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Chapter 15: Problem 64

For a certain gas-phase reaction, the fraction of products in an equilibrium mixture is increased by either increasing the temperature or by increasing the volume of the reaction vessel. (a) Is the reaction exothermic or endothermic? (b) Does the balanced chemical equation have more molecules on the reactant side or product side?

Short Answer

Expert verified
(a) The reaction is endothermic. (b) More molecules on the product side.

Step by step solution

01

Analyze Temperature Effect

Increasing temperature favors the formation of products, suggesting that the reaction is endothermic. In an endothermic reaction, heat is absorbed, so higher temperatures drive the equilibrium towards the product side to absorb excess heat.
02

Analyze Volume Change Effect

Increasing the volume of the reaction vessel shifts the equilibrium towards the side with more gas molecules. Given that the fraction of products increases when volume is increased, the product side must have more moles of gas compared to the reactant side.
03

Determine Reaction Heat Nature

Based on the temperature effect, the reaction is endothermic, as the equilibrium shifts to products with increased temperature.
04

Determine Molecule Distribution

Based on the volume effect, the balanced chemical equation has more gas molecules on the product side than the reactant side.

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Key Concepts

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

Endothermic Reactions
Endothermic reactions are processes in which the system absorbs heat from its surroundings. During these reactions, the energy required to break the bonds in the reactants is greater than the energy released when new bonds are formed in the products. This results in a net absorption of energy that feels as if the system is consuming heat.
Some key characteristics of endothermic reactions include:
  • Heat absorption: They absorb heat, which makes them cool down the environment unless an external heat source is provided.
  • Temperature-driven equilibrium: In an endothermic reaction, increasing the temperature will shift the equilibrium toward the products because additional heat drives more reactants to convert into products.
Understanding this behavior is crucial in predicting how reactions will respond to changes in temperature, which directly ties into Le Chatelier鈥檚 Principle and the overall dynamics of chemical equilibrium.
Le Chatelier's Principle
Le Chatelier's Principle is a fundamental rule used to predict the effect of a change in conditions on chemical equilibria. Essentially, it states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change. This principle helps chemists determine which direction a reaction will progress under different circumstances.
Some important aspects include:
  • Temperature changes: If the temperature of a system is increased, the equilibrium will shift towards the direction that absorbs heat. For endothermic reactions, this means shifting towards the products.
  • Pressure and volume changes: Increasing the volume of a gas-phase system decreases its overall pressure, causing the equilibrium to shift toward the side with more gas molecules (as pressure adjustment is easier this way).
  • Concentration changes: Adding more reactants will shift the equilibrium to favor more products, and vice versa.
Applying Le Chatelier's Principle is crucial in understanding how external changes influence the position of equilibrium, making it a powerful tool for predicting and controlling chemical reactions.
Gas Laws
Gas laws are essential principles that describe the behavior of gases in response to changes in conditions such as pressure, volume, and temperature. They are crucial in understanding reactions involving gases, especially at equilibrium.
Some fundamental gas laws include:
  • Boyle's Law: States that the pressure of a gas is inversely proportional to its volume at constant temperature, \( P \propto \frac{1}{V} \).
  • Charles's Law: States that the volume of a gas is directly proportional to its temperature at constant pressure, \( V \propto T \).
  • Avogadro's Law: Indicates that the volume of a gas is directly proportional to the number of moles at constant temperature and pressure, \( V \propto n \).
These laws help in understanding the effects of changes in conditions on the behavior of gases. They are especially helpful in interpreting Le Chatelier鈥檚 Principle as applied to changes in pressure and volume in chemical equilibria.

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Most popular questions from this chapter

The following equilibria were measured at \(823 \mathrm{~K}\) : $$ \begin{array}{l} \mathrm{CoO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=67 \\ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=0.14 \end{array} $$ (a) Use these equilibria to calculate the equilibrium constant, \(K_{c}\), for the reaction \(\mathrm{CoO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(s)\) \(+\mathrm{CO}_{2}(g)\) at \(823 \mathrm{~K} .(\mathbf{b})\) Based on your answer to part (a), would you say that carbon monoxide is a stronger or weaker reducing agent than \(\mathrm{H}_{2}\) at \(T=823 \mathrm{~K} ?(\mathbf{c})\) If you were to place \(5.00 \mathrm{~g}\) of \(\mathrm{CoO}(s)\) in a sealed tube with a volume of \(250 \mathrm{~mL}\) that contains \(\mathrm{CO}(g)\) at a pressure of \(101.3 \mathrm{kPa}\) and a temperature of \(298 \mathrm{~K},\) what is the concentration of the CO gas? Assume there is no reaction at this temperature and that the CO behaves as an ideal gas (you can neglect the volume of the solid). \((\mathbf{d})\) If the reaction vessel from part (c) is heated to \(823 \mathrm{~K}\) and allowed to come to equilibrium, how much \(\operatorname{CoO}(s)\) remains?

Consider the reaction $$ \mathrm{CaSO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) $$ At \(25^{\circ} \mathrm{C}\), the equilibrium constant is \(K_{c}=2.4 \times 10^{-5}\) for this reaction. (a) If excess \(\operatorname{CaSO}_{4}(s)\) is mixed with water at \(25^{\circ} \mathrm{C}\) to produce a saturated solution of \(\mathrm{CaSO}_{4}\), what are the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{SO}_{4}{ }^{2-}\) ? (b) If the resulting solution has a volume of \(1.4 \mathrm{~L}\), what is the minimum mass of \(\operatorname{CaSO}_{4}(s)\) needed to achieve equilibrium?

At \(800 \mathrm{~K},\) the equilibrium constant for \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)\) is \(K_{c}=3.1 \times 10^{-5}\). If an equilibrium mixture in a 5.00-L vessel contains \(30.5 \mathrm{mg}\) of \(\mathrm{I}(g)\), how many grams of \(\mathrm{I}_{2}\) are in the mixture?

At \(1285^{\circ} \mathrm{C},\) the equilibrium constant for the reaction \(\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)\) is \(K_{c}=1.04 \times 10^{-3} .\) A 1.00-L vessel containing an equilibrium mixture of the gases has \(1.50 \mathrm{~g}\) \(\mathrm{Br}_{2}(g)\) in it. What is the mass of \(\operatorname{Br}(g)\) in the vessel?

Write the expression for \(K_{c}\) for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous. (a) \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)\) (b) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\) (c) \(\mathrm{CO}_{2}(g)+\mathrm{C}(s) \rightleftharpoons 2 \mathrm{CO}(g)\) (d) \(\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\) (e) \(\mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q)\) (f) \(\mathrm{Fe}^{2+}(a q)+\mathrm{Zn}(s) \rightleftharpoons \mathrm{Zn}^{2+}(a q)+\mathrm{Fe}(s)\) (g) \(\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}^{2-}(a q)\)
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