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The decomposition of \(\mathrm{Ca}(\mathrm{OH})_{2}(s)\) into \(\mathrm{CaO}(s)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\) at constant pressure requires the addition of 109 \(\mathrm{kJ}\) of heat per mole of \(\mathrm{Ca}(\mathrm{OH})_{2}\) . (a) Write a balanced thermochemical equation for the reaction. (b) Draw an enthalpy diagram for the reaction.

Short Answer

Expert verified
The thermochemical equation for the decomposition of calcium hydroxide is \(Ca(OH)_2(s) + 109 kJ \rightarrow CaO(s) + H_2O(g)\), and the enthalpy diagram demonstrates that the reaction is endothermic with an enthalpy change of \(+109 kJ\). The diagram shows the reactants at a lower enthalpy level and the products at a higher level with a vertical arrow indicating the change in enthalpy.

Step by step solution

01

Write the balanced chemical equation.

To write the balanced chemical equation for the decomposition of calcium hydroxide, we need to show that the number of atoms for each element is equal on both sides of the equation: Ca(OH)₂(s) → CaO(s) + H₂O(g) This equation is already balanced since there is one calcium, two oxygen, and two hydrogen atoms on both sides. Step 2. Write the thermochemical equation for the reaction.
02

Write the thermochemical equation.

We are given that the reaction requires 109 kJ of heat per mole of Ca(OH)₂. The heat added to the reaction is positive because it is an endothermic reaction (the system absorbs energy from its surroundings). The thermochemical equation will include this heat value: Ca(OH)₂(s) + 109 kJ → CaO(s) + H₂O(g) Step 3. Draw an enthalpy diagram for the reaction.
03

Draw the enthalpy diagram.

To draw an enthalpy diagram for the reaction, we need to represent the differences in enthalpy (ΔH) between the reactants and products. For an endothermic reaction, the enthalpy of the products is higher than that of the reactants. The diagram will show the reactants on a lower enthalpy level, and the products on a higher enthalpy level with a vertical arrow between them indicating the change in enthalpy (+109 kJ). The enthalpy diagram should look like this: Reactants: Ca(OH)₂(s) | | +109 kJ ↓ Products: CaO(s) + H₂O(g) In conclusion, the thermochemical equation for the decomposition of calcium hydroxide is Ca(OH)₂(s) + 109 kJ → CaO(s) + H₂O(g), and the enthalpy diagram demonstrates that the reaction is endothermic with an enthalpy change of +109 kJ.

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Key Concepts

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

Understanding Enthalpy Diagrams
An enthalpy diagram is a visual representation of the enthalpy changes during a chemical reaction. It helps us understand whether a reaction is endothermic or exothermic. In an enthalpy diagram, the vertical axis typically represents the enthalpy (H), which is essentially the total heat content of a system.

For an endothermic reaction like the decomposition of calcium hydroxide, the diagram features the reactants at a lower position compared to the products. The heat absorbed by the system is depicted by a vertical arrow pointing upwards, indicating an increase in enthalpy from reactants to products, which is specifically shown as +109 kJ for this reaction.

This straightforward visual tool allows us to quickly gauge the energy changes in a reaction, providing clarity on how much energy is needed to progress from reactants to products.
Exploring Endothermic Reactions
Endothermic reactions are those in which a system absorbs energy from its surroundings in the form of heat. This absorption leads to an overall increase in the internal energy of the products compared to the reactants.

In the context of calcium hydroxide decomposition, 109 kJ of heat energy is required for the reaction to proceed, making it an endothermic process. Such reactions are characterized by:
  • Positive enthalpy change (\(ΔH > 0\)), indicating energy absorption.
  • Reactants being lower in energy compared to products.
  • External energy input needed for the reaction to occur.
Understanding these key features helps in predicting reaction behavior and energy requirements, which are crucial for processes like this decomposition.
Calcium Hydroxide Decomposition
The decomposition of calcium hydroxide (\( ext{Ca(OH)}_{2} ext{(s)}\)) involves its breakdown into calcium oxide (\( ext{CaO(s)}\)) and water vapor (\( ext{H}_{2} ext{O(g)}\)). The process can be represented by the equation: \[ ext{Ca(OH)}_{2} ext{(s)} ightarrow ext{CaO(s)} + ext{H}_{2} ext{O(g)}\].

This reaction requires an input of heat to break the bonds in calcium hydroxide, leading to a bond reformation in the products. Because it is endothermic, the system takes in 109 kJ per mole of \( ext{Ca(OH)}_{2}\), making it essential to supply sufficient energy to drive the reaction forward.

Recognizing the nature of this decomposition is useful in industrial and laboratory settings, where controlling temperature and energy input is crucial for efficient reaction management.

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Most popular questions from this chapter

How much work (in J) is involved in a chemical reaction if the volume decreases from 5.00 to 1.26 L against a constant pressure of 0.857 atm?

We can use Hess's law to calculate enthalpy changes that cannot be measured. One such reaction is the conversion of methane to ethylene: $$2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g)$$ Calculate the \(\Delta H^{\circ}\) for this reaction using the following thermochemical data: $$\begin{array}{ll}{\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-890.3 \mathrm{kJ}} \\ {\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)} & {\Delta H^{\circ}=-136.3 \mathrm{kJ}} \\ {2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-571.6 \mathrm{kJ}} \\ {2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-3120.8 \mathrm{kJ}}\end{array}$$

From the enthalpies of reaction $$\begin{aligned} \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{HF}(g) & \Delta H=-537 \mathrm{kJ} \\ \mathrm{C}(s)+2 \mathrm{F}_{2}(g) \longrightarrow \mathrm{CF}_{4}(g) & \Delta H=-680 \mathrm{kJ} \\ 2 \mathrm{C}(s)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g) & \Delta H=+52.3 \mathrm{kJ} \end{aligned}$$ calculate \(\Delta H\) for the reaction of ethylene with \(\mathrm{F}_{2} :\) $$\mathrm{C}_{2} \mathrm{H}_{4}(g)+6 \mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{CF}_{4}(g)+4 \mathrm{HF}(g)$$

Complete combustion of 1 mol of acetone \(\left(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}\right)\) liberates \(1790 \mathrm{kJ} :\) $$\begin{aligned} \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}(l)+4 \mathrm{O}_{2}(g) \longrightarrow & 3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) \\ & \quad \quad \quad \quad \quad \quad \quad \Delta H^{\circ}=-1790 \mathrm{kJ} \end{aligned}$$ Using this information together with the standard enthalpies of formation of \(\mathrm{O}_{2}(g), \mathrm{CO}_{2}(g),\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) from Appendix \(\mathrm{C},\) calculate the standard enthalpy of formation of acetone.

At \(20^{\circ} \mathrm{C}\) (approximately room temperature) the average velocity of \(\mathrm{N}_{2}\) molecules in air is 1050 \(\mathrm{mph}\) . (a) What is the average speed in \(\mathrm{m} / \mathrm{s} ?(\mathbf{b})\) What is the kinetic energy (in J) of an \(\mathrm{N}_{2}\) molecule moving at this speed? (c) What is the total kinetic energy of 1 mol of \(\mathrm{N}_{2}\) molecules moving at this speed?

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