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What will be the enthalpy change of conversion of graphite into diamond? \(\begin{aligned} \text { Given } & C_{\text {graphite' }}, \Delta_{\text {comb }} H=&-391.25 \mathrm{k} \mathrm{F} ; \\ & \mathrm{C}_{\text {diamond }}, \Delta_{\text {comb }} H &=-393.12 \mathrm{~kJ} \end{aligned}\) (a) zero (b) \(-391.25 \mathrm{k} \mid\) (c) \(-393.12 \mathrm{~kJ}\) (d) \(-1.87 \mathrm{~kJ}\)

Short Answer

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-1.87 kJ

Step by step solution

01

Understand the Enthalpy Change of Conversion

The enthalpy change of conversion from graphite to diamond can be determined by calculating the difference in enthalpy of formation for both allotropes of carbon. Here, combustion enthalpies for graphite and diamond are provided and we need to find the difference.
02

Calculate the Enthalpy Change of Conversion

Subtract the combustion enthalpy of graphite from the combustion enthalpy of diamond to obtain the enthalpy change of conversion. So, the change in enthalpy \( \Delta H \) is: \[ \Delta H = \Delta_{\text{comb}} H_{\text{diamond}} - \Delta_{\text{comb}} H_{\text{graphite}} \]
03

Perform the Calculation

Using the given data, \[ \Delta H = (-393.12 \text{ kJ}) - (-391.25 \text{ kJ}) = -1.87 \text{ kJ} \] The enthalpy change of conversion of graphite into diamond is -1.87 kJ.

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Key Concepts

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

Enthalpy of Formation
At the heart of chemical thermodynamics is the enthalpy of formation, which is the total heat content, or enthalpy, of a substance when it's formed from its constituent elements in their standard states. It's a crucial concept as it helps chemists understand the energy changes that occur during chemical reactions.

In lay terms, if you think of a chemical reaction as a method of building a product from specific raw materials, the enthalpy of formation would be analogous to the amount of heat either absorbed or released during that building process. This value can be either positive or negative, indicating that the reaction is endothermic (absorbs heat) or exothermic (releases heat) respectively.

For instance, if you were to form water from hydrogen and oxygen gas, the reaction would release a significant amount of heat, giving water a negative enthalpy of formation. Understanding these values helps predict whether a reaction will occur spontaneously and what kind of energy management will be required.
Allotropes of Carbon
Carbon is a fascinating element that exists in various forms called allotropes. Each allotrope has its own unique set of properties despite being made of the same carbon atoms. The most well-known allotropes of carbon are graphite and diamond.

Graphite, used in pencil leads, is composed of layers of carbon atoms arranged in a hexagonal lattice. These layers can slide over each other, which makes graphite soft and a good lubricant.

Diamond

Diamond, on the other hand, is famed for its hardness and is made of carbon atoms covalently bonded in a tetrahedral arrangement, creating a three-dimensional network.

The transformation of graphite into diamond is a matter of rearranging these carbon atoms into a different structural network, and it requires extremely high pressure and temperatures, akin to conditions found deep within the Earth's mantle. This transformation is not only a physical change but also involves changes in enthalpy, as seen in the step-by-step solution.
Combustion Enthalpy
Combustion enthalpy, also known as the heat of combustion, is the energy change when a substance combusts in the presence of oxygen. It's a specific type of enthalpy change and is always negative because combustion is an exothermic reaction — meaning it releases energy in the form of heat.

Take, for example, burning a piece of wood. The wood (typically containing carbon-based compounds) combines with oxygen in the air, producing carbon dioxide, water, and a considerable amount of heat. This heat is the combustion enthalpy of the wood.

In chemical equations and thermodynamic calculations, the combustion enthalpy helps us understand how much heat energy can be obtained by burning a certain quantity of a substance, which is crucial for fields like energy production and engineering.
Chemical Thermodynamics
Chemical thermodynamics is the branch of chemistry that deals with the relationships between heat, work, and energy within chemical reactions. It provides a framework for understanding the energy transfers that accompany chemical processes.

The discipline relies on laws of thermodynamics to explain how and why chemical reactions occur and the direction they take. These laws describe the conservation of energy, the inevitable increase in entropy, and the behavior of substances at absolute zero temperature.

Importantly, thermodynamic principles allow chemists to predict the spontaneity of reactions, optimize chemical processes, design energy-efficient systems, and much more. It helps answer not just 'how much?' but 'will it?' and 'how?' about a reaction, encompassing both the quantitative and qualitative aspects of chemistry.

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Most popular questions from this chapter

For a reaction, \(\mathrm{CaCO}_{3(\mathrm{~s})} \rightarrow \mathrm{CaO}_{(0)}+\mathrm{CO}_{2(g)}\) \(\Delta_{j} H^{\circ}(\mathrm{CaO})=-635.1 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(\Delta_{f} H^{\circ}\left(\mathrm{CO}_{2}\right)=-393.5 \mathrm{k} \mathrm{J} \mathrm{mol}^{-1}\) and \(\Delta_{f} H^{\circ}\left(\mathrm{CaCO}_{3}\right)=-1206.9 \mathrm{k} \mathrm{J} \mathrm{mol}^{-1}\) Which of the following is a correct statement? (a) A large amount of heat is evolved during the decomposition of \(\mathrm{CaCO}_{3}\). (b) Decomposition of \(\mathrm{CaCO}_{3}\) is an endothermic process and heat is provided for decomposition. (c) The amount of heat evolved cannot be calculated from the data provided. (d) \(\Delta_{r} H^{\circ}=\Sigma \Delta_{f} H^{\circ}\) (reactants) \(-\Sigma \Delta_{f} H^{\circ}\) (products)

At what temperature liquid water will be in equilibrium with water vapour? \(\Delta H_{v a p}=40.73 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta S_{v a p}=0.109 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) (a) \(282.4 \mathrm{~K}\) (b) \(373.6 \mathrm{~K}\) (c) \(100 \mathrm{~K}\) (d) \(400 \mathrm{~K}\)

Read the following statements regarding spontaneity of a process and mark the appropriate choice. (i) When enthalpy factor is absent then randomness factor decides spontaneity of a process. (ii) When randomness factor is absent then enthalpy factor decides spontaneity of a process. (iii) When both the factors take place simultaneously, the magnitude of both of factors decide spontaneity of a process. (a) Statements (i) and (ii) are correct and (iii) is incorrect. (b) Statement (iii) is correct, (i) and (ii) are incorrect. (c) Statements (i), (ii) and (iii) are correct. (d) Statements (i), (ii) and (iii) are incorrect.

Match the column I with column II and mark the appropriate choice. $$ \begin{array}{|l|l|l|l|} \hline \text { (A) } & \mathrm{H}_{2(\mathrm{~g})}+\mathrm{Br}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{HBr}_{(g)} & \text { (i) } & \Delta H=\Delta U-2 R T \\ \hline \text { (B) } & \mathrm{PCl}_{5(g)} \rightarrow \mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(g)} & \text { (ii) } & \Delta H=\Delta U+3 R T \\ \hline \text { (C) } & \mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)} \rightarrow 2 \mathrm{NH}_{3(g)} & \text { (iii) } & \Delta H=\Delta U \\ \hline \text { (D) } & \begin{array}{l} 2 \mathrm{~N}_{2} \mathrm{O}_{5(g)} \rightarrow 4 \mathrm{NO}_{2(g)} \\ +\mathrm{O}_{2(g)} \end{array} & \text { (iv) } & \Delta H=\Delta U+R T \\ \hline \end{array} $$ (a) (A) \(\rightarrow\) (iii), (B) \(\rightarrow\) (i), (C) \(\rightarrow\) (ii), (D) \(\rightarrow\) (iv) (b) (A) \(\rightarrow\) (iii), (B) \(\rightarrow\) (iv), (C) \(\rightarrow\) (i), (D) \(\rightarrow\) (ii) (c) (A) \(\rightarrow\) (ii), (B) \(\rightarrow\) (i), (C) \(\rightarrow\) (iv), (D) \(\rightarrow\) (iii) (d) (A) \(\rightarrow\) (iv), (B) \(\rightarrow\) (ii), (C) \(\rightarrow\) (i), (D) \(\rightarrow\) (iii)

For the reaction: \(\mathrm{H}_{2(g)}+\mathrm{Cl}_{2(g)} \rightarrow 2 \mathrm{HCl} ;\) \(\Delta H=-44 \mathrm{kcal}\) What is the enthalpy of decomposition of HCl? (a) \(+44 \mathrm{kcal} / \mathrm{mol}\) (b) - \(44 \mathrm{kcal} / \mathrm{mol}\) (c) \(-22 \mathrm{kcal} / \mathrm{mol}\) (d) \(+22 \mathrm{kcal} / \mathrm{mol}\)

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