Problem 96 - A Complete Resource Book in Chemistry for JEE Main

Question

The \(\Delta \mathrm{H}_{\mathrm{f}}^{\circ}\) for \(\mathrm{CO}_{2}(\mathrm{~g}), \mathrm{CO}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) are \(-393.5\), \(-110.5\) and \(-241.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively. The standard enthalpy change (in \(\mathrm{kJ}\) ) for the reaction \(\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) is (a) \(524.1\) (b) \(41.2\) (c) \(-262.5\) (d) \(-41.2\)

Step-by-Step Solution

Answer

The standard enthalpy change for the reaction is 41.2 kJ/mol, option (b).

Step 1: Identify Given Values

We have the standard enthalpy of formation (\( \Delta \mathrm{H}_{\mathrm{f}}^{\circ} \)) for \( \mathrm{CO}_2(\mathrm{g}) = -393.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \), \( \mathrm{CO}(\mathrm{g}) = -110.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \), and \( \mathrm{H}_2 \mathrm{O}(\mathrm{g}) = -241.8 \mathrm{~kJ} \mathrm{~mol}^{-1} \). These values will be used to calculate the enthalpy change for the reaction.

Step 2: Write Reaction Equation

The reaction given is: \( \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \). The goal is to find the standard enthalpy change for this reaction.

Step 3: Use Sum of Formation Enthalpies Formula

The standard enthalpy change of any reaction (\( \Delta H^{\circ} \)) can be calculated as: \( \Delta H^{\circ} = \sum \Delta H_{f,\text{products}}^{\circ} - \sum \Delta H_{f,\text{reactants}}^{\circ} \).

Step 4: Calculate Sum of Formation Enthalpies for Products

For the products \( \mathrm{CO}(\mathrm{g}) \) and \( \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \), the sum is \(-110.5 + (-241.8) = -352.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \).

Step 5: Calculate Sum of Formation Enthalpies for Reactants

For the reactants \( \mathrm{CO}_{2}(\mathrm{~g}) \) and \( \mathrm{H}_{2}(\mathrm{~g}) \), the sum is \(-393.5 + 0 = -393.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \). Note that \( \Delta \mathrm{H}_{\mathrm{f}}^{\circ} \) for any element in its standard state, like \( \mathrm{H}_2 \), is \( 0 \).

Step 6: Determine Enthalpy Change of Reaction

Substitute into the formula: \( \Delta H^{\circ} = -352.3 - (-393.5) \).

Step 7: Final Calculation

Calculate: \(-352.3 + 393.5 = 41.2 \mathrm{~kJ} \mathrm{~mol}^{-1} \).

Step 8: Select the Closest Answer

The calculated standard enthalpy change is \( 41.2 \), which corresponds to option (b) \( 41.2 \).

Key Concepts

Standard Enthalpy of Formation

The standard enthalpy of formation, denoted as \( \Delta \mathrm{H}_{\mathrm{f}}^{\circ} \), is a crucial concept in chemical thermodynamics. It represents the change in enthalpy when one mole of a compound is formed from its elements in their standard states. In simpler terms, it tells us how much energy is released or absorbed when a compound is made from substances in their most stable forms at 1 bar pressure and a specified temperature, typically 298 K (25°C).

Understanding \( \Delta \mathrm{H}_{\mathrm{f}}^{\circ} \) helps us predict the stability of compounds. Lower or more negative values generally indicate a compound is more stable compared to its elements. For instance, in our example, carbon dioxide (\( \mathrm{CO}_2 \)) and water (\( \mathrm{H}_2 \mathrm{O} \)) have negative enthalpy values, meaning they release energy when formed. Carbon monoxide (\( \mathrm{CO} \)), having less negative enthalpy than \( \mathrm{CO}_2 \), means less energy is released during its formation.

To calculate the standard enthalpy change of a reaction, you sum up the standard enthalpy of formation values of products and subtract those of the reactants. This method assumes all reactions occur under standard conditions and uses tabulated \( \Delta \mathrm{H}_{\mathrm{f}}^{\circ} \) values for accuracy. This simple formula makes it possible to predict enthalpy changes without running complex experiments in a lab. It shows the elegance of thermodynamics in predicting the behavior and energy changes in chemical reactions.

Hess's Law

Hess's Law is a fundamental principle in chemical thermodynamics that simplifies the way we calculate enthalpy changes in reactions. As per Hess's Law, the total enthalpy change for a given reaction is the same regardless of the pathway taken, as long as the initial and final conditions are the same.

This principle is invaluable for calculating the standard enthalpy change of complex reactions. By breaking down a reaction into multiple steps with known enthalpy changes, you can calculate the overall change by simply adding these steps together.

• This method is beneficial when a reaction cannot be measured directly in a laboratory, as it allows the use of known reactions to infer changes for less accessible reactions.

Let's consider our exercise. We want to find the enthalpy change for the reaction:
\( \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \).
Using the formula derived from Hess’s Law, we sum the enthalpy of formation of the products \(-110.5 + (-241.8)\) and subtract the sum for the reactants \(-393.5 + 0\), yielding \(41.2 \mathrm{~kJ} \mathrm{~mol}^{-1}\). This concept highlights the power of Hess's Law in transforming our approach to enthalpy calculations.

Chemical Thermodynamics

Chemical thermodynamics investigates the energy changes that occur during chemical reactions and how they relate to the physical properties and transformations of substances. It provides a scientific framework that allows chemists to understand how energy is transferred and conserved in chemical processes.

A fundamental component of chemical thermodynamics is the concept of enthalpy, which is a measure of heat energy in a system at constant pressure. We often look at changes in enthalpy to determine if a reaction is exothermic (releasing heat) or endothermic (absorbing heat).

• Exothermic reactions, like the formation of \( \mathrm{H}_2 \mathrm{O} \), release heat, making them favorable in terms of energy conservation.
• Endothermic reactions require an input of energy to proceed, which can be understood by observing changes in the system’s enthalpy.

Our exercise emphasizes the importance of accurately calculating these enthalpy changes. Understanding these changes helps predict reaction spontaneity, efficiency, and energy requirements in practical applications such as fuel combustion and environmental processes. Overall, chemical thermodynamics bridges theoretical concepts with real-world applications, supporting innovations in energy management and materials science.