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

Short Answer

Expert verified
For Reaction A (\(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)\)) with \(K_{c}=1.5 \times 10^{-10}\), the equilibrium mixture contains mostly reactants since \(K_{c}\) is much smaller than 1. For Reaction B (\(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2\mathrm{SO}_{3}(g)\)) with \(K_{p}=2.5 \times 10^{9}\), the equilibrium mixture contains mostly products since \(K_{p}\) is much larger than 1.

Step by step solution

01

Reaction A: Determining if equilibrium favors reactants or products

Given reaction A: \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)\). The equilibrium constant is \(K_{c}=1.5 \times 10^{-10}\). Since \(K_{c}\) is much smaller than 1, this reaction is reactant-favored. Therefore, the equilibrium mixture contains mostly reactants - nitrogen and oxygen gases.
02

Reaction B: Determining if equilibrium favors reactants or products

Given reaction B: \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2\mathrm{SO}_{3}(g)\). The equilibrium constant is \(K_{p}=2.5 \times 10^{9}\). Since \(K_{p}\) is much larger than 1, this reaction is product-favored. Therefore, the equilibrium mixture contains mostly products - sulfur trioxide gas. In conclusion, for Reaction A, the equilibrium mixture contains mostly reactants while in Reaction B, the equilibrium mixture contains mostly products.

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Key Concepts

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

Equilibrium Constants
The concept of equilibrium constants is a fundamental aspect of chemical equilibrium. An equilibrium constant is a numerical value that reflects the ratio of concentrations of products to reactants at equilibrium. This constant is represented by either \( K_c \) for concentration in molarity or \( K_p \) for partial pressures in gaseous systems.

When analyzing a chemical equation, the equilibrium constant helps to determine the relative amounts of products and reactants in the mixture at equilibrium. If \( K \) is much greater than 1, it suggests that the reaction heavily favors the formation of products. Conversely, if \( K \) is much less than 1, it indicates a greater concentration of reactants at equilibrium.
  • Large \( K \) value: more products than reactants
  • Small \( K \) value: more reactants than products
Understanding the magnitude of \( K \) allows chemists to predict the chemical behavior of reactions.
Reaction Favorability
Reaction favorability describes whether a given chemical reaction is more likely to proceed towards the formation of products or the preservation of reactants. It is largely dependent on the equilibrium constant value.

In a chemical reaction, the direction of favorability is evident from the comparison of \( K \) values:
  • \( K \) greater than 1: Reaction is product-favored, resulting in a larger quantity of products
  • \( K \) less than 1: Reaction is reactant-favored, producing fewer products
Favorable reactions tend to go to completion toward the side with lower chemical potential. In practical terms, this indicates whether more of the starting materials remain in the reaction vessel or if they convert significantly into new substances.

The conditions under which these reactions occur, like temperature and pressure, can also influence the equilibrium but are not solely responsible for driving the direction of favorability.
Gaseous Reactions
Gaseous reactions are chemical processes that involve substances in the gas phase. Such reactions are particularly influenced by changes in volume, pressure, and temperature due to the nature of gases.

For gaseous equilibria, the use of \( K_p \) (equilibrium constant in terms of pressure) is common because gases behave according to the ideal gas law; hence, their concentrations are directly related to pressure. In scenarios where gaseous reaction equilibrium is concerned:
  • Partial pressures of the gases are key in calculating \( K_p \).
  • Changes in pressure affect equilibrium positions based on Le Chatelier's principle — increasing pressure will shift the equilibrium towards the side with fewer moles of gas.
Gaseous reactions are a window into understanding dynamic equilibria since the conditions can be easily manipulated, offering insights into both kinetics as well as thermodynamics.

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Most popular questions from this chapter

At \(1200 \mathrm{~K}\), the approximate temperature of automobile exhaust gases (Figure 15.15\(), K_{p}\) for the reaction $$2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)$$ is about \(1 \times 10^{-13}\). Assuming that the exhaust gas (total pressure 1 atm \()\) contains \(0.2 \% \mathrm{CO}, 12 \% \mathrm{CO}_{2},\) and \(3 \% \mathrm{O}_{2}\) by volume, is the system at equilibrium with respect to the \(\mathrm{CO}_{2}\) reaction? Based on your conclusion, would the CO concentration in the exhaust be decreased or increased by a catalyst that speeds up the \(\mathrm{CO}_{2}\) reaction?

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 \(\mathrm{C}\).

For the equilibrium $$2 \mathrm{IBr}(g) \rightleftharpoons \mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g)$$ \(K_{p}=8.5 \times 10^{-3}\) at \(150^{\circ} \mathrm{C}\). If 0.025 atm of IBr is placed in a 2.0-L container, what is the partial pressure of all substances after equilibrium is reached?

Explain why we normally exclude solvents from liquid-phase reactions in equilibrium-constant expressions.

(a) How is a reaction quotient used to determine whether a system is at equilibrium? (b) If \(Q_{c}>K_{c}\), how must the reaction proceed to reach equilibrium? (c) At the start of a certain reaction, only reactants are present; no products have been formed. What is the value of \(Q_{c}\) at this point in the reaction?

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