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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 favored at high temperatures due to its endothermic nature.

Step by step solution

01

Understanding the Reaction

The decomposition of water to produce hydrogen and oxygen is shown as a reversible reaction. The enthalpy change \( \Delta H^{\circ} \) is \( 484 \text{ kJ} \), which indicates that the reaction is endothermic.
02

Impact of Temperature on Endothermic Reactions

In endothermic reactions, heat is absorbed. According to Le Chatelier's principle, increasing the temperature favors the endothermic direction of the reaction because the system shifts to absorb the added heat.
03

Temperature Favorability Analysis

Since the decomposition of water is an endothermic process, increasing temperature will favor the decomposition into hydrogen and oxygen as the system will absorb more energy.
04

Conclusion

Therefore, the decomposition of water is likely more favorable at high temperatures as it supports the absorption of energy required for the reaction.

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Key Concepts

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

endothermic reactions
In chemistry, reactions are often categorized based on their energy changes. One key category is the endothermic reaction, where the system absorbs energy from its surroundings. Unlike exothermic reactions that release energy, endothermic reactions need an input of energy to proceed. This energy absorption is typically reflected in a positive enthalpy change (ΔH). When you see a positive enthalpy value, like the 484 kJ for the decomposition of water, it signals an endothermic process.
  • Energy must be supplied for the reaction to occur.
  • The absorbed energy breaks the chemical bonds in the reactants.
Endothermic reactions are often found in processes that require heating, like the melting of ice or the evaporation of water. Understanding these reactions is crucial because they explain why certain processes require heat to move forward, highlighting the intrinsic link between energy and chemical transformations.
Le Chatelier's principle
Le Chatelier's principle is a fundamental concept in chemical equilibrium that helps predict how a change in conditions affects a reaction. When a system at equilibrium is subjected to a change in temperature, pressure, or concentration, the system shifts to counteract that change. This principle is very helpful in determining the direction of a reaction when external conditions are altered.
  • For endothermic reactions, increasing temperature causes the system to absorb heat, thus shifting the equilibrium towards the products.
  • For exothermic reactions, decreasing temperature favors the production of more products to release heat.
Applying this principle to the decomposition of water, we can predict that higher temperatures would drive the reaction towards the formation of hydrogen and oxygen, as the reaction absorbs further energy to reach a new equilibrium state.
enthalpy change
Enthalpy change is a concept that represents the heat content variation during a reaction at constant pressure. For chemical reactions, enthalpy change is denoted by (ΔH) and informs whether the process absorbs or releases heat. In endothermic reactions, as seen in the water decomposition equation 2H₂O(g) ⇌ 2H₂(g) + O₂(g), ΔH = 484 kJ, which means 484 kJ of energy are absorbed.
  • A positive ΔH means the reaction is endothermic (energy is absorbed).
  • A negative ΔH means the reaction is exothermic (energy is released).
Understanding enthalpy change helps specify the energy requirements or releases during reactions and is a valuable predictor of reaction conditions, such as whether heat must be supplied or removed to encourage the formation of products.
temperature effects on reactions
Temperature is a crucial factor affecting chemical reactions because it influences the kinetic energy of molecules. Raising the temperature provides more energy for the reactants, allowing successful collisions and reactions to happen more frequently. In the context of chemical equilibrium, temperature shifts the balance between reactants and products.
  • In endothermic reactions, increasing temperature favors the production of products, as higher kinetic energy allows the reaction to absorb more heat.
  • In exothermic reactions, lowering the temperature favors the formation of products by releasing heat.
For the hydrogen and oxygen production from water, high temperatures make the decomposition favorable due to increased energy input, enabling the reaction to proceed more readily. This insight is essential for industrial processes that rely on temperature-modulated chemical reactions, optimizing energy use and efficiency.

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Most popular questions from this chapter

Phosgene, \(\mathrm{COCl}_{2}\), used in the manufacture of polyurethane plastics, is prepared from \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\). $$ \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}_{2}(g) $$ An equilibrium mixture at \(395^{\circ} \mathrm{C}\) contains \(0.012 \mathrm{~mol} \mathrm{CO}\) and \(0.025 \mathrm{~mol} \mathrm{Cl}_{2}\) per liter, as well as \(\mathrm{COCl}_{2}\). If \(K_{c}\) at \(395^{\circ} \mathrm{C}\) is \(1.23 \times 10^{3}\), what is the concentration of \(\mathrm{COCl}_{2} ?\)

Methanol, \(\mathrm{CH}_{3} \mathrm{OH}\), is manufactured industrially by the reaction $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$ A gaseous mixture at \(500 \mathrm{~K}\) is \(0.020 \mathrm{M} \mathrm{CH}_{3} \mathrm{OH}, 0.10 \mathrm{M} \mathrm{CO}\), and \(0.10 M \mathrm{H}_{2} .\) What will be the direction of reaction if this mixture goes to equilibrium? The equilibrium constant \(K_{c}\) equals \(10.5\) at \(500 \mathrm{~K}\).

Phosphorus(V) chloride, \(\mathrm{PCl}_{5}\), dissociates on heating to give phosphorus(III) chloride, \(\mathrm{PCl}_{3}\), and chlorine. $$ \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$ A closed \(2.00-\mathrm{L}\) vessel initially contains \(0.0100 \mathrm{~mol} \mathrm{PCl}_{5}\). What is the total pressure at \(250^{\circ} \mathrm{C}\) when equilibrium is achieved? The value of \(K_{c}\) at \(250^{\circ} \mathrm{C}\) is \(4.15 \times 10^{-2}\).

The equilibrium constant \(K_{c}\) for the equation $$ \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) $$ at \(900^{\circ} \mathrm{C}\) is \(27.8\). What is the value of \(K_{c}\) for the following equation? $$ \frac{1}{2} \mathrm{CS}_{2}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \frac{1}{2} \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{~S}(g) $$

A 4.00-L vessel contained \(0.0148\) mol of phosphorus trichloride, \(0.0126 \mathrm{~mol}\) of phosphorus pentachloride, and \(0.0870 \mathrm{~mol}\) of chlorine at \(230^{\circ} \mathrm{C}\) in an equilibrium mixture. Calculate the value of \(K_{c}\) for the reaction $$ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) $$

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