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Chapter 5: Problem 23

Standard molar enthalpy of formation of \(\mathrm{CO}_{2}\) is equal to (a) zero (b) the standard molar enthalpy of combustion of gaseous carbon (c) the sum of standard molar enthalpies of formation of \(\mathrm{CO}\) and \(\mathrm{O}_{2}\) (d) the standard molar enthalpy of combustion of carbon (graphite)

Short Answer

Expert verified
The standard molar enthalpy of formation of \(\mathrm{CO}_{2}\) is equal to the standard molar enthalpy of combustion of carbon (graphite), which is options (b) and (d) since they represent the same concept.

Step by step solution

01

Understand the Standard Molar Enthalpy of Formation

The standard molar enthalpy of formation of a compound is defined as the enthalpy change when one mole of the compound is formed from its elements in their standard states. The standard state of a substance is its physical state at 1 atmosphere of pressure and a specified temperature (usually 25°C or 298.15K). By convention, the standard molar enthalpy of formation for any element in its most stable form is zero.
02

Evaluate Answer Choices

(a) The standard molar enthalpy of formation of \(\mathrm{CO}_{2}\) is not zero because \(\mathrm{CO}_{2}\) is a compound, not an element in its most stable form.(b) The standard molar enthalpy of formation of a compound is indeed equivalent to the standard molar enthalpy of combustion of its elements. Therefore, for \(\mathrm{CO}_{2}\), it would be equal to the enthalpy of combustion of carbon (in its most stable form, graphite) plus the associated oxygen needed for the formation of \(\mathrm{CO}_{2}\) from its elements.(c) This choice is incorrect because the standard molar enthalpy of formation of \(\mathrm{CO}_{2}\) is not related to the enthalpy of formation of \(\mathrm{CO}\) and \(\mathrm{O}_{2}\) as they are different compounds.(d) This is a restatement of option (b) and refers to the same concept, that the standard molar enthalpy of formation of \(\mathrm{CO}_{2}\) is equal to the standard molar enthalpy of combustion of carbon (in most stable form, graphite), not combined with other substances like \(\mathrm{O}_{2}\).
03

Select the Correct Answer

Considering the definition of standard molar enthalpy of formation and the explanations above, the correct answer is (b) the standard molar enthalpy of combustion of gaseous carbon, which involves the combustion of carbon to form \(\mathrm{CO}_{2}\). Additionally, choice (d) is the same as choice (b) since it refers to the combustion of carbon graphite to form \(\mathrm{CO}_{2}\). Therefore, the correct answer is either (b) or (d) as they are equivalent statements.

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Key Concepts

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

Enthalpy Change
Enthalpy change is a measurement of energy transfer in a chemical reaction that occurs at constant pressure. It's symbolized by \( \Delta H \). When a reaction releases heat, it is exothermic and has a negative enthalpy change. Conversely, when a reaction absorbs heat, it is endothermic, indicated by a positive enthalpy change.

Understanding enthalpy change is crucial for predicting whether reactions will occur spontaneously and for calculating the energy efficiency of reactions. In the context of the exercise, the enthalpy change represents the energy difference when gaseous carbon, in its standard state, reacts with oxygen to form carbon dioxide (\(\mathrm{CO}_2\)).
Chemical Thermodynamics
Chemical thermodynamics is a branch of physical chemistry that deals with the relationship between chemical reactions and thermodynamic laws. The key principles include the laws of thermodynamics which govern the flow and conversion of energy.

One core concept is that the energy of the universe is constant and that energy cannot be created or destroyed — only transformed. This principle helps scientists understand reactions like the combustion of carbon to form \(\mathrm{CO}_2\), where chemical energy is converted to thermal energy. Chemical thermodynamics also provides insights into reaction spontaneity, system equilibria, and predicts the direction of chemical changes.
Combustion Reactions
Combustion reactions are a type of chemical reaction where a substance combines with oxygen to release heat and light. Typically, these reactions involve a hydrocarbon fuel and oxygen from the air, and they produce carbon dioxide and water as byproducts.

In the exercise, the combustion of carbon (graphite) to form carbon dioxide is an example of a combustion reaction. The standard molar enthalpy of combustion measures the energy released when one mole of a substance undergoes complete combustion under standard conditions. This value is crucial in calculating the enthalpy change associated with reactions and assessing their energy performance.
Physical Chemistry
Physical chemistry is the study of how matter behaves on a molecular and atomic level and how chemical reactions occur. Combining principles of physics and chemistry, it provides the framework for understanding chemical reactions' mechanisms, rates, and energy transfers.

Understanding physical chemistry principles allows us to dissect complex processes like the formation of \(\mathrm{CO}_2\) from carbon and oxygen. It also explains how measurements such as the standard molar enthalpy of formation are intrinsic to not only academic exercises but also practical applications in industries like materials science and energy production.

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Most popular questions from this chapter

Use the following data to calculate the enthalpy of hydration for caesium iodide and caesium hydroxide, respectively: $$\begin{array}{ccc} \hline \text { Compound } & \begin{array}{c} \text { Lattice energy } \\ \text { (kJ/mol) } \end{array} & \begin{array}{c} \Delta \boldsymbol{H}_{\text {Solution }} \\ \text { (kJ/mol) } \end{array} \\ \hline \text { CsI } & +604 & +33 \\ \text { CsOH } & +724 & -72 \\ \hline \end{array}$$ (a) \(-571 \mathrm{~kJ} / \mathrm{mol}\) and \(-796 \mathrm{~kJ} / \mathrm{mol}\) (b) \(637 \mathrm{~kJ} / \mathrm{mol}\) and \(652 \mathrm{~kJ} / \mathrm{mol}\) (c) \(-637 \mathrm{~kJ} / \mathrm{mol}\) and \(-652 \mathrm{~kJ} / \mathrm{mol}\) (d) \(571 \mathrm{~kJ} / \mathrm{mol}\) and \(796 \mathrm{~kJ} / \mathrm{mol}\)

The enthalpy of hydrogenation of benzene is \(-51.0\) kcal/mol. If enthalpy of hydrogenation of 1,4 -cyclohexadiene and cyclohexene is \(-58 \mathrm{kcal} / \mathrm{mol}\) and \(-29 \mathrm{kcal} / \mathrm{mol}\), respectively, what is the resonance energy of benzene? (a) \(29 \mathrm{kcal} / \mathrm{mole}\) (b) \(36 \mathrm{kcal} / \mathrm{mole}\) (c) \(58 \mathrm{kcal} / \mathrm{mole}\) (d) \(7 \mathrm{kcal} / \mathrm{mole}\)

Equal volumes of one molar hydrochloric acid and one molar sulphuric acid are neutralized completely by dilute \(\mathrm{NaOH}\) solution by which \(X\) and \(Y\) kcal of heat are liberated, respectively. Which of the following is true? (a) \(X=Y\) (b) \(2 X=Y\) (c) \(X=2 Y\) (d) none of these

The \(\Delta G^{\circ}\) values for the hydrolysis of creatine phosphate (creatine-P) and glucose-6-phosphate \((\mathrm{G}-6-\mathrm{P})\) are (i) Creatine-P \(+\mathrm{H}_{2} \mathrm{O} \rightarrow\) Creatine \(+\mathrm{P}\) \(\Delta G^{\mathrm{o}}=-29.2 \mathrm{~kJ}\) (ii) \(\mathrm{G}-6-\mathrm{P}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{G}+\mathrm{P} ; \Delta G^{\mathrm{o}}=-12.4 \mathrm{~kJ}\) \(\Delta G^{0}\) for the reaction: \(\mathrm{G}-6-\mathrm{P}+\) Creatine \(\rightarrow \mathrm{G}+\) Creatine- \(\mathrm{P}\), is (a) \(+16.8 \mathrm{~kJ}\) (b) \(-16.8 \mathrm{~kJ}\) (c) \(-41.6 \mathrm{~kJ}\) (d) \(+41.6 \mathrm{~kJ}\)

Calculate \(\Delta_{\mathrm{f}} H\) for \(\mathrm{ZnSO}_{4}(\mathrm{~s})\) from the following data: \(\mathrm{ZnS}(\mathrm{s}) \rightarrow \mathrm{Zn}(\mathrm{s})+\mathrm{S}\) (rhombic), \(\Delta H_{1}\) \(=44 \mathrm{kcal} / \mathrm{mol}\) \(2 \mathrm{ZnS}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{ZnO}(\mathrm{s})+2 \mathrm{SO}_{2}(\mathrm{~g})\) \(\Delta H_{2}=-221.88 \mathrm{kcal} / \mathrm{mol}\) \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{SO}_{3}(\mathrm{~g}), \quad \Delta H_{3}\) \(=-46.88 \mathrm{kcal} / \mathrm{mol}\) \(\mathrm{ZnSO}_{4}(\mathrm{~s}) \rightarrow \mathrm{ZnO}(\mathrm{s})+\mathrm{SO}_{3}(\mathrm{~g}), \Delta H_{4}\) \(=55.1 \mathrm{kcal} / \mathrm{mol}\) (a) \(-233.48 \mathrm{kcal} / \mathrm{mol}\) (b) \(-343.48 \mathrm{kcal} / \mathrm{mol}\) (c) \(-434.84 \mathrm{kcal} / \mathrm{mol}\) (d) \(-311.53 \mathrm{kcal} / \mathrm{mol}\)
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