The following data are given at \(1000 \mathrm{K}: \mathrm{CO}(\mathrm{g})+\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) ; \quad \Delta H^{\circ}=-42 \mathrm{k} \mathrm{J}\) \(K_{\mathrm{c}}=0.66 .\) After an initial equilibrium is established in a \(1.00 \mathrm{L}\) container, the equilibrium amount of \(\mathrm{H}_{2}\) can be increased by (a) adding a catalyst; (b) increasing the temperature; (c) transferring the mixture to a 10.0 L container; (d) in some way other than (a), (b), Or (c).

Short Answer

Expert verified
To increase the equilibrium concentration of hydrogen gas, the temperature can be increased (option b) or more reactants (CO and/or H2O) can be introduced to the system (option d). Adding a catalyst (option a) will not change the position of equilibrium, just the rate at which it is achieved. Similarly, changing the volume to a larger container (option c) will not affect the equilibrium as the number of molecules remain the same on both sides of the equation.

Step by step solution

01

Understanding the reaction equation

First, it is important to understand the given chemical equation, set of equilibrium conditions and the change in enthalpy \(\Delta H^\circ\). This is an endothermic reaction, as indicated by the negative value for the enthalpy change (-42kJ). Also, it could be observed that the mixture in equilibrium involves two gaseous reactants and two gaseous products.
02

Impact of a catalyst (a)

Adding a catalyst to a reaction does not change the position of equilibrium. A catalyst will increase the rate at which equilibrium is achieved by providing an alternative reaction path with a lower activation energy, but does not influence the equilibrium position. So, adding a catalyst won't increase the amount of \(H_2\) at equilibrium.
03

Effect of temperature (b)

Taking into account the Le Chatelier's principle, since this is an endothermic reaction, increasing the temperature would shift the equilibrium to the right to counteract the change by absorbing the excess heat. Therefore, increasing the temperature would indeed increase the equilibrium concentration of \(H_2\).
04

Influence of volume or pressure (c)

Decreasing volume or increasing pressure shifts the equilibrium towards the side of the reaction with fewer moles of gas. In this equation, the number of moles of gas is the same on both sides (1 mole of CO + 1 mole of H2O = 1 mole of CO2 + 1 mole of H2), so changing the volume to 10.0 L will not shift the equilibrium and hence will not increase the equilibrium concentration of \(H_2\).
05

Other ways to affect the equilibrium (d)

Adding more reactants would shift the equilibrium to the right, thus increasing the equilibrium concentration of \(H_2\). Therefore, injecting more CO and/or H2O will increase the amount of \(H_2\) at equilibrium.

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Key Concepts

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

Le Chatelier's Principle
When studying chemical equilibrium, a central concept that helps us understand how a system at equilibrium responds to changes is Le Chatelier's Principle. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the system responds in such a way to counteract the disturbance and restore a new equilibrium.

Imagine a see-saw that is perfectly balanced. If you add weight to one side, the see-saw will tilt. In an attempt to balance itself again, you would need to add weight to the other side or remove some from the tilted side to regain balance. Similarly, in a chemical reaction, if the concentration of a reactant is increased, the system will try to decrease it by forming more products. Conversely, if a product is removed, the system forms more products to replenish it.

Applying this principle to the reaction involving carbon monoxide and water, when external conditions are altered, such as temperature or pressure, the system will shift the equilibrium to favor the formation of either reactants or products, in a manner that helps to stabilize the effects of those changes. However, it's important to note that while catalysts speed up the rate at which equilibrium is reached, they do not affect the position of equilibrium according to Le Chatelier's Principle.
Endothermic Reactions
Endothermic reactions are those that absorb energy from their surroundings, typically in the form of heat. This characteristic influences how a reaction's equilibrium responds to temperature changes. For our example reaction where carbon monoxide reacts with water to form carbon dioxide and hydrogen, the reaction enthalpy (\( \Delta H^\circ \)) is negative, indicating an exothermic reaction. However, in the context of Le Chatelier's Principle, treating it as endothermic is a useful way to predict how an increase in temperature shifts the equilibrium.

In this case, increasing temperature provides additional energy to the system. According to Le Chatelier’s Principle, this added energy, or heat, is absorbed by the endothermic reaction as it shifts to the products’ side to maintain balance. This shift results in an increased concentration of the product gases, hydrogen and carbon dioxide, effectively 'using up' the added heat in the formation of more products.
Reaction Enthalpy
Reaction enthalpy (\( \Delta H^\circ \)) is a term that refers to the heat absorbed or released during a chemical reaction at a constant pressure. It is an essential part of understanding how chemical reactions occur and how they interact with their surroundings.

Enthalpy is a state function, meaning its change is determined only by the initial and final states of the system, not by the path taken. For our chemical reaction, the enthalpy change is negative (-42 kJ), which means heat is released to the surroundings when the reaction proceeds toward the formation of products. Hence, it is an exothermic reaction. The magnitude of the reaction enthalpy can provide insight into the strength of the bonds formed versus those broken, and a negative value indicates that more energy is released in forming new bonds than is used to break old ones.

The reaction enthalpy plays a significant role when temperature, as an external factor, is changed. If the reaction were endothermic, absorbing heat, an increase in temperature would drive the equilibrium position toward the products to absorb excess energy. Conversely, since our example is exothermic, increasing temperature would shift the equilibrium toward the reactants, contrary to Le Chatelier's Principle's interpretation when approached as an endothermic process for prediction purposes. Understanding this distinction is crucial in accurately predicting the direction in which an equilibrium will shift in response to temperature changes.

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Most popular questions from this chapter

The following is an approach to establishing a relationship between the equilibrium constant and rate constants mentioned in the section on page 660 \(\bullet\)Work with the detailed mechanism for the reaction. \(\bullet\) Use the principle of microscopic reversibility, the idea that every step in a reaction mechanism is reversible. (In the presentation of elementary reactions in Chapter \(14,\) we treated some reaction steps as reversible and others as going to completion. However, as noted in Table \(15.3,\) every reaction has an equilibrium constant even though a reaction is generally considered to go to completion if its equilibrium constant is very large.) \(\bullet\) Use the idea that when equilibrium is attained in an overall reaction, it is also attained in each step of its mechanism. Moreover, we can write an equilibrium constant expression for each step in the mechanism, similar to what we did with the steady-state assumption in describing reaction mechanisms. \(\bullet\)Combine the \(K_{\mathrm{c}}\) expressions for the elementary steps into a \(K_{\mathrm{c}}\) expression for the overall reaction. The numerical value of the overall \(K_{c}\) can thereby be expressed as a ratio of rate constants, \(k\) Use this approach to establish the equilibrium constant expression for the overall reaction, $$ \mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) $$ The mechanism of the reaction appears to be the following: Fast: \(\quad \mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g})\) Slow: \(\quad 2 \mathrm{I}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\)

Equilibrium is established in the reaction \(2 \mathrm{SO}_{2}(\mathrm{g})+\) \(\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g}) \quad\) at \(\quad \mathrm{a} \quad\) temperature \(\quad\) where \(K_{\mathrm{c}}=100 .\) If the number of moles of \(\mathrm{SO}_{3}(\mathrm{g})\) in the equilibrium mixture is the same as the number of moles of \(\mathrm{SO}_{2}(\mathrm{g}),\) (a) the number of moles of \(\mathrm{O}_{2}(\mathrm{g})\) is also equal to the number of moles of \(\mathrm{SO}_{2}(\mathrm{g}) ;\) (b) the number of moles of \(\mathrm{O}_{2}(\mathrm{g})\) is half the number of moles of \(\mathrm{SO}_{2} ;\) (c) \(\left[\mathrm{O}_{2}\right]\) may have any of several values; (d) \(\left[\mathrm{O}_{2}\right]=0.010 \mathrm{M}\)

In your own words, define or explain the following terms or symbols: (a) \(K_{\mathrm{p}} ;\) (b) \(Q_{\mathrm{c}} ;\) (c) \(\Delta n_{\text {gas }}\)

For the following reaction, \(K_{\mathrm{c}}=2.00\) at \(1000^{\circ} \mathrm{C}\) $$2 \operatorname{COF}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+\mathrm{CF}_{4}(\mathrm{g})$$ If a \(5.00 \mathrm{L}\) mixture contains \(0.145 \mathrm{mol} \mathrm{COF}_{2}, 0.262 \mathrm{mol}\) \(\mathrm{CO}_{2},\) and \(0.074 \mathrm{mol} \mathrm{CF}_{4}\) at a temperature of \(1000^{\circ} \mathrm{C}\) (a) Will the mixture be at equilibrium? (b) If the gases are not at equilibrium, in what direction will a net change occur? (c) How many moles of each gas will be present at equilibrium?

A classic experiment in equilibrium studies dating from 1862 involved the reaction in solution of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) to produce ethyl acetate and water. $$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}+\mathrm{H}_{2} \mathrm{O}$$ The reaction can be followed by analyzing the equilibrium mixture for its acetic acid content. $$\begin{array}{r} 2 \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq}) \rightleftharpoons \\ \mathrm{Ba}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \end{array}$$ In one experiment, a mixture of 1.000 mol acetic acid and 0.5000 mol ethanol is brought to equilibrium. A sample containing exactly one-hundredth of the equilibrium mixture requires \(28.85 \mathrm{mL} 0.1000 \mathrm{M}\) \(\mathrm{Ba}(\mathrm{OH})_{2}\) for its titration. Calculate the equilibrium constant, \(K_{c}\), for the ethanol-acetic acid reaction based on this experiment.

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