Question

The equilibrium constant for the reaction $$\begin{array}{r} 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) \\ \text { is } K_{c}=1.3 \times 10^{-2} \text {at } 1000 \mathrm{~K} \text { . (a) At this tempera } \end{array}$$ (a) At this temperature does the equilibrium favor \(\mathrm{NO}\) and \(\mathrm{Br}_{2}\), or does it favor NOBr? (b) Calcu- $$\begin{array}{l} \text { late } K_{c} \text { for } 2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \text { . } \\ K_{c} \text { for } \mathrm{NOBr}(g) \rightarrow \mathrm{NO}(g)+\frac{1}{2} \mathrm{Br}_{2}(g) \end{array}$$ (c) Calculate

Short answer

(a) At the given temperature, the equilibrium favors the reactants, NO and Br2, since \(K_c = 1.3 \times 10^{-2} < 1\). (b) The equilibrium constant for the reverse reaction, \(2\,\mathrm{NOBr}(g) \rightleftharpoons \,2\,\mathrm{NO}(g) +\,\mathrm{Br}_{2}(g)\), is approximately \(K_c^{\prime} = 76.9\). (c) The equilibrium constant for the reaction \(\mathrm{NOBr}(g) \rightarrow \mathrm{NO}(g) + \frac{1}{2} \mathrm{Br}_{2}(g)\) is approximately \(K_c^{\prime\prime} = 8.77\).

Step-by-step solution

Determine the favorability at the given temperature

We're given the equilibrium constant, \(K_c = 1.3 \times 10^{-2}\). If \(K_c > 1\), the reaction favors the products, while if \(K_c < 1\), the reaction favors the reactants. Since \(K_c = 1.3 \times 10^{-2} < 1\), the reaction favors the reactants, which are \(\mathrm{NO}\) and \(\mathrm{Br}_{2}\).

Calculate the equilibrium constant for the reverse reaction

To find the equilibrium constant for the reverse reaction, \(2\,\mathrm{NOBr}(g) \rightleftharpoons \,2\,\mathrm{NO}(g) +\,\mathrm{Br}_{2}(g)\), we simply take the reciprocal of the given \(K_c\): $$K_c^{\prime}=\frac{1}{K_c}=\frac{1}{1.3\times10^{-2}}$$

Calculate the reciprocal value of \(K_c\)

Now we can plug in the value of \(K_c\) to find its reciprocal: $$K_c^{\prime}=\frac{1}{1.3\times10^{-2}}\approx 76.9$$ So, \(K_c'\) for the reverse reaction is approximately 76.9.

Determine the equilibrium constant for the half reaction

In order to find the equilibrium constant \(K_c^{\prime\prime}\) for \(\mathrm{NOBr}(g) \rightarrow \mathrm{NO}(g) + \frac{1}{2} \mathrm{Br}_{2}(g)\), we need to find the square root of the equilibrium constant for the reverse reaction, \(K_c^{\prime}\). Let's calculate: $$K_c^{\prime\prime} = \sqrt{K_c^{\prime}} = \sqrt{76.9} \approx 8.77$$ So, the equilibrium constant \(K_c^{\prime\prime}\) for the reaction \(\mathrm{NOBr}(g) \rightarrow \mathrm{NO}(g) + \frac{1}{2} \mathrm{Br}_{2}(g)\) is approximately 8.77.

Key concepts

Understanding Chemical Equilibrium

Chemical equilibrium is a fundamental principle in chemistry that occurs when the rates of the forward and reverse reactions are equal, leading to no net change in the concentration of reactants and products over time. Imagine a busy street where the number of cars entering is the same as the number of cars leaving; the total number of cars on the street remains constant, much like the concentrations of reactants and products at equilibrium.

Let's apply this to our problem with the reaction \[2 \text{NO}(g) + \text{Br}_2(g) \rightleftharpoons 2 \text{NOBr}(g)\] and its equilibrium constant \(K_c = 1.3 \times 10^{-2}\) at 1000 K. Since the value of \(K_c\) is less than 1, it suggests that, in this dynamic balance, the concentration of the reactants, NO and Br2, would be higher than the concentration of the product, NOBr, when the system has reached equilibrium at this temperature.

It's essential to understand that an equilibrium constant is a snapshot of a reaction at a specific moment and temperature, providing insight into the concentrations of the reacting species when the reaction has settled into a state of equilibrium.

Deciphering Reaction Favorability

Reaction favorability is often assessed using the equilibrium constant (\(K_c\)), which provides a clue to the extent of a chemical reaction at a particular temperature. A value of \(K_c > 1\) means that at equilibrium, the concentration of products is greater than that of reactants, implying the reaction 'favors' the formation of products. Conversely, a \(K_c < 1\) indicates that the reactants are favored.

In our exercise, since \(K_c\) is significantly less than 1, this tells us the reaction mixture comprises mainly of the reactants rather than the products at 1000 K. However, it's important to note that a low \(K_c\) doesn't mean the reaction won't happen at all—it simply reflects the position of equilibrium under the given conditions.

Applying Le Chatelier's Principle

Le Chatelier's Principle is the chemist's compass for predicting how a system at equilibrium reacts to changes in concentration, pressure, or temperature. It states that if an external change is applied to a system at equilibrium, the system adjusts itself to counteract the effect of the change and re-establishes equilibrium.

For instance, if we were to increase the concentration of NO or Br2 in our given reaction, Le Chatelier’s Principle guides us to expect a shift in the position of equilibrium to favor the production of NOBr in an attempt to reduce the stress caused by the addition of more reactants. Similarly, a change in temperature would shift the equilibrium to absorb the added heat or release it, depending on whether the reaction is exothermic or endothermic.

Understanding this principle is crucial, as it helps chemists to manipulate the conditions to steer reactions towards a desired product or optimize yield in industrial processes.