Question

For the reaction $$2 \mathrm{NBr}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{Br}_{2}(g)$$ the system at equilibrium at a particular temperature is analyzed, and the following concentrations are found: \(\left[\mathrm{NBr}_{3(g)\right]=2.07 \times 10^{-3} \mathrm{M} ; \quad\left[\mathrm{N}_{2}(g)\right]=\) \(4.11 \times 10^{-2} M ;\left[\mathrm{Br}_{2}(g)\right]=1.06 \times 10^{-3} \mathrm{M} .\) Calculate the value of \(K\) for the reaction at this temperature.

Short answer

The value of the equilibrium constant K for this reaction at the given temperature is \(1.08 \times 10^{2}\).

Step-by-step solution

1. Write the equation for the equilibrium constant, K

For a general reaction, the equilibrium constant K can be defined as: \[K = \frac{\text{[Products]}}{\text{[Reactants]}}\] In this case, the reaction is given by: \[2 \mathrm{NBr}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g) + 3 \mathrm{Br}_{2}(g)\] From this reaction, we can deduce the equilibrium constant expression as follows: \[K = \frac{[\mathrm{N}_{2}]^{1}[\mathrm{Br}_{2}]^{3}}{[\mathrm{NBr}_{3}]^{2}}\]

2. Substitute the given concentrations into the equilibrium constant expression

Now, we need to substitute the given concentrations into the equilibrium constant expression to find the value of K. Given: - \([\mathrm{NBr}_{3}] = 2.07 \times 10^{-3} \mathrm{M}\) - \([\mathrm{N}_{2}] = 4.11 \times 10^{-2} \mathrm{M}\) - \([\mathrm{Br}_{2}] = 1.06 \times 10^{-3} \mathrm{M}\) Substituting these values into the equilibrium constant expression: \[K = \frac{(4.11 \times 10^{-2})^{1}(1.06 \times 10^{-3})^{3}}{(2.07 \times 10^{-3})^{2}}\]

3. Calculate the value of K

Now, we will perform the calculations to find the value of K. \[K = \frac{(4.11 \times 10^{-2})(1.06 \times 10^{-3})^{3}}{(2.07 \times 10^{-3})^{2}} = \frac{4.11(1.19\times10^{-9})}{4.28\times10^{-6}} = 1.08 \times 10^{2}\] Therefore, the value of the equilibrium constant K for this reaction at the given temperature is \(1.08 \times 10^{2}\).

Key concepts

Chemical Equilibrium

Understanding chemical equilibrium is critical for grasping how reactions occur in a closed system. It's the state where the rates of the forward and reverse reactions are equal, leading to no observable changes in the concentrations of reactants and products over time. In the given exercise,
the equilibrium of the decomposition of NBr_3 to N_2 and Br_2 is considered. At equilibrium, both the breaking of bonds in NBr_3 and the formation of NBr_3 from N_2 and Br_2 happen at the same rate. This state can be reached from any starting point, whether you begin with reactants or products.
It's crucial to note that equilibrium doesn't mean that the amounts of reactants and products are equal. Rather, their rates of formation are what's balanced. Different reactions have different equilibrium positions—a concept quantitatively described by the equilibrium constant (K).

Reaction Quotient

When a reaction hasn't yet reached equilibrium, we use the reaction quotient (Q) to determine the direction the reaction will proceed to achieve equilibrium. It is calculated using the same formula as the equilibrium constant (K). However, it doesn't require the system to be at equilibrium. Instead, it involves the initial concentrations or partial pressures of the reactants and products.

Comparing Q to K

• If Q < K, the reaction will proceed in the forward direction to produce more products.
• If Q > K, the reaction will go in reverse to produce more reactants.
• If Q = K, the system is already at equilibrium.

The reaction quotient is useful for predicting the effect of changing conditions on the position of equilibrium, as seen in the application of Le Chatelier's principle.

Equilibrium Constant Expression

The equilibrium constant expression represents the ratio of product concentrations to reactant concentrations at equilibrium, with each concentration raised to a power corresponding to its coefficient in the balanced chemical equation. For the reaction in our exercise, the formula in a general form looks like this:
\[\begin{equation}K = \frac{[\text{Products}]}{[\text{Reactants}]}\end{equation}\]
In the specific example, we have:

• The product N_2 appearing once in the balanced equation, thus its concentration is raised to the first power.
• The product Br_2 appears thrice, raising its concentration to the third power.
• The reactant NBr_3 has a coefficient of two, so its concentration is squared in the denominator.

This expression must be numerically evaluated at equilibrium to calculate the equilibrium constant (K). The value of K is a ratio that can suggest the extent of the reaction, but it doesn't give any information about the speed of the reaction.

Le Chatelier's Principle

Le Chatelier's principle is a rule of thumb that describes how a system at equilibrium responds to disturbances. It states that if a system at equilibrium is disturbed, the system will adjust itself in such a way as to counteract the disturbance and restore a new equilibrium state.
Some typical changes that can affect the equilibrium include alterations in concentration, temperature, and pressure. For instance, increasing the concentration of reactants pushes the equilibrium towards more products, in response to the principle. In the context of the given reaction, adding more NBr_3 would shift the equilibrium to the right to produce more N_2 and Br_2. Similarly, reducing the temperature for an exothermic reaction will shift the equilibrium towards the products to release heat and offset the temperature change. By understanding this principle, chemists can manipulate the conditions to favor the production of a particular substance.