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Consider the endothermic reaction: $$\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{I}_{2}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{I}_{2}(g)$$ If you were trying to maximize the amount of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{I}_{2}\) produced, which tactic might you try? Assume that the reaction mixture reaches equilibrium. \begin{equation} \begin{array}{l}{\text { a. decreasing the reaction volume }} \\ {\text { b. removing I }_{2} \text { from the reaction mixture }} \\ {\text { c. raising the reaction temperature }} \\ {\text { d. adding } C_{2} H_{4} \text { to the reaction mixture }}\end{array} \end{equation}

Short Answer

Expert verified
To maximize the amount of C2H4I2 produced, you should raise the reaction temperature (c), as it is an endothermic reaction.

Step by step solution

01

Analyze the reaction

Identify that the given reaction is endothermic, which implies that it absorbs heat. The general principle by Le Chatelier's is that a system at equilibrium will react to counteract any imposed change.
02

Examine the effect of volume decrease

Decreasing the volume of the reaction will increase the pressure and favor the formation of fewer moles of gas. This reaction, however, shows the same number of moles on both sides (1 mole of C2H4 and 1 mole of I2 produce 1 mole of C2H4I2), so a change in volume won't shift the position of equilibrium much.
03

Assess the effect of removing I2

Removing one of the reactants (I2) from the system would, by Le Chatelier's principle, shift the equilibrium to the left to produce more of the reactant, thus reducing the amount of product C2H4I2.
04

Evaluate the effect of temperature increase

Raising the temperature of an endothermic reaction will shift the equilibrium to the right, favoring the formation of the product. This is because the system will counteract the added heat by favoring the reaction which absorbs heat (the endothermic direction).
05

Consider adding more C2H4

Adding more of the reactant C2H4 will shift the equilibrium to the right, according to Le Chatelier's principle, to produce more product C2H4I2.
06

Select the best tactic

To maximize the amount of C2H4I2 produced, the best tactic would be raising the reaction temperature since the reaction is endothermic. Adding more C2H4 would also shift the equilibrium towards more product, but increasing temperature directly utilizes the nature of the reaction.

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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 fundamental concept in chemistry, particularly when studying dynamic equilibria in reversible reactions. It states that if a system at chemical equilibrium experiences a change in concentration, temperature, volume, or pressure, the equilibrium will shift to counteract the imposed change and a new equilibrium will be established.

For instance, when there is an increase in the concentration of reactants, the system shifts in the direction that consumes the excess reactants and produces more products. Conversely, a decrease in reactant concentration leads to an increased production of reactants from the products. Similarly, a change in pressure or volume in a system involving gases can cause the equilibrium to shift in the direction that mitigates this change, depending on the number of moles of gaseous substances on each side of the balanced equation.

Application in the Exercise

In the given exercise, adding more of the reactant \( C_{2}H_{4} \) or reducing the concentration of \( I_{2} \) would shift the equilibrium position according to Le Chatelier's principle. The system would attempt to counteract these changes by producing more product, which in this case is \( C_{2}H_{4}I_{2} \).
Chemical Equilibrium
Chemical equilibrium occurs in a reversible reaction when the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of reactants and products. It is important to note that this does not mean the reactants and products are in equal concentrations, but rather that their concentrations have stabilized in a particular ratio.

At equilibrium, the system's macroscopic properties, such as color, density, and concentration, remain constant over time. This can be represented by an equilibrium constant (\( K_{eq} \) for a given temperature), which is a numerical value derived from the ratio of product concentrations to reactant concentrations, each raised to the power of their coefficients in the balanced chemical equation.

Exercise Context

For the reaction in our exercise, the establishment of equilibrium between \( C_{2}H_{4} \) (g), \( I_{2} \) (g), and \( C_{2}H_{4}I_{2} \) (g) means there will be a specific ratio of these substances' concentrations. If we manipulate the conditions—such as temperature or concentrations—then the position of the chemical equilibrium will shift to either produce more product or more reactants, while still reaching a point of balance over time.
Reaction Temperature Effects
Temperature is a critical factor affecting chemical equilibrium. For endothermic reactions, which absorb energy, increasing the temperature favors the forward reaction and produces more products. Conversely, decreasing the temperature favors the backward reaction, producing more reactants.

This is related to the concept of exothermic reactions (which release energy), where an increase in temperature would favor the reverse reaction, hence decreasing the product formation. It's important to recognize that changing the temperature of a system effectively changes the value of the equilibrium constant (\( K_{eq} \) for the reaction at the new temperature.

Implications for the Given Example

As the textbook exercise involves an endothermic reaction, raising the temperature would make the system respond by shifting the equilibrium to the right (towards more product), as per the Le Chatelier's principle. This tactic serves to maximize the amount of \( C_{2}H_{4}I_{2} \) produced. Thus, selecting the option to raise the reaction temperature would be the most efficient strategy to increase product yields, by exploiting the thermal characteristics of the endothermic reaction.

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Most popular questions from this chapter

Coal, which is primarily carbon, can be converted to natural gas, primari- ly \(\mathrm{CH}_{4, \text { by the exothermic reaction: }}\) $$\mathrm{C}(s)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)$$ Which disturbance favors \(\mathrm{CH}_{4}\) at equilibrium? \begin{equation}\begin{array}{l}{\text { a. adding more } \mathrm{C} \text { to the reaction mixture }} \\ {\text { b. adding more } \mathrm{H}_{2} \text { to the reaction mixture }} \\ {\text { c. raising the temperature of the reaction mixture }} \\ {\text { d. } \text { lowering the volume of the reaction mixture }} \\ {\text { e. adding a catalyst to the reaction mixture }} \\ {\text { f. adding neon gas to the reaction mixture }}\end{array}\end{equation}

The reaction \(\mathrm{CO}_{2}(g)+\mathrm{C}(s) \rightleftharpoons 2 \mathrm{CO}(g)\) has \(\mathrm{K}_{\mathrm{p}}=5.78\) at 1200 \(\mathrm{K}\) \begin{equation}\begin{array}{l}{\text { a. Calculate the total pressure at equilibrium when } 4.45 \text { g of } \mathrm{CO}_{2} \text { is in- }} \\\ {\text { troduced into a } 10.0 \text { L container and heated to } 1200 \mathrm{K} \text { in the presence }} \\ {\text { of } 2.00 \mathrm{g} \text { of graphite. }} \\ {\text { b. Repeat the calculation of part a in the presence of 0.50 } \mathrm{g} \text { of graphite. }}\end{array}\end{equation}

Consider the reaction: $$\mathrm{NH}_{4} \mathrm{HS}(s) \Longrightarrow \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{S}(g)$$ At a certain temperature, \(K_{\mathrm{c}}=8.5 \times 10^{-3} .\) A reaction mixture at this temperature containing solid \(\mathrm{NH}_{4} \mathrm{HS}\) has \(\left[\mathrm{NH}_{3}\right]=0.166 \mathrm{M}\) and \(\left[\mathrm{H}_{2} \mathrm{S}\right]=0.166 \mathrm{M} .\) Will more of the solid form, or will some of the existing solid decompose as equilibrium is reached?

For the reaction \(2 \mathrm{A}(g) \rightleftharpoons \mathrm{B}(g)+2 \mathrm{C}(g),\) a reaction vessel initially contains only \(\mathrm{A}\) at a pressure of \(P_{\mathrm{A}}=225 \mathrm{mm} \mathrm{Hg} .\) At equilibrium,\(P_{\mathrm{A}}=55 \mathrm{mmHg}\) . Calculate the value of \(K_{\mathrm{P}}\) . (Assume no changes in vol- ume or temperature.)

What is the definition of the reaction quotient \((Q)\) for a reaction? What does \(Q\) measure?

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