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A diamond can be considered a giant all-carbon supermolecule in which almost every carbon atom is bonded to four other carbons. When a diamond cutter cleaves (splits) a diamond, carbon-carbon bonds must be broken. Is the cleavage (splitting) of a diamond endothermic or exothermic? Explain.

Short Answer

Expert verified
The cleavage of a diamond is endothermic because it requires energy to break carbon-carbon bonds.

Step by step solution

01

Understanding Endothermic vs Exothermic

First, let's understand the difference between endothermic and exothermic processes. Endothermic processes absorb energy from their surroundings, usually in the form of heat, while exothermic processes release energy.
02

Analyzing Diamond Bonding

A diamond consists of carbon atoms each covalently bonded to four other carbon atoms. These covalent bonds are strong and require energy to break.
03

Energy Requirement for Bond Breaking

To cleave a diamond, carbon-carbon bonds must be broken. Since energy is required to break these strong covalent bonds, the process must absorb energy.
04

Conclusion on Process Nature

Because the cleavage process absorbs energy to break the bonds, it is an endothermic process.

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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 all about absorbing energy from the surroundings. These processes need a supply of energy to occur. Most commonly, this energy comes in the form of heat. When you mix ingredients in a chemical reaction and the surroundings feel cooler, it's likely an endothermic reaction.
Examples of endothermic processes include:
  • Melting ice cubes
  • Evaporation of water
  • Photosynthesis in plants
In the context of cleaving a diamond, the bonds between carbon atoms need extra energy to be broken. The system absorbs this energy from the external environment, making diamond cleavage an endothermic process. This absorption of energy distinguishes it as endothermic rather than releasing energy like in exothermic reactions.
Covalent Bonds
Covalent bonds form when atoms share pairs of electrons. These bonds create a stable balance of attractive and repulsive forces between atoms. Covalent bonds are particularly common in organic compounds.
Here's what makes covalent bonds special:
  • They involve the sharing of electrons between atoms.
  • They are generally strong and require substantial energy to break.
  • They can occur between atoms of the same or different elements.
In a diamond, each carbon atom forms covalent bonds with four other carbon atoms, creating a robust network. This sharing of electrons leads to the characteristic strength and hardness of diamonds. When a diamond is cleaved, these strong covalent bonds must be overcome, necessitating a high energy input.
Diamond Structure
Diamonds have a unique and extraordinary crystal structure. They are made from carbon atoms, which are one of the most versatile elements. In a diamond, each carbon atom is bonded to four other carbon atoms with covalent bonds, forming a tetrahedral lattice. This three-dimensional network results in a remarkably stable and hard structure.
Key aspects of diamond's structure include:
  • Each carbon atom forms four strong covalent bonds.
  • The arrangement is in a repeating pattern called a crystal lattice.
  • This structure contributes to a diamond's extreme hardness and optical properties.
Understanding this structure helps explain why diamonds are tough to cleave and why breaking their bonds requires significant energy. This necessity for energy absorption during bond breakage in the diamond lattice results in endothermic cleavage.

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Most popular questions from this chapter

Given the thermochemical expression \(\mathrm{H}_{2} \mathrm{O}(\mathrm{s}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\ell)\) $$ \Delta_{\mathrm{r}} H^{\circ}=6.0 \mathrm{~kJ} / \mathrm{mol} $$ calculate what quantity of energy is transferred to the surroundings when (a) 34.2 mol liquid water freezes. (b) \(100.0 \mathrm{~g}\) liquid water freezes.

What is required for heat transfer of energy from one sample of matter to another to occur?

Calculate the quantity of energy, in joules, required to raise the temperature of \(454 \mathrm{~g}\) tin from room temperature, \(25.0^{\circ} \mathrm{C}\), to its melting point, \(231.9{ }^{\circ} \mathrm{C},\) and then melt the tin at that temperature. (The specific heat capacity of tin is \(0.227 \mathrm{~J} \mathrm{~g}^{-1} \mathrm{~K}^{-1}\), and the enthalpy of fusion of this metal is \(59.2 \mathrm{~J} / \mathrm{g} .\) )

When a \(13.0-\mathrm{g}\) sample of \(\mathrm{NaOH}(\mathrm{s})\) dissolves in \(400.0 \mathrm{~mL}\) water in a coffee cup calorimeter, the temperature of the water changes from \(22.6^{\circ} \mathrm{C}\) to \(30.7^{\circ} \mathrm{C}\). Assuming that the specific heat capacity of the solution is the same as for water, calculate (a) The heat transfer from system to surroundings. (b) \(\Delta_{\mathrm{r}} H\) for the reaction. $$ \mathrm{NaOH}(\mathrm{s}) \longrightarrow \mathrm{Na}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) $$

Design an experiment to directly measure the reaction enthalpy for this reaction $$ 2 \mathrm{NaOH}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\ell) $$ Describe the apparatus and how the experiment would be carried out.

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