Problem 156
Question
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) \(676.5 \mathrm{~kJ}\)
Step-by-Step Solution
Verified Answer
The enthalpy of formation of carbon monoxide is
(b)
-110.5
\text{kJ/mol}.
1Step 1: 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 \(-283 \text{ kJ/mol} \).
2Step 2: 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.
3Step 3: 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 \(+283 \text{ kJ/mol} \).
4Step 4: 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, \( \Delta H_f = -393.5 \text{ kJ/mol} + 283 \text{ kJ/mol} = -110.5 \text{ kJ/mol} \).
Key Concepts
Hess's Law: A Fundamental PrincipleEnthalpy of Combustion and Its ImportanceUnderstanding Carbon Monoxide Formation
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.
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.
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:
- \( C(s) + \frac{1}{2}O_2(g) \rightarrow CO(g) \)
- Partial oxidation in processes like steel manufacturing.
- As a precursor in organic synthesis.
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