The following graph represents the yield of the compound \(\mathrm{AB}\) at equilibrium in the reaction \(\mathrm{A}(g)+\mathrm{B}(g) \longrightarrow \mathrm{AB}(g)\) at two different pressures, \(x\) and \(y\), as a function of temperature. (a) Is this reaction exothermic or endothermic? (b) Is \(P=x\) greater or smaller than $P=y ?

Short Answer

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(a) The reaction is exothermic. (b) P=x is greater than P=y.

Step by step solution

01

Identify Le Chatelier's Principle

Le Chatelier's principle states that if a change is made to a system in equilibrium, the system will adjust to counteract that change and reestablish equilibrium. In this exercise, we need to understand how changes in pressure and temperature affect the position of equilibrium and the reaction yield.
02

Analyze the effect of temperature on the reaction

According to the graph, as the temperature increases, the equilibrium yield of compound AB decreases. This behavior suggests that the forward reaction (A + B -> AB) releases heat energy while the reverse reaction (AB -> A + B) absorbs heat energy. In other words, raising the temperature of this reaction promotes the endothermic reverse reaction, meaning the forward reaction is exothermic. Thus, with the information from the graph: (a) The reaction is exothermic.
03

Analyze pressure changes by Le Chatelier's Principle

According to Le Chatelier's principle, increasing pressure for a reaction in a closed system will shift the equilibrium toward the side of the reaction with fewer moles of gas. In this case, the forward reaction (A + B -> AB) reduces the total mole count of gas species, implying that an increase in pressure leads to an increase in the yield of compound AB. Considering any of the pressures P=x or P=y shown in the graph, and noting that temperatures are fixed: (b) P=x must be greater than P=y because the yield of compound AB is higher at the same temperature, which indicates the equilibrium has shifted more in favor of the forward reaction, in response to the increased pressure.

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Key Concepts

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

Le Chatelier's Principle
Le Chatelier's principle is a fascinating concept in chemistry that helps predict how a reaction will respond to changes in its environment. When a system at equilibrium is disturbed, it adjusts to diminish the change and restore balance. This principle applies to various changes, including concentration, pressure, and temperature.

Consider a seesaw balanced perfectly on a fulcrum—an equilibrium of sorts. If a weight is added to one side, the seesaw will tip in that direction. Similarly, in a chemical reaction, if the concentration of a reactant is increased, the system will 'tip' to produce more product to counteract the change. If pressure or volume is altered in a reaction involving gases, the reaction will shift towards the side that helps reduce the pressure change, according to the number of gas molecules involved. Le Chatelier's principle is not only a guiding theory but also an essential tool for chemists to optimize the conditions for chemical reactions to get the desired yield.
Exothermic Reactions
To get a grasp on exothermic reactions, imagine warming your hands by a campfire. The fire releases heat to the surroundings; similarly, in an exothermic reaction, energy is released in the form of heat. These reactions can occur everywhere, from the combustion in car engines to the cellular respiration in our bodies.

Chemically, exothermic reactions are characterized by the release of heat as the bonds in the products are stronger—hence more stable—than the bonds in the reactants. The excess energy is let out into the environment. This is significant because the temperature of the system increases unless the heat is removed. An everyday example is burning natural gas (methane) on your stove to cook food. The methane reacts with oxygen to produce carbon dioxide, water, and heat—quite essential for preparing that delicious meal.
Reaction Yield and Temperature
Let's dive into the relationship between reaction yield and temperature. This connection is pivotal in chemistry, especially in the industrial synthesis of compounds. By manipulating temperature, chemists can nudge a reaction to produce more of a desired product.

The yield of a chemical reaction refers to the amount of product formed under certain conditions. Temperature, being a measure of thermal energy, greatly influences the speed and extent of reactions. For exothermic reactions, an increase in temperature can shift the equilibrium towards the reactants due to Le Chatelier's principle, resulting in a lower yield of the product. The opposite holds true for endothermic reactions; higher temperatures can raise the yield. The trick lies in finding the optimal temperature to maximize the reaction yield while maintaining efficiency and cost-effectiveness. This is crucial in industries where the scale of production can mean the difference between profit and loss.

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Most popular questions from this chapter

Silver chloride, \(\mathrm{AgCl}(s)\), is an "insoluble" strong electrolyte. (a) Write the equation for the dissolution of \(\mathrm{AgCl}(s)\) in \(\mathrm{H}_{2} \mathrm{O}(l)\) (b) Write the expression for \(K_{c}\) for the reaction in part (a). (c) Based on the thermochemical data in Appendix \(\mathrm{C}\) and Le Châtelier's principle, predict whether the solubility of \(\mathrm{AgCl}\) in \(\mathrm{H}_{2} \mathrm{O}\) increases or decreases with increasing temperature. (d) The equilibrium constant for the dissolution of \(\mathrm{AgCl}\) in water is \(1.6 \times 10^{-10}\) at \(25^{\circ} \mathrm{C}\). In addition, \(\mathrm{Ag}^{+}(a q)\) can react with \(\mathrm{Cl}^{-}(a q)\) according to the reaction $$\mathrm{Ag}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{AgCl}_{2}^{-}(a q)$$ where \(K_{c}=1.8 \times 10^{5}\) at \(25^{\circ} \mathrm{C}\). Although \(\mathrm{AgCl}\) is "not soluble" in water, the complex \(\mathrm{AgCl}_{2}^{-}\) is soluble. At \(25^{\circ} \mathrm{C},\) is the solubility of AgCl in a \(0.100 M\) NaCl solution greater than the solubility of AgCl in pure water, due to the formation of soluble \(\mathrm{AgCl}_{2}^{-}\) ions? Or is the \(\mathrm{AgCl}\) solubility in \(0.100 \mathrm{M} \mathrm{NaCl}\) less than in pure water because of a Le Châtelier-type argument? Justify your answer with calculations.

In Section 11.5 we defined the vapor pressure of a liquid in terms of an equilibrium. (a) Write the equation representing the equilibrium between liquid water and water vapor and the corresponding expression for \(K_{p} .\) (b) By using data in Appendix \(\mathrm{B}\), give the value of \(K_{p}\) for this reaction at \(30^{\circ} \mathrm{C} .(\mathrm{c})\) What is the value of \(K_{p}\) for any liquid in equilibrium with its vapor at the normal boiling point of the liquid?

At \(900^{\circ} \mathrm{C}, K_{c}=0.0108\) for the reaction $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$ A mixture of \(\mathrm{CaCO}_{3}, \mathrm{CaO},\) and \(\mathrm{CO}_{2}\) is placed in a \(10.0-\mathrm{L}\) vessel at \(900^{\circ} \mathrm{C}\). For the following mixtures, will the amount of \(\mathrm{CaCO}_{3}\) increase, decrease, or remain the same as the system approaches equilibrium? (a) \(15.0 \mathrm{~g} \mathrm{CaCO}_{3}, 15.0 \mathrm{~g} \mathrm{CaO},\) and \(4.25 \mathrm{~g} \mathrm{CO}_{2}\) (b) \(2.50 \mathrm{~g} \mathrm{CaCO}_{3}, 25.0 \mathrm{~g} \mathrm{CaO},\) and \(5.66 \mathrm{~g} \mathrm{CO}_{2}\) (c) \(30.5 \mathrm{~g} \mathrm{CaCO}_{3}, 25.5 \mathrm{~g} \mathrm{CaO},\) and \(6.48 \mathrm{~g} \mathrm{CO}_{2}\)

A \(0.831-\mathrm{g}\) sample of \(\mathrm{SO}_{3}\) is placed in a 1.00 - \(\mathrm{L}\) container and heated to \(1100 \mathrm{~K}\). The \(\mathrm{SO}_{3}\) decomposes to \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) : $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$ At equilibrium the total pressure in the container is \(1.300 \mathrm{~atm}\). Find the values of \(K_{p}\) and \(K_{c}\) for this reaction at \(1100 \mathrm{~K}\).

Consider the hypothetical reaction \(\mathrm{A}(g)+2 \mathrm{~B}(g) \rightleftharpoons\) \(2 \mathrm{C}(g),\) for which \(K_{c}=0.25\) at a certain temperature. A \(1.00-\mathrm{L}\) reaction vessel is loaded with \(1.00 \mathrm{~mol}\) of compound \(\mathrm{C}\), which is allowed to reach equilibrium. Let the variable \(x\) represent the number of \(\mathrm{mol} / \mathrm{L}\) of compound A present at equilibrium. (a) In terms of \(x,\) what are the equilibrium concentrations of compounds \(\mathrm{B}\) and \(\mathrm{C} ?\) (b) What limits must be placed on the value of \(x\) so that all concentrations are positive? (c) By putting the equilibrium concentrations (in terms of \(x\) ) into the equilibrium-constant expression, derive an equation that can be solved for \(x\). (d) The equation from part (c) is a cubic equation (one that has the form \(\left.a x^{3}+b x^{2}+c x+d=0\right)\). In general, cubic equations cannot be solved in closed form. However, you can estimate the solution by plotting the cubic equation in the allowed range of \(x\) that you specified in part (b). The point at which the cubic equation crosses the \(x\) -axis is the solution. (e) From the plot in part (d), estimate the equilibrium concentrations of \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\).

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