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Chapter 14: Problem 78

One way of preparing hydrogen is by the decomposition of water. $$ 2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) ; \Delta H^{\circ}=484 \mathrm{~kJ} $$ Would you expect the decomposition to be favorable at high or low temperature? Explain.

Short Answer

Expert verified
The decomposition is more favorable at high temperatures.

Step by step solution

01

Understand the Reaction

The given reaction is the decomposition of water into hydrogen and oxygen gases. It is described by the equation: \[ 2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g) + \mathrm{O}_{2}(g) \] with an enthalpy change \( \Delta H^{\circ} = 484 \text{ kJ} \). This positive \( \Delta H^{\circ} \) value indicates that the reaction is endothermic, meaning it absorbs heat.
02

Consider the Effect of Temperature on an Endothermic Reaction

For endothermic reactions, according to Le Chatelier's Principle, increasing the temperature shifts the equilibrium toward the products, making the reaction more favorable. This is because the system absorbs the extra heat, which helps drive the endothermic reaction forward.
03

Determine Temperature Preference

Since the decomposition of water is endothermic and requires heat, it will be more favorable at higher temperatures. At higher temperatures, the additional energy provided can overcome the activation energy barrier and drive the reaction towards the formation of hydrogen and oxygen gases.

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Key Concepts

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

Endothermic Reaction
An endothermic reaction is a type of chemical process that involves the absorption of heat from its surroundings. In these reactions, the energy required to break the bonds in the reactants is greater than the energy released when new bonds form in the products. As a result, endothermic reactions have a positive enthalpy change (\( \Delta H^{\circ} > 0 \)).

In the context of the decomposition of water:
  • The reaction absorbs energy from the environment to break the bonds in water molecules.
  • This process forms hydrogen and oxygen gases.
  • The positive enthalpy change, \( \Delta H^{\circ} = 484 \text{ kJ} \) , confirms the reaction is endothermic.
Endothermic reactions often require an external source of energy, such as heat, to proceed effectively. Increasing the temperature can provide the necessary energy to drive these reactions forward.
Le Chatelier's Principle
Le Chatelier's Principle helps us understand how systems at equilibrium respond to changes in their environment. It states that if an external change is imposed on a system at equilibrium, the system will adjust itself to counteract that change and restore a new equilibrium.

For the decomposition of water, which is an endothermic process:
  • If the temperature is increased, the system reacts by absorbing more heat, which favors the formation of products (hydrogen and oxygen gases).
  • This shift in equilibrium towards the products occurs because the reaction uses the heat energy to drive the decomposition process forward.
Le Chatelier's Principle is key in predicting how a chemical reaction at equilibrium will adjust under changes in temperature, pressure, or concentration.
Enthalpy Change
Enthalpy change (\( \Delta H^{\circ} \)) is a measure of the total energy exchange in a chemical reaction, typically evaluated at constant pressure. It represents the difference in energy between the reactants and the products.

For water decomposition:
  • The enthalpy change is \( 484 \text{ kJ} \) , which indicates the energy absorbed during the reaction when two moles of water decompose into hydrogen and oxygen.
  • Positive enthalpy changes are characteristic of endothermic reactions.
Understanding enthalpy change is crucial in predicting whether a reaction will require or release energy, affecting its feasibility and conditions under which it is preferred.
Equilibrium Shift
An equilibrium shift occurs when a change in conditions causes the equilibrium position of a chemical reaction to move. According to Le Chatelier鈥檚 Principle, a system will shift to oppose any changes in its equilibrium state.

In the decomposition of water:
  • As temperature increases, the reaction absorbs more thermal energy, which shifts the equilibrium towards the formation of hydrogen and oxygen gases.
  • This shift favors the products of the endothermic reaction at higher temperatures.
An equilibrium shift can be influenced by changes in concentration, pressure, or temperature, and understanding these shifts helps chemists control reaction conditions for desired outcomes.

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Most popular questions from this chapter

The following equilibrium was studied by analyzing the equilibrium mixture for the amount of \(\mathrm{HCl}\) produced. $$ \mathrm{LaCl}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{LaOCl}(s)+2 \mathrm{HCl}(g) $$ A vessel whose volume was \(1.25 \mathrm{~L}\) was filled with \(0.0125 \mathrm{~mol}\) of lanthanum(III) chloride, \(\mathrm{LaCl}_{3}\), and \(0.0250 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\). After the mixture came to equilibrium in the closed vessel at \(619^{\circ} \mathrm{C}\), the gaseous mixture was removed and dissolved in more water. Sufficient silver ion was added to precipitate the chloride ion completely as silver chloride. If \(3.59 \mathrm{~g} \mathrm{AgCl}\) was obtained, what is the value of \(K_{c}\) at \(619^{\circ} \mathrm{C} ?\)

A mixture of carbon monoxide, hydrogen, and methanol, \(\mathrm{CH}_{3} \mathrm{OH},\) is at equilibrium according to the equation $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$ At \(290^{\circ} \mathrm{C},\) the mixture is \(0.034 \mathrm{M} \mathrm{CO}, 0.450 \mathrm{M} \mathrm{H}_{2},\) and \(0.00023 \mathrm{M} \mathrm{CH}_{3} \mathrm{OH}\). What is \(K_{c}\) for this reaction at \(290^{\circ} \mathrm{C} ?\)

Calculate the composition of the gaseous mixture obtained when \(1.25 \mathrm{~mol}\) of carbon dioxide is exposed to hot carbon at \(800^{\circ} \mathrm{C}\) in a 1.25 - \(\mathrm{L}\) vessel. The equilibrium constant \(K_{c}\) at \(800^{\circ} \mathrm{C}\) is 14.0 for the reaction $$ \mathrm{CO}_{2}(g)+\mathrm{C}(s) \rightleftharpoons 2 \mathrm{CO}(g) $$

A chemist put 1.18 mol of substance \(A\) and 2.85 mol of substance \(\mathrm{B}\) into a 10.0 - \(\mathrm{L}\) flask, which she then closed. A and B react by the following equation: $$ \mathrm{A}(g)+2 \mathrm{~B}(g) \rightleftharpoons 3 \mathrm{C}(g)+\mathrm{D}(g) $$ She found that the equilibrium mixture at \(25^{\circ} \mathrm{C}\) contained \(0.376 \mathrm{~mol}\) of \(\mathrm{D}\). How many moles of \(\mathrm{B}\) are in the flask at equilibrium at \(25^{\circ} \mathrm{C} ?\) a. \(2.47 \mathrm{~mol}\) b. \(3.60 \mathrm{~mol}\) c. \(2.52 \mathrm{~mol}\) d. \(2.10 \mathrm{~mol}\) e. \(2.41 \mathrm{~mol}\)

Nitrogen monoxide, \(\mathrm{NO},\) is formed in automobile exhaust by the reaction of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) (from air). $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ The equilibrium constant \(K_{c}\) is 0.0025 at \(2127^{\circ} \mathrm{C}\). If an equilibrium mixture at this temperature contains \(0.023 \mathrm{~mol} \mathrm{~N}_{2}\) and \(0.031 \mathrm{~mol} \mathrm{O}_{2}\) per liter, what is the concentration of \(\mathrm{NO}\) ?
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