Problem 121 - Chemistry The Science in Context

Question

The standard enthalpies of formation of benzene \(C_{6} \mathrm{H}_{6}(\ell)\) \(\mathrm{CO}_{2}(g),\) and \(\mathrm{H}_{2} \mathrm{O}(\ell)\) are \(49.0,-394,\) and \(-286 \mathrm{kJ} / \mathrm{mol}\) respectively. Use this information to calculate the standard enthalpy of combustion of \(\mathrm{C}_{6} \mathrm{H}_{6}(\ell)\)

Step-by-Step Solution

Answer

Question: Calculate the standard enthalpy of combustion for benzene, given the standard enthalpy of formation for benzene (\(C_6H_6(\ell)\)) is 49 kJ/mol, carbon dioxide (\(CO_2(g)\)) is -394 kJ/mol, and water (\(H_2O(\ell)\)) is -286 kJ/mol. Answer: The standard enthalpy of combustion for benzene is -2809 kJ/mol.

Step 1: Write the combustion reaction

The combustion reaction of benzene is: \(C_6H_6(\ell) + \frac{15}{2}O_2(g) \rightarrow 6CO_2(g) + 3H_2O(\ell)\)

Step 2: Calculate the enthalpy change of products

Sum of the enthalpies of products: \(6 \times \Delta H_{CO_2} + 3 \times \Delta H_{H_2O} = 6 \times (-394\, kJ/mol) + 3 \times (-286\, kJ/mol) = -2760 \,kJ\)

Step 3: Calculate the enthalpy change of reactants

Sum of the enthalpies of reactants (we don't consider \(O_2\) because its enthalpy of formation is 0): \(\Delta H_{benzene} = 49\, kJ/mol\)

Step 4: Calculate the standard enthalpy of combustion

Applying Hess's Law: \(\Delta H_{combustion} = \Delta H_{products} - \Delta H_{reactants} = -2760\, kJ - 49\, kJ = -2809 \,kJ/mol\) The standard enthalpy of combustion for benzene is -2809 kJ/mol.

Key Concepts

Enthalpy of Formation

Enthalpy of formation refers to the heat change when one mole of a compound forms from its elements in their standard states. These standard states are at 1 atm and usually 25°C. For example, the enthalpy of formation for

• Carbon dioxide (\( \text{CO}_2(g) \) ): \(-394 \, \text{kJ/mol}\)
• Water (\( \text{H}_2\text{O}(\ell) \) ): \(-286 \, \text{kJ/mol}\)
• Benzene (\( \text{C}_6\text{H}_6(\ell) \) ): 49 \( \text{kJ/mol} \)

These values indicate whether the process absorbs or releases energy.
The negative sign reflects an exothermic process. An exothermic reaction releases heat, while a positive sign shows an endothermic process, which absorbs heat.
Understanding these values is crucial when calculating enthalpy changes in a reaction.

Hess's Law

Hess's Law is an essential principle in chemistry. It states that the total enthalpy change in a reaction is the same, no matter how it is carried out, as long as the initial and final conditions are the same.
This allows us to calculate the enthalpy change of a complex reaction using simpler steps or known reactions.

• To find the enthalpy of combustion of benzene, we calculate all steps involved.
• The products and reactants' standard enthalpies of formation help determine this value.

By applying Hess's Law:
The enthalpy of combustion equals the sum of the products' enthalpies minus the reactants' enthalpies. This provides the energy change when benzene combusts completely.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. During combustion, substances like benzene react with oxygen to form carbon dioxide and water.
This reaction releases energy, typically seen in the form of heat and light. The combustion of benzene can be shown in the equation:

• \( \text{C}_6\text{H}_6(\ell) + \frac{15}{2}\text{O}_2(g) \rightarrow 6\text{CO}_2(g) + 3\text{H}_2\text{O}(\ell) \)

In this equation:

• \( \text{C}_6\text{H}_6(\ell) \) represents benzene
• \( \text{O}_2(g) \) is oxygen
• The products are carbon dioxide and water

Understanding this reaction helps in calculating the enthalpy changes.It's a practical example of how chemical reactions are essential in energy exchange processes.