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Chapter 15: Problem 17

When the following reactions come to equilibrium, does the equilibrium mixture contain mostly reactants or mostly products? $$\begin{array}{ll}{\text { (a) } \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)} & {K_{c}=1.5 \times 10^{-10}} \\ {\text { (b) } 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)} & {K_{p}=2.5 \times 10^{9}}\end{array}$$

Short Answer

Expert verified
In conclusion, the equilibrium mixtures for the given reactions will contain mostly reactants for reaction (a) \( (\mathrm{N}_{2} \text{ and } \mathrm{O}_{2})\) due to the small Kc value and mostly products for reaction (b) \( (\mathrm{SO}_{3})\) due to the large Kp value.

Step by step solution

01

(a) N2(g) + O2(g) 鈬� 2 NO(g), Kc = 1.5 脳 10^(-10)

: The equilibrium constant Kc for this reaction is very small \( (K_c = 1.5 \times 10^{-10}) \). Since the value of Kc is much less than 1, the equilibrium favors the reactants over the products. Therefore, the equilibrium mixture for this reaction will contain mostly reactants (N2 and O2).
02

(b) 2 SO2(g) + O2(g) 鈬� 2 SO3(g), Kp = 2.5 脳 10^9

: For this reaction, the equilibrium constant Kp is very large \( (K_p = 2.5 \times 10^9) \). Since the value of Kp is much greater than 1, the equilibrium favors the products over the reactants. So, the equilibrium mixture will mainly consist of the product (SO3). In conclusion, for reaction (a) the equilibrium mixture will contain mostly reactants, whereas for reaction (b) the equilibrium mixture will contain mostly products.

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Key Concepts

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

Equilibrium Constant
When understanding chemical equilibrium, the role of the **equilibrium constant** is crucial. It quantifies the balance between the concentrations of reactants and products at equilibrium.
The equilibrium constant, represented as either \(K_c\) for concentrations or \(K_p\) for partial pressures, provides insight into which direction the reaction favors once equilibrium is reached.
For reaction (a), \(K_c = 1.5 \times 10^{-10}\) indicates a very small value. This tells us the reaction forms very little product compared to the reactants, leaning the equilibrium towards reactants. On the other hand, reaction (b), with \(K_p = 2.5 \times 10^9\), shows a large equilibrium constant, pointing towards being product-favored, meaning more products are present at equilibrium. This information is pivotal in predicting the composition of the equilibrium mixture in chemical reactions.
Favorable Reactions
**Favorable reactions** refer to those that are more inclined to proceed in a particular direction when reaching equilibrium. The equilibrium constant helps determine the direction in which a reaction is favorable.
- If \(K\) is large (greater than 1), the products at equilibrium are favored, as seen in reaction (b). This indicates a **product-favored** reaction, where more products are generated than reactants at equilibrium.
- Conversely, if \(K\) is small (less than 1), such as in reaction (a), the reactants are favored. This means the reaction will not yield much product, maintaining a higher concentration of reactants.Understanding which reactions are product- or reactant-favored guides the expectations around reaction yields and the feasibility of a process under certain conditions.
Reactants vs Products
In a chemical reaction, the presence of **reactants and products** at equilibrium is a balancing act. Whether the mixture contains more reactants or products can be deduced using the equilibrium constant.
- For reaction (a), with a small \(K_c\), **mostly reactants** (\(N_2\) and \(O_2\)) remain. This occurs because the formation of products (\(NO\)) is minimal when equilibrium is reached.
- For reaction (b), a large \(K_p\) value suggests the presence of **mostly products** (\(SO_3\)). The system adjusts to ensure a high production of the product to reach equilibrium.Analyzing the ratio of products to reactants offers a clear picture of how reactions behave in equilibrium. This balance is essential in determining reaction conditions, efficiency, and potential yield in practical applications.

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Most popular questions from this chapter

The reaction of an organic acid with an alcohol, in organic solvent, to produce an ester and water is commonly done in the pharmaceutical industry. This reaction is catalyzed by strong acid (usually \(\mathrm{H}_{2} \mathrm{SO}_{4} ) .\) A simple example is the reaction of acetic acid with ethyl alcohol to produce ethyl acetate and water: $$\begin{aligned} \mathrm{CH}_{3} \mathrm{COOH}(s o l v)+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(s o l v) & \rightleftharpoons \\ \mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}(\mathrm{solv}) &+\mathrm{H}_{2} \mathrm{O}(\text {solv}) \end{aligned}$$ where \(^{a}(s o l v)^{\prime \prime}\) indicates that all reactants and products are in solution but not an aqueous solution. The equilibrium constant for this reaction at \(55^{\circ} \mathrm{C}\) is 6.68 . A pharmaceutical chemist makes up 15.0 \(\mathrm{L}\) of a solution that is initially 0.275 \(\mathrm{M}\) in acetic acid and 3.85\(M\) in ethanol. At equilibrium, how many grams of ethyl acetate are formed?

A flask is charged with 1.500 atm of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and 1.00 atm \(\mathrm{NO}_{2}(g)\) at \(25^{\circ} \mathrm{C},\) and the following equilibrium is achieved: $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ After equilibrium is reached, the partial pressure of \(\mathrm{NO}_{2}\) is 0.512 atm. (a) What is the equilibrium partial pressure of \(\mathrm{N}_{2} \mathrm{O}_{4} ?\) (b) Calculate the value of \(K_{p}\) for the reaction. (c) Calculate \(K_{c}\) for the reaction.

\(\mathrm{At} 900 \mathrm{K},\) the following reaction has \(K_{p}=0.345 :\) $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ In an equilibrium mixture the partial pressures of \(S \mathrm{O}_{2}\) and \(\mathrm{O}_{2}\) are 0.135 atm and 0.455 atm, respectively. What is the equilibrium partial pressure of \(\mathrm{SO}_{3}\) in the mixture?

At \(373 \mathrm{K}, K_{p}=0.416\) for the equilibrium $$2 \operatorname{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ If the pressures of \(\mathrm{NOBr}(g)\) and \(\mathrm{NO}(g)\) are equal, what is the equilibrium pressure of \(\mathrm{Br}_{2}(g) ?\)

Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons\) \(\mathrm{PCl}_{5}(g) . \mathrm{A} 7.5-\mathrm{L}\) gas vessel is charged with a mixture of \(\mathrm{PCl}_{3}(g)\) and \(\mathrm{Cl}_{2}(g),\) which is allowed to equilibrate at 450 K. At equilibrium the partial pressures of the three gases are \(P_{\mathrm{PCl}_{3}}=0.124 \mathrm{atm}, P_{\mathrm{Cl}_{2}}=0.157 \mathrm{atm},\) and \(P_{\mathrm{PCl}_{\mathrm{s}}}=1.30 \mathrm{atm}\) (a) What is the value of \(K_{p}\) at this temperature? (b) Does the equilibrium favor reactants or products? (c) Calculate \(K_{c}\) for this reaction at 450 \(\mathrm{K}\)
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