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The enthalpies of combustion of carbon and carbon monoxide are \(-393.5\) and \(-283 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively. The enthalpy of formation of carbon monoxide per mole is (a) \(-676.5 \mathrm{~kJ}\) (b) \(-110.5 \mathrm{~kJ}\) (c) \(110.5 \mathrm{~kJ}\) (d) \(676.5 \mathrm{~kJ}\)

Short Answer

Expert verified
The enthalpy of formation of carbon monoxide is (b) -110.5 \text{kJ/mol}.

Step by step solution

01

Write Combustion Reactions

The combustion reaction for carbon is: \( C(s) + O_2(g) \rightarrow CO_2(g) \) with the enthalpy change of \(-393.5 \text{ kJ/mol} \). For carbon monoxide, the reaction is: \( 2CO(g) + O_2(g) \rightarrow 2CO_2(g) \) with the enthalpy change of \(-566 \text{ kJ/mol} \), as the combustion of one mole of carbon monoxide releases \(-283 \text{ kJ/mol} \).
02

Write Formation Reaction for CO

The formation reaction of carbon monoxide from its elements is: \( C(s) + \frac{1}{2}O_2(g) \rightarrow CO(g) \). The enthalpy change for this reaction is what we are solving for.
03

Use Hess’s Law

By Hess's Law, the enthalpy change for the formation of CO can be determined by manipulating the given combustion reactions. Reverse the carbon monoxide combustion reaction: \( CO_2(g) \rightarrow CO(g) + \frac{1}{2}O_2(g) \), which has an enthalpy change of \(+283 \text{ kJ/mol} \).
04

Calculate Enthalpy Change for CO Formation

Combine the reactions from Steps 1 and 3 to derive the equation: \( C(s) + \frac{1}{2}O_2(g) \rightarrow CO(g) \). Thus, \( \Delta H_f = -393.5 \text{ kJ/mol} + 283 \text{ kJ/mol} = -110.5 \text{ kJ/mol} \).

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Key Concepts

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

Hess's Law: A Fundamental Principle
Hess's Law is a powerful tool used in thermodynamics to determine the enthalpy change of a reaction. It states that the total enthalpy change in a chemical reaction is the same, no matter how the reaction is carried out in steps. Here's how it works:
  • Each step of a reaction has an associated enthalpy change. These are added up to find the total enthalpy change.
  • Because enthalpy is a state function, it depends only on the initial and final states, not on the path taken.
  • This principle allows the calculation of enthalpy changes by using known reaction enthalpies.
In the context of calculating the enthalpy of formation for carbon monoxide, Hess's Law is used to sum up the enthalpies of combustion of carbon and carbon monoxide. By doing so, it becomes clear that the reaction pathway doesn't alter the overall energy change. This principle helps clarify that no matter the reaction steps, the enthalpy change remains constant.
Enthalpy of Combustion and Its Importance
Enthalpy of combustion refers to the heat released when one mole of a substance combusts completely with oxygen. It is typically expressed in kilojoules per mole (kJ/mol) and is always a negative value because combustion releases energy. Let's break it down:
  • Combustion reactions involve a substance reacting with oxygen to form oxides, releasing energy.
  • For carbon, the combustion process produces carbon dioxide: \( C(s) + O_2(g) \rightarrow CO_2(g) \), releasing \(-393.5 \text{ kJ/mol}\).
  • The combustion of carbon monoxide releases \(-283 \text{ kJ/mol}\) and forms carbon dioxide.
The enthalpy of combustion is crucial as it helps determine the efficiency and energy yield of fuels. This value allows chemists to compare different fuels and understand their performance during chemical reactions. By knowing these values, we can predict and manipulate energy changes in various chemical processes.
Understanding Carbon Monoxide Formation
Carbon monoxide (CO) is formed when carbon is not completely oxidized, resulting in the partial combustion of carbon. Its formation can be represented by the reaction:
  • \( C(s) + \frac{1}{2}O_2(g) \rightarrow CO(g) \)
Carbon monoxide is both a significant industrial compound and a notorious pollutant. Its formation is pivotal in many industrial processes such as:
  • Partial oxidation in processes like steel manufacturing.
  • As a precursor in organic synthesis.
Even though it has practical uses, carbon monoxide is hazardous to health, known for binding to hemoglobin more effectively than oxygen, inhibiting oxygen transport in the body. Therefore, understanding its formation not only aids in chemical production but also highlights the importance of safety measures in industries where it is generated. Through Hess's Law and by understanding the combustion reactions, we can also calculate the enthalpy of formation for CO, obtaining essential insights into its chemical behavior.

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Most popular questions from this chapter

The work done by a system is 10 joule, when 40 joule heat is supplied to it. What is the increase in internal energy of system? (a) \(30 \mathrm{~J}\) (b) \(50 \mathrm{~J}\) (c) \(40 \mathrm{~J}\) (d) \(20 \mathrm{~J}\)

Two moles of an ideal gas are compressed at \(300 \mathrm{~K}\) from a pressure of 1 atm to a pressure of 2 atm. The change in free energy is (a) \(5.46 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(2.46 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(3.46 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(8.46 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

If at \(298 \mathrm{~K}\) the bond energies of \(\mathrm{C}-\mathrm{H}, \mathrm{C}-\mathrm{C}, \mathrm{C}=\mathrm{C}\) and \(\mathrm{H}-\mathrm{H}\) bonds are respectively \(414,347,615\) and \(435 \mathrm{~kJ} \mathrm{~mol}^{-1}\), the value of enthalpy change for the reaction \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{3}(\mathrm{~g})\) at \(298 \mathrm{~K}\) will be (a) \(+250 \mathrm{~kJ}\) (b) \(-250 \mathrm{~kJ}\) (c) \(+125 \mathrm{~kJ}\) (d) \(-125 \mathrm{~kJ}\)

The internal energy change when a system goes from state \(\mathrm{A}\) to \(\mathrm{B}\) is \(40 \mathrm{~kJ} / \mathrm{mol}\). If the system goes from \(\mathrm{A}\) to B by a reversible path and returns to state A by an irreversible path what would be the net change in internal energy? (a) \(40 \mathrm{~kJ}\) (b) \(>40 \mathrm{~kJ}\) (c) \(<40 \mathrm{~kJ}\) (d) zero

In the reaction: \(\mathrm{CO}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})\) the change in \(\Delta \mathrm{S}^{\circ}\) is (given \(\mathrm{S}^{\circ}\) for \(\mathrm{CO}, \mathrm{O}_{2}\) and \(\mathrm{CO}_{2}\) are \(197.6,205.3\) and \(213.6 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) respectively) (a) \(-78.6 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (b) \(-50 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (c) \(-86.5 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (d) \(-30 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\)

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