Both graphite and diamond burn. \(\mathrm{C}(s, \text { diamond })+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)\) For the conversion of graphite to diamond: \(\mathbf{C}(s, \text { graphite }) \longrightarrow \mathbf{C}(s, \text { diamond })\) \(\Delta H^{\circ}=1.90 \mathrm{kJ}\) Which produces more heat, the combustion of graphite or the combustion of diamond?

Short Answer

Expert verified
The combustion of graphite produces more heat than the combustion of diamond.

Step by step solution

01

Identify the Chemical Reactions

Recognize that the given reactions are the combustion reactions for diamond and graphite. Both reactions convert carbon to carbon dioxide (CO2) using oxygen.
02

Analyze the Thermochemical Equations

Understand that the combustion reactions will release energy (exothermic), and the conversion from graphite to diamond is endothermic, absorbing 1.90 kJ.
03

Compare the Heat of Combustion

Since the only difference between the combustion of graphite and diamond is the initial form of carbon, and the conversion from graphite to diamond requires an absorption of heat, the combustion of graphite will release more heat compared to diamond. To combust diamond, one must first input the energy to convert graphite to diamond, then combust it, leading to a lower net release of heat.

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Key Concepts

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

Chemical Reactions
Understanding the fundamentals of chemical reactions is crucial in analyzing combustion processes. A chemical reaction is a process where one set of chemical substances (reactants) is transformed into another (products). This is accompanied by the making and breaking of chemical bonds, leading to a change in the properties of substances.

For example, when we consider the combustion of carbon forms such as graphite and diamond, the chemical reaction involves carbon (\textbf{C}) and oxygen gas (\textbf{O}\(_2\)), the reactants, being converted to carbon dioxide (CO\(_2\)), the product. Despite both graphite and diamond being forms of carbon, the structure and stability of their bonds vary, which in turn affects the amount of energy released during the combustion process. It's this difference in energy release that answers the question about which form of carbon, graphite or diamond, combusts to produce more heat.
Thermochemical Equations
Thermochemical equations provide a way to understand the energy changes during chemical reactions. They not only show the stoichiometry of the reactions but also indicate whether heat is absorbed or released, and in what quantity. In such equations, the heat of the reaction is treated as a product (for exothermic reactions) or a reactant (for endothermic reactions).

Take the conversion of graphite to diamond for instance, the thermochemical equation is expressed as:
\textbf{C}(s, \text{ graphite }) \rightarrow \textbf{C}(s, \text{ diamond }) \[\Delta H^\circ = 1.90 \text{kJ}\]

This indicates that transforming graphite into diamond requires absorption of 1.90 kJ of energy, making it an endothermic process. In context, the combustion of graphite and diamond must account for this energy difference to accurately determine which releases more heat.
Heat of Combustion
The heat of combustion is the total amount of heat that is released when a substance undergoes complete combustion with oxygen under standard conditions. It is intimately tied to the structure of the molecule being combusted: the stronger the bonds within the reactant molecules, the less heat will be released upon combustion. Conversely, weaker bonds in the reactants typically mean more heat will be released when those bonds are broken.

Drawing on our specific example, the combustion of diamond would involve breaking stronger covalent bonds compared to graphite. Therefore, graphite combustion releases more heat because converting between these forms requires additional energy input. For such comparisons, the heat of combustion is a critical concept that allows us to predict which substances release more energy and thus has practical significance in fields such as energy production and material science.

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Most popular questions from this chapter

Calculate the standard molar enthalpy of formation of \(\mathrm{NO}(g)\) from the following data: \(\mathrm{N}_{2}(g)+2 \mathrm{O}_{2} \longrightarrow 2 \mathrm{NO}_{2}(g) \quad \Delta H^{\circ}=66.4 \mathrm{kJ}\) \(2 \mathrm{NO}(g)+\mathrm{O}_{2} \longrightarrow 2 \mathrm{NO}_{2}(g) \quad \Delta H^{\circ}=-114.1 \mathrm{kJ}\)

The addition of 3.15 g of \(\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}\) to a solution of \(1.52 \mathrm{g}\) of \(\mathrm{NH}_{4} \mathrm{SCN}\) in \(100 \mathrm{g}\) of water in a calorimeter caused the temperature to fall by \(3.1^{\circ} \mathrm{C} .\) Assuming the specific heat of the solution and products is \(4.20 \mathrm{J} / \mathrm{g}^{\circ} \mathrm{C}\) calculate the approximate amount of heat absorbed by the reaction, which can be represented by the following equation: $$\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}(s)+2 \mathrm{NH}_{4} \mathrm{SCN}(a q) \longrightarrow \mathrm{Ba}(\mathrm{SCN})_{2}(a q)+2 \mathrm{NH}_{3}(a q)+10 \mathrm{H}_{2} \mathrm{O}(I)$$

How much heat, in joules and in calories, must be added to a 75.0-g iron block with a specific heat of 0.449 J/g "C to increase its temperature from \(25^{\circ} \mathrm{C}\) to its melting temperature of \(1535^{\circ} \mathrm{C}\) ?

A 45-g aluminum spoon (specific heat 0.88 J/g ^ C) at 24 ^ C is placed in 180 mL (180 g) of coffee at 85 ^ C and the temperature of the two become equal. (a) What is the final temperature when the two become equal? Assume that coffee has the same specific heat as water. (b) The first time a student solved this problem she got an answer of \(88^{\circ} \mathrm{C}\). Explain why this is clearly an incorrect answer.

Calculate \(\Delta H^{\circ}\) for the process \(\mathrm{Zn}(s)+\mathrm{S}(s)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{ZnSO}_{4}(s)\) from the following information: \(\mathrm{Zn}(s)+\mathrm{S}(s) \longrightarrow \mathrm{ZnS}(s) \quad \Delta H^{\circ}=-206.0 \mathrm{kJ}\) \(\mathrm{ZnS}(s)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{ZnSO}_{4}(s) \quad \Delta H^{\circ}=-776.8 \mathrm{kJ}\)

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