added to the system.
2. Ammonia (NH₃) is removed from the container.
3. The temperature is increased.
4. The volume of the container is decreased.

b) If the equilibrium constant K for this reaction is 0.50 at a certain temperature, what does this value suggest about the relative amounts of reactants and products at equilibrium?

Answer 1

For part (d), decreasing the volume (keeping all else the same) will increase the concentration (number per volume) of each species (because they are gases). Therefore, Le Chatelier's principle says the system will react so as to undo the change and will try to reduce the concentration of particles. By shifting away from reactants (4 moles on the LHS) to products (only 2 moles on the RHS), all the atoms can be regrouped into fewer molecules, leaving a smaller number of particles in the reaction chamber ( nnnn/V --> nn/V ), reducing the overall concentration of gas particles back down towards the original value.

Answer 2

When it comes to Le Châtelier’s Principle, the name of the game is move towards equilibrium after a perturbation. If we add more of something, it should decompose or be used up in a reaction, and if we remove something, it should be formed more rapidly.

a1) When we add more hydrogen gas we have moved from equilibrium to being reactants heavy. The system will then move away from this by forming more product more rapidly.

b1) By removing ammonia from the container, we have again biased our chemical system to reactants instead of being at equilibrium. To make up for this, we form more ammonia.

c1) When it comes to temperature, we can think of it as product as it is produced from the forward reaction. By adding more heat, we are biasing the system to products, and we will start to form our reactants.



d1) Volume has a bit trickier of a relationship when it comes to equilibrium. Since we are in the gas phase, we need to look at how many moles of gas is present on each side of the reaction in order to determine what direction will move. Here, on the left side we have four moles of gas, and on the right side we only have two moles of gas. If we reduce the volume of our reaction chamber, it is a lot easier to fit more of our two moles of products than it is to keep our four moles of reactants, thus, the reaction will form more products. If the inverse were true, we had fewer moles on the reactants side of things, then we would form more reactants.

For our equilibrium constant, we have to first remember what a rate constant is. If we have the reaction

aA + bB -> cC + dD

we end up with

K = ([C]^c[D]^d)/([A]^a[B]^b).

Here, our expression for the equilibrium would be

0.5 = ([NH₃]²)/([N₂][H₂]³)

From this, we can tell that we are going to have significantly more moles of reactants than we are going to have products, as a K constant less than one necessitates a greater number of reactants in order to get so low.

Answer 3