Question

Consider the following statements: "Consider the reaction $\mathrm{A}(g)+\mathrm{B}(g)\rightleftharpoons \mathrm{C}(g),$ for which at equilibrium $[\mathrm{A}]=2M,[\mathrm{B}]=1\mathrm{M},$ and $[\mathrm{C}]=4\mathrm{M}.$ To a $1-\mathrm{L}$ con- tainer of the system at equilibrium you add 3 moles of B. A possible equilibrium condition is $[\mathrm{A}]=1\mathrm{M}$ $[\mathrm{B}]=3M,$ and $[\mathrm{C}]=6M,$ because in both cases, $K=$ 2." Indicate everything you think is correct in these statements, and everything that is incorrect. Correct the incorrect statements, and explain.

Short answer

In the given statements, the initial equilibrium constant calculation is correct, as is the fact that K remains the same in both cases. However, the new equilibrium concentration of B should be 4 M instead of the given 3 M, as 3 moles of B are added to the system.

Step-by-step solution

Calculate the initial equilibrium constant (K)

We are given the initial equilibrium concentrations: [A] = 2 M [B] = 1 M [C] = 4 M For the reaction A(g) + B(g) ⇌ C(g), we can write the expression for the equilibrium constant (K) as: $$K=\frac{[C]}{[A][B]}$$ Now, plug the initial equilibrium concentrations into the expression to calculate K: $$K=\frac{4}{(2)(1)}=2$$

Analyze the new equilibrium condition provided in the statement

We're given new equilibrium concentrations after the addition of 3 moles of B: [A] = 1 M [B] = 3 M [C] = 6 M Calculate the new value of K based on the given concentrations: $$K^{\prime }=\frac{[C]}{[A][B]}=\frac{6}{(1)(3)}=2$$

Compare the initial and new equilibrium constants

Now, we compare the initial equilibrium constant K and the new equilibrium constant K': $$K=2$$ $$K^{\prime }=2$$ Since K = K', the statement is correct that K remains the same in both cases.

Analyze the statement about adding 3 moles of B

It's stated that 3 moles of B are added to the 1 L container when the system is at equilibrium. The total volume remains 1 L.

Verify if the new equilibrium concentrations are correct

Before adding 3 moles of B, [B] = 1 M When we add 3 moles of B to the 1 L container, the new concentration of B becomes: [B] = 1 M + 3 M = 4 M However, in the new equilibrium condition provided in the statement, it's given that [B] = 3 M, which is incorrect as it should be 4 M. To get the correct new equilibrium concentrations, use the reaction quotient (Q) and the value of K that was calculated earlier, and then solve for the new equilibrium concentrations using the ICE table method. In conclusion, most of the initial statements are correct, only the final concentration of B should be 4 M instead of 3 M.

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as K, is a fundamental concept in the study of chemical equilibrium. It is a numerical value that signifies the ratio of the concentrations of products to reactants for a reversible reaction at equilibrium, each raised to the power of their respective stoichiometric coefficients. In our exercise, the reaction is balanced as $\text{A}(g)+\text{B}(g)\rightleftharpoons \text{C}(g)$, and the equilibrium constant expression is $$K=\frac{[C]}{[A][B]}$$. The values given in the exercise yield an equilibrium constant, K, of 2. This value of K does not change unless there is a change in temperature; it is immutable regardless of the initial concentrations of the reactants and products.
The key takeaway here is that the equilibrium constant provides insight into the position of equilibrium and the tendency of a reaction to form products or reactants. A large K value indicates a mixture rich in products at equilibrium, while a small K value suggests a reactant-favored mixture.

Reaction Quotient

The reaction quotient, represented as Q, is a measure used to determine the direction in which a reaction will proceed to reach equilibrium. It has the same form as the equilibrium constant expression but is calculated using the concentrations of the reactants and products at any point in time, not necessarily at equilibrium. For the reaction in question, the expression for Q is the same as that for K: $$Q=\frac{[C]}{[A][B]}$$.
By comparing the values of Q and K, one can predict the reaction's behavior. If $Q<K$, the reaction will proceed in the forward direction to produce more products and reach equilibrium. Conversely, if $Q>K$, the reaction will shift in reverse to generate more reactants. If $Q=K$, the system is already at equilibrium and no shift is expected. In the exercise, after adding 3 moles of B, a new reaction quotient should be calculated to predict the shift needed to re-establish equilibrium.

ICE Table Method

The ICE table method is a systematic approach to solving equilibrium problems. 'ICE' stands for Initial, Change, and Equilibrium, reflecting the concentration stages a system goes through when moving toward equilibrium. This method is extraordinarily helpful when incorporating changes to a system at equilibrium, such as adding extra reactant or product.
For the given scenario, after we add 3 moles of $\text{B}$ to the system, we need to create an ICE table to determine the changes in concentrations of $\text{A}$, $\text{B}$, and $\text{C}$ as the system shifts to a new equilibrium. The addition of $\text{B}$ will be reflected in the 'Change' row, typically showing decreases in the concentrations of $\text{A}$ and $\text{B}$ and an increase in $\text{C}$.
By error, the provided exercise stated that the concentration of $\text{B}$ would be 3 M after adding 3 moles to a 1 L container, which is not correct. The concentration should actually be 4 M. Using the ICE table, we would calculate the equilibrium concentrations from these new initial concentrations, and adjust the amounts accordingly, considering the stoichiometry of the reaction and the previously established K value.