Problem 107 - A Complete Resource Book in Chemistry for JEE Main

Question

At \(300 \mathrm{~K}\), the standard enthalpies of formation of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}(\mathrm{s}), \mathrm{CO},(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}\) (s) are \(-408,-393\) and \(-286 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively. Calculate the heat of combustion of benzoic acid at constant volume. (a) \(-3296 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(-3200 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(-3201 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(-3603 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The heat of combustion is option (c) \(-3201\, \mathrm{kJ}\, \mathrm{mol}^{-1}\).

Step 1: Write the balanced chemical equation for the combustion

The balanced chemical equation for the combustion of benzoic acid (\(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}(\mathrm{s})\)) is:\[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}(\mathrm{s}) + \frac{15}{2}\mathrm{O}_2(\mathrm{g}) \rightarrow 7\mathrm{CO}_2(\mathrm{g}) + 3\mathrm{H}_2\mathrm{O}(\mathrm{l})\]

Step 2: Recall the formula for the heat of combustion

The heat of combustion at constant volume can be calculated using the enthalpy of formation values:\[\Delta H_{comb} = \sum \Delta H_{f,\text{products}} - \sum \Delta H_{f,\text{reactants}}\]where \(\Delta H_{f}\) denotes the standard enthalpy of formation.

Step 3: Calculate the enthalpy change for products

The standard enthalpy of formation for \(\mathrm{CO}_2(\mathrm{g})\) is \(-393\, \mathrm{kJ}\,\mathrm{mol}^{-1}\), and for \(\mathrm{H}_2\mathrm{O}(\mathrm{l})\) is \(-286\, \mathrm{kJ}\,\mathrm{mol}^{-1}\). Thus:\[\Delta H_{f,\text{products}} = 7(-393) + 3(-286) = -2751 - 858 = -3609\, \mathrm{kJ}\,\]

Step 4: Calculate the enthalpy change for reactants

The standard enthalpy of formation for \(\mathrm{C}_6\mathrm{H}_5\mathrm{COOH}(\mathrm{s})\) is given as \(-408\, \mathrm{kJ}\,\mathrm{mol}^{-1}\). The enthalpy of formation for \(\mathrm{O}_2\) as an element in its standard state is \(0\, \mathrm{kJ}\,\mathrm{mol}^{-1}\). Thus:\[\Delta H_{f,\text{reactants}} = -408 + \frac{15}{2}(0) = -408\, \mathrm{kJ}\]

Step 5: Calculate the heat of combustion

Substitute the values calculated in Steps 3 and 4 into the formula from Step 2:\[\Delta H_{comb} = -3609 - (-408) = -3609 + 408 = -3201\, \mathrm{kJ}\]

Step 6: Match answer with given options

The calculated heat of combustion \(-3201\,\mathrm{kJ/mol}\) matches option (c). Therefore, the correct answer is:\(\text{(c) } -3201\,\mathrm{kJ/mol}\)

Key Concepts

Enthalpy of Formation

The enthalpy of formation is a crucial concept in chemistry, especially when dealing with chemical reactions like combustion. It refers to the heat change that occurs when one mole of a compound is formed from its elements in their standard states. These are usually measured under standard conditions, denoted by standard enthalpies of formation, \(\Delta H_{f}^{°}\), with units typically expressed in kilojoules per mole \(\mathrm{kJ/mol}\). By using these values, we can calculate the overall enthalpy change for a reaction.

• For example, the standard enthalpy of formation for carbon dioxide, \(\mathrm{CO}_2\), is typically \(−393\, \mathrm{kJ/mol}\).
• The values are critical inputs when calculating the heat of combustion, where we consider the formation and breaking of chemical bonds.

From these values, the enthalpy change for a reaction can be computed using formulas like the one found in the original exercise's step-by-step solution. This helps us understand the energy dynamics of chemical reactions and predict the thermal outcomes of substance interactions in various environments.

Combustion Reactions

Combustion reactions are a type of chemical reaction where a substance reacts with oxygen to produce heat and light, often resulting in the formation of carbon dioxide and water as products. This is an exothermic reaction, meaning it releases energy.

• Combustion is a common reaction in both industrial processes and everyday life, such as in engines and burning fuels.
• For the combustion of benzoic acid, as illustrated in the problem, the balanced chemical equation is crucial: \\[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}(\mathrm{s}) + \frac{15}{2} \mathrm{O}_2(\mathrm{g}) \rightarrow 7 \mathrm{CO}_2(\mathrm{g}) + 3 \mathrm{H}_2 \mathrm{O}(\mathrm{l})\]\

Here, benzoic acid combines with oxygen to give seven moles of carbon dioxide and three moles of water, showcasing the stoichiometry that balances the equation. By calculating the heat of combustion, we can assess the energy released during such reactions.

Thermochemistry

Thermochemistry is the branch of chemistry that examines the heat energy associated with chemical reactions and phase changes. This discipline is fundamental to understanding how energy is absorbed or released in reactions.

• Key objectives include measuring the energy change, designing energy-efficient processes, and predicting the direction of reactions.
• It employs several key principles, such as conservation of energy and Hess's law, which states that total energy change is the same regardless of how a reaction occurs.

The practical application of thermochemistry in solving problems like the heat of combustion of benzoic acid involves using standard thermodynamic tables and understanding the enthalpy changes associated with formation and decomposition.

Standard Enthalpy Change

The standard enthalpy change is the enthalpy change when all reactants and products are in their standard states. It provides a point of reference for measuring how much heat is absorbed or released during a chemical reaction under these conditions.

• This concept is essential for chemists to compare reactions and make predictions about reaction behavior.
• Symbolically represented as \(\Delta H^{°}\), it offers a consistent framework to calculate the difference in energy content from reactants to products.

In the context of combustion, the standard enthalpy change helps us understand the energetic efficiency and potential energy releases in different chemical processes, aiding in the development of safer and more effective energy systems.