Problem 64 - Chemistry The Molecular Science

Question

\(K_{\mathrm{p}}\) for this reaction is 0.16 at \(25^{\circ} \mathrm{C}:\) $$ 2 \operatorname{NOBr}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{~g}) \quad \Delta_{\mathrm{r}} H^{\circ}=16.3 \mathrm{~kJ} / \mathrm{mol} $$ Predict the effect of each change on the position of the equilibrium; that is, state which way the equilibrium shifts (left, right, or no change) when each change is made. Assume constant-volume conditions for parts (a), (b), and (c). (a) Add more \(\mathrm{Br}_{2}(\mathrm{~g})\). (b) Remove some \(\mathrm{NOBr}(\mathrm{g})\). (c) Decrease the temperature. (d) Increase the container volume.

Step-by-Step Solution

Answer

(a) Shift left. (b) Shift left. (c) Shift left. (d) Shift right.

Step 1: Analyze the Reaction and Equilibrium Constant

The given reaction is \(2 \operatorname{NOBr}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{~g})\). The equilibrium constant \(K_{\mathrm{p}}\) is related to the concentrations of the products and reactants at equilibrium. It is stated that \(K_{\mathrm{p}} = 0.16\) at \(25^{\circ} \mathrm{C}\). This information indicates the balance between products and reactants under the given conditions.

Step 2: Consider Adding More Br2 (Effect on Equilibrium)

Adding more \(\mathrm{Br}_{2}(\mathrm{~g})\) increases the concentration of one of the products. According to Le Chatelier's principle, the system will shift to counteract the change, which means it will shift to the left, favoring the formation of reactants \(2 \operatorname{NOBr}(\mathrm{g})\) .

Step 3: Consider Removing Some NOBr (Effect on Equilibrium)

Removing \(\mathrm{NOBr}(\mathrm{g})\) will decrease the concentration of this reactant. According to Le Chatelier's principle, the system will shift to the left to increase the concentration of \(\mathrm{NOBr}(\mathrm{g})\), thus favoring the formation of \(2 \operatorname{NOBr}(\mathrm{g})\).

Step 4: Decrease Temperature (Effect on Equilibrium)

The reaction has a positive \(\Delta_{\mathrm{r}} H^{\circ}\) of \(16.3 \mathrm{~kJ} / \mathrm{mol}\), indicating it is endothermic. Reducing the temperature will result in a shift towards the exothermic side, which is the left, favoring the reactants \(2 \operatorname{NOBr}(\mathrm{g})\).

Step 5: Increase Container Volume (Effect on Equilibrium)

Increasing the container volume decreases the total pressure of the system. The reaction involves 3 moles of gas on the right (products) and 2 moles on the left (reactants). To counteract this decrease in pressure, the equilibrium will shift towards the side with more moles of gas to increase pressure, which is the right.

Key Concepts

Le Chatelier's Principle

Le Chatelier's Principle is a fundamental concept used to predict how a change in conditions can affect the position of equilibrium in a chemical reaction. In simple terms, if a system at equilibrium experiences a disturbance, such as a change in concentration, pressure, or temperature, it will adjust itself to minimize that disturbance and regain equilibrium.
For instance, in the exercise, adding more \( \mathrm{Br}_2(\mathrm{g}) \) results in a shift towards forming more reactants, NOBr. This is because the system acts to reduce the effect of added \( \mathrm{Br}_2 \). Similarly, if \( \mathrm{NOBr}(\mathrm{g}) \) is removed, the reaction will shift to produce more \( \mathrm{NOBr} \) by forming reactants.

• Adding reactants or removing products typically shifts the equilibrium to the right (towards more products).
• Adding products or removing reactants usually shifts it to the left (towards more reactants).

Understanding this principle is crucial for predicting and controlling chemical reactions under various conditions.

Equilibrium Constant

The equilibrium constant, often designated as \( K \), is a number that describes the ratio of concentrations of products to reactants at equilibrium for a given reaction at a specific temperature. In the given exercise, the equilibrium constant \( K_{\mathrm{p}} = 0.16 \) at \( 25^{\circ} \mathrm{C} \) shows that, at equilibrium, the concentration of the reactants is higher than that of the products.
An equilibrium constant:

• Larger than 1 indicates that at equilibrium, products are favored.
• Smaller than 1 suggests that reactants are favored, as in the exercise.

It's important to note that only changes in temperature can alter the value of \( K \). This is due to the fact that changing concentration or pressure can shift equilibrium positions, but won't change the \( K \) value itself. Understanding \( K \) helps in predicting the direction of reaction shifts and determining the extent to which a reaction will proceed.

Endothermic Reaction

An endothermic reaction is one that absorbs heat from its surroundings, which is indicated by a positive change in enthalpy (\( \Delta_{\mathrm{r}} H^{\circ} \)). The given exercise shows an enthalpy change of \( +16.3 \mathrm{~kJ/mol} \), confirming that the reaction consumes heat.
Endothermic reactions are sensitive to temperature changes. According to Le Chatelier's Principle, when the temperature decreases, the equilibrium will shift towards the side that generates heat to compensate for the loss; hence, it shifts left, towards the reactants, in this case.

• Heat can be considered a reactant for endothermic reactions.
• An increase in temperature will shift the equilibrium to the right, favoring the endothermic process.
• A decrease in temperature shifts equilibrium towards the left.

Understanding whether a reaction is endothermic or exothermic is key to predicting how temperature changes will affect equilibrium.