Problem 8 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Given \(\mathrm{C}_{\text {(graphite) }}+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}) ; \Delta_{r} \mathrm{H}^{\circ}=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(1) ; \Delta_{r} \mathrm{H}^{\circ}=-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(\mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(1) \rightarrow \mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g})\) \(\Delta_{r} \mathrm{H}^{\circ}=+890.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\) Based on the above thermochemical equations, the value of \(\Delta_{r} \mathrm{H}^{\circ}\) at \(298 \mathrm{~K}\) for the reaction \(\mathrm{C}_{\text {(graphite) }}+2 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{CH}_{4}(\mathrm{~g})\) will be : [Main 2017] (a) \(+74.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(+144.0 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(-74.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(-144.0 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The correct answer is (c) \(-74.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\).

Step 1: Write Down the Given Reactions

First, list each of the given thermochemical equations with their enthalpy changes. This helps in identifying how to rearrange them to obtain the target reaction. \[\mathrm{C}_{\text {(graphite) }}+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}) ; \Delta_{r} \mathrm{H}^{\circ}=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\] \[\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(1) ; \Delta_{r} \mathrm{H}^{\circ}=-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\] \[\mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(1) \rightarrow \mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}); \Delta_{r} \mathrm{H}^{\circ}=+890.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\]

Step 2: Write the Target Reaction

The target reaction whose enthalpy change we need to determine is:\[\mathrm{C}_{\text {(graphite) }}+2 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{CH}_{4}(\mathrm{~g})\]. We will manipulate the given reactions to achieve this target reaction.

Step 3: Reverse the Third Reaction

We need to reverse the third reaction to align products/reactants correctly with the target reaction. After reversing, enthalpy changes sign to:\[\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(1); \Delta_{r} \mathrm{H}^{\circ}=-890.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\].

Step 4: Combine All Reactions

Now, combine all reactions to achieve the target reaction: 1. \(\mathrm{C}_{\text {(graphite) }}+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}); \Delta_{r} \mathrm{H}^{\circ}=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\) 2. 2 times \(\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(1); \Delta_{r} \mathrm{H}^{\circ}=-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1} \times 2 = -571.6 \mathrm{~kJ} \) 3. Reverse of \(\mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(1) \rightarrow \mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}); \Delta_{r} \mathrm{H}^{\circ}=-890.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\) The sum of these reactions results in the target reaction.

Step 5: Calculate the Overall Enthalpy Change

Add up the enthalpy changes of the combined reactions to find the overall enthalpy change for the target reaction: \(-393.5 -571.6 -890.3 = -74.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\).

Step 6: Identify the Correct Answer

Based on our calculations, the enthalpy change for the reaction \(\mathrm{C}_{\text {(graphite) }}+2 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{CH}_{4}(\mathrm{~g})\) is \(-74.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\), which corresponds to option (c).

Key Concepts

Thermochemical Equations

Thermochemical equations are a fundamental part of understanding chemical reactions. They are chemical equations that include the enthalpy change, which tells us whether a reaction absorbs or releases heat. This enthalpy change is often denoted as \( \Delta H \). In a thermochemical equation, not only are the reactants and products specified, but the heat effect—whether exothermic or endothermic—is also included.

Consider the reaction:

• \( \mathrm{C}_{(\text{graphite})} + \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{g}) ; \Delta_{r} \mathrm{H}^{\circ} = -393.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \)

The negative sign indicates that the reaction is exothermic, releasing 393.5 kJ of energy per mole of carbon dioxide formed. Thermochemical equations can be manipulated similarly to algebraic equations, by reversing, combining, and scaling, which is key for applying Hess's Law.
Understanding how thermochemical equations represent both the chemical and energetic changes is crucial for solving problems involving heat transfer in reactions.

Hess's Law

Hess's Law is an essential concept that allows us to predict the heat of any chemical reaction based on known heats of related reactions. This law states that the total enthalpy change for a reaction is the same, no matter how many steps the reaction is carried out in. It is based on the principle of conservation of energy.

To apply Hess's Law, one must:

• Identify the target chemical reaction and its desired enthalpy change.
• Use known thermochemical equations to manipulate reactions to form the target reaction. This may involve reversing reactions (which changes the sign of \(\Delta H\)) and multiplying reactions by coefficients to balance with the target reaction.
• Sum the enthalpy changes of the manipulated reactions to obtain the overall \(\Delta H\) for the target reaction.

By using Hess's Law, you can effectively calculate the enthalpy of complex reactions without direct measurement. This is particularly useful when direct measurement is difficult or impossible. It highlights the idea that enthalpy is a state function, meaning it depends only on the initial and final states of the system, not on the path taken to get there.

Heat of Reaction

The heat of reaction, or enthalpy change \( \Delta H \), is the amount of heat absorbed or released during a chemical reaction. It can be determined experimentally or calculated using Hess's Law and thermochemical equations.
The heat of reaction tells us about the energy changes that occur during a reaction and can indicate whether a reaction is exothermic or endothermic:

• Exothermic reactions have negative \( \Delta H \) values, indicating that the reaction releases energy to the surroundings, often increasing the temperature.
• Endothermic reactions have positive \( \Delta H \) values, which means they absorb energy from the surroundings, usually causing a temperature drop.

In calculations involving Hess's Law, the heat of reaction plays a central role. For example, by carefully combining given reactions to derive the target reaction, we can predict its \( \Delta H \) through the algebraic sum of the known heats of the component reactions. This allows chemists to estimate the energy changes associated with reactions that are hard to study directly, providing a crucial insight into reaction feasibility and spontaneity.