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Chapter 15: Problem 5

Ethene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) reacts with halogens \(\left(\mathrm{X}_{2}\right)\) by the following reaction: $$ \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{X}_{2}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{X}_{2}(g) $$ The following figures represent the concentrations at equilibrium at the same temperature when \(\mathrm{X}_{2}\) is \(\mathrm{Cl}_{2}\) (green), \(\mathrm{Br}_{2}\) (brown), and \(\mathrm{I}_{2}\) (purple). List the equilibria from smallest to largest equilibrium constant. [Section 15.3]

Short Answer

Expert verified
Based on the color representations in the figures, the order of the equilibria from smallest to largest equilibrium constant is: Chlorine (Cl鈧�) < Bromine (Br鈧�) < Iodine (I鈧�).

Step by step solution

01

To list the equilibrium constants from smallest to largest, we first need to identify the concentrations of the reactants and products in each reaction (based on the given color code), as follows: 1. For Chlorine (Cl鈧�), green: $$[\text{C}_{2}\text{H}_{4}], [\text{Cl}_{2}], [\text{C}_{2}\text{H}_{4}\text{Cl}_{2}]$$ 2. For Bromine (Br鈧�), brown: $$[\text{C}_{2}\text{H}_{4}], [\text{Br}_{2}], [\text{C}_{2}\text{H}_{4}\text{Br}_{2}]$$ 3. For Iodine (I鈧�), purple: $$[\text{C}_{2}\text{H}_{4}], [\text{I}_{2}], [\text{C}_{2}\text{H}_{4}\text{I}_{2}]$$ #Step 2: Compare the K values for each reaction.#

Now that we have the concentrations, we can perform a comparison of the K values for the 3 reactions. In each case, K is calculated using the formula mentioned earlier: For Chlorine (Cl鈧�): $$ K_{Cl_{2}} = \frac{[\text{C}_{2}\text{H}_{4} \text{Cl}_{2}]}{[\text{C}_{2}\text{H}_{4}] [\text{Cl}_{2}]} $$ For Bromine (Br鈧�): $$ K_{Br_{2}} = \frac{[\text{C}_{2}\text{H}_{4} \text{Br}_{2}]}{[\text{C}_{2}\text{H}_{4}] [\text{Br}_{2}]} $$ For Iodine (I鈧�): $$ K_{I_{2}} = \frac{[\text{C}_{2}\text{H}_{4} \text{I}_{2}]}{[\text{C}_{2}\text{H}_{4}] [\text{I}_{2}]} $$ From the given information, we can directly compare the values of the numerator (product's concentration) and denominator (reactants' concentration). We can make these comparisons based on the color representation in the figures. Considering the figures, we can see the proportions of product and reactants as follows: - Chlorine (Cl鈧�) forms the least amount of product. - Bromine (Br鈧�) forms a moderate amount of the product. - Iodine (I鈧�) forms the most amount of product. So we can conclude the relative values for K: $$ K_{Cl_{2}} < K_{Br_{2}} < K_{I_{2}} $$ Therefore, the order of the equilibria is: Chlorine (Cl鈧�) < Bromine (Br鈧�) < Iodine (I鈧�).

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Key Concepts

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

Reaction Quotient
The reaction quotient, denoted by Q, is a critical concept in understanding chemical reactions in a dynamic state. It measures the relative amounts of products and reactants present during a reaction at any given point in time, before the state of equilibrium is reached. The expression for the reaction quotient is similar to the equilibrium constant expression but is calculated using the current concentrations of the reactants and products, regardless of whether the system is at equilibrium.

The reaction quotient can help predict the direction in which a reaction will proceed to reach equilibrium. To summarize:
  • If Q < K (the equilibrium constant), the reaction will proceed in the forward direction to produce more products.
  • If Q > K, the reaction will shift backward, making more reactants.
  • If Q = K, the system is already at equilibrium, and no net reaction occurs.
This concept is especially useful when analyzing reactions such as the halogenation of ethene, where the reaction may not yet have reached equilibrium, and changes in conditions can alter the reaction's course.
Equilibrium Expressions
Equilibrium expressions are mathematical representations of the state of a chemical reaction at equilibrium. They calculate the equilibrium constant, K, which quantifies the ratio of the concentrations of products to reactants at this specific point. Each equilibrium expression is tailored to its reaction, involving the reactants' and products' concentrations raised to the power of their stoichiometric coefficients in the balanced chemical equation.

For a general reaction \(aA + bB \rightleftharpoons cC + dD\), the equilibrium constant expression would be:
\[K = \frac{[C]^c[D]^d}{[A]^a[B]^b}\]
Equilibrium expressions do not include solids or pure liquids since their concentrations do not vary. Understanding how to write and calculate these expressions is vital for comparing the position of different equilibria, as in the case of halogenation reactions with various halogens (Cl鈧�, Br鈧�, I鈧�).
Halogenation of Ethene
The halogenation of ethene is a classic example of an addition reaction where halogen molecules, such as chlorine (Cl鈧�), bromine (Br鈧�), or iodine (I鈧�), add across the double bond of ethene (C鈧侶鈧�), resulting in dihalogenated ethanes. This type of chemical reaction is crucially dependent on the halogen involved, as it influences both the kinetics and the thermodynamic stability of the product.

In these reactions, ethene acts as a nucleophile and the halogen as an electrophile. The mechanism typically proceeds through a cyclic halonium ion intermediate before the final product forms. Interestingly, the equilibrium constant for each halogen-ethene reaction varies, which has practical implications in synthetic chemistry, where the desired product's yield is a major concern. The reactivity order for the halogenation of ethene generally follows the trend I鈧� > Br鈧� > 颁濒鈧�.
Le Chatelier's Principle
Le Chatelier's principle is a pivotal concept in chemical equilibrium, encapsulating how a system at equilibrium responds to external changes such as concentration, pressure, and temperature. Simply put, the principle states that if an external stress is applied to a system at equilibrium, the system will adjust itself in such a way as to partially oppose the change and re-establish a new equilibrium.

For instance, adding more reactants to the system will shift the equilibrium towards the products to counteract the increased concentration of reactants. Conversely, removing products will favor the forward reaction to replace the lost products. Also, a change in temperature or pressure can significantly impact the position of the equilibrium. Le Chatelier's principle gives us a qualitative tool to predict these shifts, which is essential in optimizing industrial processes and understanding the behavior of reactions under various conditions.

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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 Ch芒telier's principle, predict whether the solubility of \(\mathrm{AgCl}\) in \(\mathrm{H}_{2} \mathrm{O}\) increases or decreases with increasing temperature.

A \(0.831-g\) sample of \(\mathrm{SO}_{3}\) is placed in a 1.00-Lcontainer 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 \(1100 \mathrm{~K}\).

Nitric oxide (NO) reacts readily with chlorine gas as follows: $$ 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g) $$ At \(700 \mathrm{~K}\) the equilibrium constant \(K_{p}\) for this reaction is \(0.26\). Predict the behavior of each of the following mixtures at this temperature: (a) \(P_{\mathrm{NO}}=0.15 \mathrm{~atm}, P_{\mathrm{Cl}_{2}}=0.31 \mathrm{~atm}\), and \(P_{\mathrm{NOCl}}=0.11 \mathrm{~atm} ;\) (b) \(\mathrm{P}_{\mathrm{NO}}=0.12 \mathrm{~atm}, P_{\mathrm{Cl}_{2}}=0.10 \mathrm{~atm}\) and \(\quad P_{\mathrm{NOCl}}=0.050 \mathrm{~atm} ; \quad\) (c) \(\quad P_{\mathrm{NO}}=0.15 \mathrm{~atm}\), \(P_{\mathrm{C}_{2}}=0.20 \mathrm{~atm}\), and \(P_{\mathrm{NOCl}}=5.10 \times 10^{-3} \mathrm{~atm}\)

For the reaction \(\mathrm{I}_{2}+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{IBr}(g), K_{c}=280 \mathrm{at}\) \(150^{\circ} \mathrm{C}\). Suppose that \(0.500 \mathrm{~mol}\) IBr in a 1.00-L flask is allowed to reach equilibrium at \(150^{\circ} \mathrm{C}\). What are the equilibrium concentrations of \(\mathrm{IBr}, \mathrm{I}_{2}\), and \(\mathrm{Br}_{2}\) ?

Consider the hypothetical reaction \(\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(\mathrm{~g})\). A flask is charged with \(0.75\) atm of pure \(A\), after which it is allowed to reach equilibrium at \(0{ }^{\circ} \mathrm{C}\). At equilibrium the partial pressure of A is \(0.36\) atm. (a) What is the total pressure in the flask at equilibrium? (b) What is the value of \(K_{p}\) ?
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