Unit 7: Topic 8 - Representations of Equilibrium

Question 1

How do we use particulate diagrams to compare equilibrium and nonequilibrium reactions?

Answer 1

We can understand equilibrium concentrations through particle diagrams. At equilibrium, K= [12]/[3][1]^2 which is 4. Reaction vessel A is Q=[14]/[3][2]^2 which is 1.167. QK, so the reaction will shift toward making more reactants.

Question 2

Here is the diagram to the question.
A) Which way has the reaction shifted?
B) What would it look like if the reaction shifted the other way?

Answer 2

Looking at the diagram, we can see that both reactants, X and Y2, have decreased while the product, 2XY, has increased. This means that the reaction has shifted in the forward direction.

If the reaction was reversed, it would look like this (many correct answers). Here, we chose to increase the reactant X by two particles. Because of the coefficients, Y2 will increase by half that amount which is one particle. On the other hand, the product XY will decrease by two because of its coefficient.

Question 3

Here is the diagram.
A) Determine the molarity of Q2, Z2, and QZ at equilibrium.
B) Write the equilibrium constant expression in terms of molarity and calculate the equilibrium constant.

Answer 3

There are 4 molecules of Q2 and Z2 and 3 molecules of QZ. However, as stated in the problem, each particle is 0.02 moles, not 1 mole. 4 particles * (0.02 mole/1 particle) = 0.08 moles for both Q2 and Z2 and 3 particles * (0.02 mole/1 particle) = 0.06 moles for QZ. However, molarity is moles/liters, and the reaction vessel is 2 liters. Thus, the molarity of Q2 and Z2 will be 0.04M while that of QZ would be 0.03M.

The equilibrium constant expression would be K= [QZ]^2 / [Q2][Z2]. When you plug in the molarities, the value of K would be 0.5625.