Problem 156 - A Complete Resource Book in Chemistry for JEE Main

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

Answer

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

Step 1: Define the Combustion Reactions

The combustion of carbon can be expressed by the equation: \[ \text{C (s) + O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H = -393.5 \text{ kJ/mol} \] The combustion of carbon monoxide is: \[ \text{CO (g) + }\frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H = -283 \text{ kJ/mol} \] Our goal is to find the enthalpy change for the formation of carbon monoxide.

Step 2: Write the Formation Reaction for CO

The formation of carbon monoxide from carbon and oxygen can be written as: \[ \text{C (s) + }\frac{1}{2}\text{O}_2(g) \rightarrow \text{CO (g)} \quad \Delta H_f \] This is the reaction whose enthalpy change we need to find.

Step 3: Use Hess's Law to Formulate the Problem

According to Hess's Law, the enthalpy change of the direct path (forming CO from C and \( \frac{1}{2} \text{O}_2\)) is the same as the indirect path, which is the difference between the combustion reactions. Thus, we can express it as: \[ \Delta H_f = \Delta H_{comb, \text{C (s)}} - \Delta H_{comb, \text{CO (g)}} \] where \( \Delta H_{comb, \text{C (s)}} = -393.5 \text{ kJ/mol} \) and \( \Delta H_{comb, \text{CO (g)}} = -283 \text{ kJ/mol} \).

Step 4: Calculate the Enthalpy of Formation

Substitute the given values into the equation: \[ \Delta H_f = (-393.5) - (-283) \] Simplify the calculation: \[ \Delta H_f = -393.5 + 283 = -110.5 \text{ kJ/mol} \] Thus, the enthalpy of formation of carbon monoxide per mole is \(-110.5 \text{ kJ}\).

Key Concepts

Hess's Law

Understanding Hess's Law is crucial for solving problems related to enthalpy changes in chemical reactions. According to Hess's Law, the total enthalpy change for a chemical reaction is the same, regardless of the pathway taken from reactants to products. This is because enthalpy is a state function. In other words, the enthalpy change depends only on the initial and final states of the system, not on how the process occurs.

This law can be especially useful when direct measurements of an enthalpy change are difficult to perform.

• Consider the overall reaction: the enthalpy change can be calculated as a sum of known enthalpy changes.
• Allows us to use known values of simpler reactions to find the enthalpy change of an unknown reaction, which simplifies calculations.

In our exercise, we applied Hess's Law to find the enthalpy of formation for carbon monoxide by considering combustion reactions of carbon and carbon monoxide. By summing and rearranging these reactions based on their given enthalpies, we found the enthalpy of a reaction whose direct measurement might be challenging.

Combustion Reactions

Combustion reactions are reactions where a substance combines with oxygen gas, releasing energy in the form of heat and light. These reactions are exothermic, meaning they release energy.

In the combustion of carbon, carbon reacts with oxygen to form carbon dioxide, as represented by the equation:\[ \text{C (s) + O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H = -393.5 \text{ kJ/mol} \]Similarly, the combustion of carbon monoxide involves CO and oxygen producing carbon dioxide:\[ \text{CO (g) + }\frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H = -283 \text{ kJ/mol} \]These reactions showcase how the combustion of substances releases significant amounts of energy.

• Able to convert chemical energy into heat energy.
• Commonly used in energy production and industry.

By understanding the combustion reactions involved, we can determine the enthalpy of formation for carbon monoxide, lending insight into how energy is transferred during these processes.

Thermochemistry

Thermochemistry is the study of the heat involved in chemical reactions. It's essential for understanding how energy is exchanged in chemical processes, which is critical for both theoretical and practical applications.

Key concepts include:

• Enthalpy (H): A measure of total energy in a thermodynamic system, considering both internal energy and energy related to pressure and volume.
• Exothermic vs Endothermic Reactions: Exothermic reactions release energy, observed as an increase in the temperature of the surroundings. Endothermic reactions absorb energy, usually causing temperature drops in the surroundings.

In our exercise, we dealt specifically with the enthalpy of formation, the change in enthalpy when one mole of a compound forms from its elements in their standard states. Thermochemistry concepts guide us in calculating and understanding these reactions, showing us how the energy flow interacts with the environment. This understanding helps solve practical challenges in fields like energy production, environmental science, and engineering.