Problem 71 - Chemistry: Principles and Reactions

Question

Consider the system $$\mathrm{A}(g)+2 \mathrm{~B}(g)+\mathrm{C}(s) \rightleftharpoons 2 \mathrm{D}(g)$$ at \(25^{\circ} \mathrm{C}\). At zero time, only \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\) are present. The reaction reaches equilibrium \(10 \mathrm{~min}\) after the reaction is initiated. Partial pressures of \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{D}\) are written as \(P_{\mathrm{A}}, P_{\mathrm{B}}\), and \(P_{\mathrm{D}}\). Answer the questions below, using LT (for is less than), GT (for is greater than), EQ (for is equal to), or MI (for more information required). (a) \(P_{\mathrm{D}}\) at \(11 \mathrm{~min}\) ________ \(P_{\mathrm{D}}\) at \(12 \mathrm{~min} .\) (b) \(P_{\mathrm{A}}\) at \(5 \mathrm{~min}\) \(P_{\mathrm{A}}\) ______ at \(7 \mathrm{~min}\) (c) \(K\) for the forward reaction ______ \(K\) for the reverse reaction. (d) At equilibrium, \(K\)______Q. (e) After the system is at equilibrium, more of gas \(\mathrm{B}\) is added. After the system returns to equilibrium, \(K\) before the addition of \(B\) \(K\) _____ after the addition of \(\mathrm{B}\). (f) The same reaction is initiated, this time with a catalyst. \(K\) for the system without a catalyst _____ \(K\) for the system with a catalyst. (g) \(K\) for the formation of one mole of \(\mathrm{D}\) \(K\) _____ for the formation of two moles of \(\mathrm{D}\). (h) The temperature of the system is increased to \(35^{\circ} \mathrm{C} . P_{\mathrm{B}}\) at equilibrium at \(25^{\circ} \mathrm{C} \longrightarrow P_{\mathrm{B}}\) _______at equilibrium at \(35^{\circ} \mathrm{C}\). (i) Ten more grams of \(\mathrm{C}\) are added to the system. \(P_{\mathrm{B}}\) before the addition of \(\mathrm{C} \quad P_{\mathrm{B}}\) _____ after the addition of \(\mathrm{C}\).

Step-by-Step Solution

Answer

Question: Compare the partial pressure of D at 11 minutes and 12 minutes. Answer: EQ Question: Compare the partial pressure of A at 5 minutes and 7 minutes. Answer: GT Question: Compare the equilibrium constant for the forward and reverse reactions. Answer: EQ Question: Compare the equilibrium constant and the reaction quotient at equilibrium. Answer: EQ Question: Compare the equilibrium constant before and after the addition of B. Answer: EQ Question: Compare the equilibrium constant for the system without and with a catalyst. Answer: EQ Question: Compare the equilibrium constant for the formation of one mole and two moles of D. Answer: EQ Question: Compare the partial pressure of B at equilibrium at 25°C and 35°C. Answer: MI Question: Compare the partial pressure of B before and after the addition of C. Answer: EQ

Step 1: Compare \(P_{\mathrm{D}}\) at 11 min and 12 min

The reaction reaches equilibrium at 10 min, so there will be no change in partial pressures after 10 min. Therefore, \(P_{\mathrm{D}}\) at 11 min = \(P_{\mathrm{D}}\) at 12 min: Answer: EQ (b)

Step 2: Compare \(P_{\mathrm{A}}\) at 5 min and 7 min

Both time points are before the system reaches equilibrium (at 10 min). As the reaction progresses, the concentration of A decreases. So, \(P_{\mathrm{A}}\) at 5 min > \(P_{\mathrm{A}}\) at 7 min: Answer: GT (c)

Step 3: Compare K for the forward and reverse reactions

The equilibrium constant (K) is the same for both forward and reverse reactions at a given temperature. Therefore, K for the forward reaction = K for the reverse reaction: Answer: EQ (d)

Step 4: Compare K and Q at equilibrium

At equilibrium, the reaction quotient (Q) has the same value as the equilibrium constant (K). Therefore, K=Q: Answer: EQ (e)

Step 5: Compare K before and after the addition of B

At a constant temperature, the value of K does not change even though the concentration of the reactants or products change. Therefore, K before the addition of B = K after the addition of B: Answer: EQ (f)

Step 6: Compare K for the system without and with a catalyst

A catalyst affects the rate at which equilibrium is reached but does not affect the equilibrium position itself. Therefore, K for the system without a catalyst = K for the system with a catalyst: Answer: EQ (g)

Step 7: Compare K for the formation of one mole and two moles of D

This question is regarding the same reaction, so the equilibrium constant remains the same. K for the formation of one mole of \(\mathrm{D}\) = K for the formation of two moles of \(\mathrm{D}\): Answer: EQ (h)

Step 8: Compare \(P_{\mathrm{B}}\) at equilibrium at 25°C and 35°C

More information is required to determine if the reaction is endothermic or exothermic, as this will dictate how the partial pressure of B is affected by a temperature change. Therefore: Answer: MI (i)

Step 9: Compare \(P_{\mathrm{B}}\) before and after the addition of C

Adding solid \(\mathrm{C}\) to the system does not change the partial pressure of the gases in equilibrium since it does not participate in the gaseous phase. Therefore, \(P_{\mathrm{B}}\) before the addition of \(\mathrm{C}\) = \(P_{\mathrm{B}}\) after the addition of \(\mathrm{C}\): Answer: EQ

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as K, is a crucial concept in chemical equilibrium. It quantifies the ratio of the concentrations of products to reactants at equilibrium for a reversible reaction. In a general reaction where aA + bB ⇌ cC + dD, the equilibrium constant K is represented as
\[ K = \frac{{[C]^c[D]^d}}{{[A]^a[B]^b}} \]
provided the substances are in the same phase. For gases, concentrations are often expressed in terms of partial pressures denoted as \( P \). In the provided exercise, the equilibrium constant would be based on the partial pressures of gases A, B, and D. An essential point to remember is that the equilibrium constant is only affected by changes in temperature. Introducing additional reactants, products or catalysts, as seen in the example, does not change the value of K. This understanding helps justify why the equilibrium constant for both forward and reverse reactions, as well as for different amounts of product formation, remains equal.

Reaction Quotient

The reaction quotient, Q, is similar to the equilibrium constant, but it applies to any point in time during the reaction, not just at equilibrium. It uses the same formula as K but with the actual concentrations (or partial pressures) at a specific moment. Comparison of Q to K tells you which direction the reaction will shift to reach equilibrium.
\[ Q = \frac{{[C]^c[D]^d}}{{[A]^a[B]^b}} \]
If Q < K, the reaction will proceed forward to produce more products. Conversely, if Q > K, the reaction will shift to form more reactants. At equilibrium, Q equals K, and there's no net change in the concentrations of reactants and products, as indicated in step 4 of the solution.

Le Chatelier's Principle

Le Chatelier's Principle predicts how a system at equilibrium will respond to changes in concentration, pressure, volume, or temperature. This principle states that if an external stress is applied to a system at equilibrium, the system will adjust to minimize the stress and establish a new equilibrium. Adding more reactant or product (as in step e), changing the pressure by altering the volume of the system, or changing the temperature can all stress the equilibrium. For example, increasing the temperature of an exothermic reaction will shift the equilibrium towards the reactants. However, the value of K will remain constant unless the temperature changes, despite the shift in position of the equilibrium. Hence, as seen in the exercise, the addition of compound B would not alter the equilibrium constant but would shift the equilibrium position to oppose the change, producing more D according to Le Chatelier's principle.

Effect of Catalyst on Equilibrium

Catalysts play a pivotal role in increasing the rate at which equilibrium is reached without affecting the position of equilibrium itself. They achieve this by providing an alternative reaction pathway that has a lower activation energy. As a result, both the forward and reverse reactions are sped up equally, allowing the system to reach equilibrium faster. For the exercise at hand, introducing a catalyst (step f) will not alter the value of the equilibrium constant K, as supported by the provided answer. This is a common misconception; while catalysts affect kinetics (the rate of reaction), they do not affect thermodynamics (the equilibrium constant regarding the concentration of reactants and products at equilibrium).