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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 water-gas shift reaction \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons\) \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\) is used industrially to produce hydrogen. The reaction enthalpy is \(\Delta H^{\circ}=-41 \mathrm{kJ}\) . (a) To increase the equilibrium yield of hydrogen would you use high or low temperature? ( b) Could you increase the equilibrium yield of hydrogen by controlling the pressure of this reaction? If so would high or low pressure favor formation of \(\mathrm{H}_{2}(g) ?\)

True or false: When the temperature of an exothermic reaction increases, the rate constant of the forward reaction decreases, which leads to a decrease in the equilibrium constant, \(K_{c} .\)

The protein hemoglobin (Hb) transports \(\mathrm{O}_{2}\) in mammalian blood. Each \(\mathrm{Hb}\) can bind 4 \(\mathrm{O}_{2}\) molecules. The equilibrium constant for the \(\mathrm{O}_{2}\) binding reaction is higher in fetal hemoglobin than in adult hemoglobin. In discussing protein oxygen-binding capacity, biochemists use a measure called the \(P 50\) value, defined as the partial pressure of oxygen at which 50\(\%\) of the protein is saturated. Fetal hemoglobin has a P50 value of 19 torr, and adult hemoglobin has a \(\mathrm{P} 50\) value of 26.8 torr. Use these data to estimate how much larger \(K_{c}\) is for the aqueous reaction \(4 \mathrm{O}_{2}(g)+\mathrm{Hb}(a q) \rightleftharpoons\left[\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)\right]\) in a fetus, compared to \(K_{c}\) for the same reaction in an adult.

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.

Write the expressions for \(K_{c}\) for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous. (a) 2 \(\mathrm{O}_{3}(g) \rightleftharpoons 3 \mathrm{O}_{2}(g)\) (b) \(\mathrm{Ti}(s)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{TiCl}_{4}(l)\) (c) \(2 \mathrm{C}_{2} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g)\) (d) \(\mathrm{C}(s)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)\) (e) \(4 \mathrm{HCl}(a q)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{Cl}_{2}(g)\) (f) \(2 \mathrm{C}_{8} \mathrm{H}_{18}(l)+25 \mathrm{O}_{2}(g) \rightleftharpoons 16 \mathrm{CO}_{2}(g)+18 \mathrm{H}_{2} \mathrm{O}(g)\) (g) \(2 \mathrm{C}_{8} \mathrm{H}_{18}(l)+25 \mathrm{O}_{2}(g) \rightleftharpoons 16 \mathrm{CO}_{2}(g)+18 \mathrm{H}_{2} \mathrm{O}(l)\)
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