Consider the potential energy diagrams for two types of reactions \(\mathrm{A} \rightleftharpoons \mathrm{B}\). In each case, answer the following questions for the system at equilibrium. (a) How would a catalyst affect the forward and reverse rates of the reaction? (b) How would a catalyst affect the energies of the reactant and product? (c) How would an increase in temperature affect the equilibrium constant? (d) If the only effect of a catalyst is to lower the activation energies for the forward and reverse reactions, show that the equilibrium constant remains unchanged if a catalyst is added to the reacting mixture.

Short Answer

Expert verified
A catalyst increases both forward and reverse reaction rates without affecting the energies of the reactants and products, nor the equilibrium constant. An increase in temperature affects the equilibrium constant based on whether the reaction is exothermic or endothermic. Despite the catalyst speeding up reactions, the equilibrium constant remains the same because it doesn’t change the ratios of product and reactant concentrations at equilibrium.

Step by step solution

01

Understanding Catalysts in Reactions

A catalyst is a substance that speeds up a chemical reaction, but is not consumed in the reaction. It achieves this by providing an alternative reaction pathway with a lower activation energy. However, a catalyst does not change the energies of the reactants and products. Therefore, the forward and reverse rates of the reaction will increase equally as the catalyst lowers the activation energy barrier for both reactions.
02

Effect of Catalysts on Energy Levels of Reactants and Products

A catalyst does not affect the energies of the reactants and products. It only lowers the energy of the transition state. Therefore, the introduction of a catalyst does not change the relative energy levels of the reactant A and product B.
03

Impact of Temperature on Equilibrium Constant

The equilibrium constant \(K\) is temperature-dependent. According to the Van't Hoff equation, if the reaction is exothermic (\( \Delta H < 0\)), an increase in temperature will reduce the equilibrium constant, shifting the equilibrium to the left (favoring the reactants). If the reaction is endothermic (\( \Delta H > 0\)), an increase in temperature will increase the equilibrium constant, shifting the equilibrium to the right (favoring the products).
04

Catalysts and the Equilibrium Constant

Even though a catalyst accelerates both the forward and reverse reactions, it does not have an effect on the equilibrium constant, \(K\). This is because the equilibrium constant is a ratio of the concentrations of the products to the reactants when the reaction is at equilibrium, and the catalyst does not change this ratio. It only enables the system to reach equilibrium faster. Therefore, \(K\) remains unchanged even if a catalyst is added to the reacting mixture.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Catalysts in Chemical Reactions
In chemical reactions, catalysts play a crucial role in influencing the rate at which reactions occur. They function by offering an alternate pathway for the reactants to transform into products, often with less energy required, which is referred to as lower activation energy. Crucially, catalysts are not consumed or altered permanently during the reaction; they facilitate the reaction without becoming part of the final products.

When a catalyst is introduced to a reaction system, both the forward and reverse reactions are accelerated equally. This parallel increase is because the catalyst lowers the activation energy for both directions of the reaction, enabling the reactants to more easily reach the transition state, which is the highest energy state that occurs during the reaction. However, since the catalyst does not affect the energy levels of the reactants or the products, there is no change in the overall energy landscape of the reaction. This means that the introduction of a catalyst will not alter the equilibrium between the reactants and products.
Activation Energy
Activation energy is the energy barrier that reactants must overcome to form products in a chemical reaction. This energy hurdle is crucial because it determines the rate at which a reaction will proceed. A high activation energy correlates with a slower reaction rate because fewer reactant molecules will have sufficient energy to reach the transition state.

To facilitate reactions, nature and humans alike employ catalysts to lower the activation energy. By decreasing this energy requirement, catalysts enable more reactant molecules to reach the transition state and hence react, even at lower temperatures. This does not, however, change the inherent stability of the reactants and products; it simply helps the reaction to occur more readily. Understanding activation energy is key to manipulating chemical processes to our advantage, such as in industrial synthesis and enzyme activity in biological systems.
Equilibrium Constant
The equilibrium constant, denoted as \(K\), is a fundamental concept in chemical thermodynamics that quantifies the ratio of product concentrations to the reactant concentrations at chemical equilibrium. Its value reflects the extent of the reaction; a large \(K\) implies a reaction that favors the formation of products, whereas a small \(K\) indicates a reaction that favors the reactants.

The equilibrium constant is experimentally determined and only changes with temperature. In the context of catalysts, while they speed up the attainment of equilibrium, they have no effect on the equilibrium constant because they do not alter the concentrations of reactants and products at equilibrium. Understanding \(K\) helps predict the direction of the reaction's shift when changes, like temperature variations, are applied to the system.
Van't Hoff Equation
The Van't Hoff equation is an essential principle in chemical thermodynamics that depicts the relationship between the equilibrium constant and temperature. Represented as \( \frac{d \ln K}{dT} = \frac{\Delta H}{R T^2} \), it mathematically expresses how changes in temperature will cause the equilibrium constant, \(K\), to change.

Understanding the Van't Hoff Equation

At the core of the Van't Hoff equation is the enthalpy change, \(\Delta H\), of the reaction. If the reaction is exothermic (releases heat, \(\Delta H < 0\)), an increase in temperature typically results in a decreased \(K\), favoring reactant formation. Conversely, for endothermic reactions (absorb heat, \(\Delta H > 0\)), the equilibrium constant increases with temperature, favoring the production of products. This equation becomes a powerful tool to predict how equilibrium will be affected by temperature fluctuations.
Effect of Temperature on Equilibrium
Temperature has a profound impact on the position of equilibrium in chemical reactions. When the temperature is altered, the equilibrium constant changes, leading to a shift in the balance of reactants and products. This phenomenon is governed by Le Châtelier's Principle, which states that a system at equilibrium will adjust to counteract any changes imposed on it.

For an exothermic reaction, an increase in temperature will shift the equilibrium to favor the reactants, as it essentially adds heat which the reaction itself produces. Conversely, for an endothermic reaction, an increase in temperature shifts the equilibrium in favor of the products since the reaction consumes heat. Predicting these shifts is critical for controlling industrial chemical processes and understanding natural phenomena. For example, in the case of climate change, temperature increases could shift the equilibrium of certain chemical reactions in the atmosphere, potentially leading to varied environmental impacts.

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Most popular questions from this chapter

At \(1130^{\circ} \mathrm{C}\) the equilibrium constant \(\left(K_{\mathrm{c}}\right)\) for the reaction $$ 2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) $$ is \(2.25 \times 10^{-4}\). If \(\left[\mathrm{H}_{2} \mathrm{~S}\right]=4.84 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]=\) \(1.50 \times 10^{-3} M,\) calculate \(\left[\mathrm{S}_{2}\right].\)

Consider the equilibrium $$ 2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ If nitrosyl bromide, NOBr, is 34 percent dissociated at \(25^{\circ} \mathrm{C}\) and the total pressure is \(0.25 \mathrm{~atm},\) calculate \(K_{P}\) and \(K_{\mathrm{c}}\) for the dissociation at this temperature.

The formation of \(\mathrm{SO}_{3}\) from \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) is an intermediate step in the manufacture of sulfuric acid, andit is also responsible for the acid rain phenomenon. The equilibrium constant \(\left(K_{P}\right)\) for the reaction $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ is 0.13 at \(830^{\circ} \mathrm{C}\). In one experiment 2.00 moles of \(\mathrm{SO}_{2}\) and 2.00 moles of \(\mathrm{O}_{2}\) were initially present in a flask. What must the total pressure be at equilibrium to have an 80.0 percent yield of \(\mathrm{SO}_{3} ?\)

Consider the reaction in a closed container: $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ Initially, 1 mole of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is present. At equilibrium, \(\alpha\) mole of \(\mathrm{N}_{2} \mathrm{O}_{4}\) has dissociated to form \(\mathrm{NO}_{2}\). (a) Derive an expression for \(K_{P}\) in terms of \(\alpha\) and \(P,\) the total pressure. (b) How does the expression in (a) help you predict the shift in equilibrium caused by an increase in \(P ?\) Does your prediction agree with Le Châtelier's principle?

At \(25^{\circ} \mathrm{C}\), the equilibrium partial pressures of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) are \(0.15 \mathrm{~atm}\) and \(0.20 \mathrm{~atm}\), respectively. If the volume is doubled at constant temperature, calculate the partial pressures of the gases when a new equilibrium is established.

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