Question

Given that \(\mathrm{CII}_{3} \mathrm{CIIO}+\frac{5}{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{2}+2 \mathrm{II}_{2} \mathrm{O} ; \Delta H=\) \(1168 \mathrm{~kJ} / \mathrm{molc} ; \mathrm{CII}_{3} \mathrm{COOII}+2 \mathrm{O}_{2} \longrightarrow 2 \mathrm{CO}_{2}\) \(+2 \mathrm{II}_{2} \mathrm{O} ; \Delta H=876 \mathrm{~kJ} / \mathrm{mole} . \Delta H\) for the reaction \(\mathrm{CII}_{3} \mathrm{CIIO}+\frac{1}{2} \mathrm{O}_{2} \longrightarrow \mathrm{CII}_{3} \mathrm{COOII}\) is (1) \(292 \mathrm{~kJ} / \mathrm{molc}\) (2) \(378 \mathrm{~kJ} / \mathrm{molc}\) (3) \(195 \mathrm{~kJ} / \mathrm{molc}\) (4) \(2044 \mathrm{~kJ} / \mathrm{molc}\)

Short answer

292 \mathrm{kJ/molc}

Step-by-step solution

Identify the Given Reactions

First, list the provided chemical reactions and their associated enthalpy changes (ΔH). Reaction 1: \[\mathrm{CII_{3}CIIO} + \frac{5}{2}\mathrm{O_{2}} \rightarrow 2\mathrm{CO_{2}} + 2\mathrm{II_{2}O} ; \Delta H = 1168\, \mathrm{kJ/molc}\]Reaction 2: \[\mathrm{CII_{3}COOII} + 2 \mathrm{O_{2}} \rightarrow 2 \mathrm{CO_{2}} + 2 \mathrm{II_{2}O} ; \Delta H = 876\, \mathrm{kJ/molc}\]The target reaction is: \[\mathrm{CII_{3}CIIO} + \frac{1}{2} \mathrm{O_{2}} \rightarrow \space \mathrm{CII_{3}COOII}\]

Reverse the Second Equation

Reverse reaction 2 to represent the formation of \(\mathrm{CII_{3}COOII}\). This changes the sign of ΔH: \[2\mathrm{CO_{2}} + 2\mathrm{II_{2}O} \rightarrow \mathrm{CII_{3}COOII} + 2\mathrm{O_{2}} ; \Delta H = -876\, \mathrm{kJ/molc}\]

Combine Reactions

Combine the reversed reaction 2 with reaction 1 in order to cancel out the common products (\(2\mathrm{CO_{2}}\) and \(2\mathrm{II_{2}O}\)). Adding these reactions gives: ###### Reaction Addition:\((\mathrm{CII_{3}CIIO} + \frac{5}{2}\mathrm{O_{2}}) + (2\mathrm{CO_{2}} + 2\mathrm{II_{2}O} \rightarrow \mathrm{CII_{3}COOII} + 2\mathrm{O_{2}}))\)

Simplify the Combined Reactions

Combine and simplify the resulting equation to obtain the desired target reaction:\[CII_{3}CIIO + \frac{5}{2}\mathrm{O_{2}} - 2\mathrm{O_{2}} \rightarrow \mathrm{CII_{3}COOII}\]Simplify to obtain:\[\mathrm{CII_{3}CIIO} + \frac{1}{2}\mathrm{O_{2}} \rightarrow \mathrm{CII_{3}COOII}\]

Calculate Enthalpy Change for Target Reaction

To find ΔH for the target reaction, subtract the enthalpy of the reversed reaction 2 from reaction 1's enthalpy:\[\Delta H = 1168\, \mathrm{kJ/molc} - 876\, \mathrm{kJ/molc} = 292\, \mathrm{kJ/molc}\]

Key concepts

Understanding Chemical Reactions

Chemical reactions involve the breaking and forming of chemical bonds, leading to the transformation of reactants into products. In the given exercise, we see two specific chemical reactions involving compounds such as \(\text{CII}_3\text{CIIO}\) and \(\text{CII}_3\text{COOII}\). These reactions can be represented as: \[\text{CII}_3\text{CIIO} + \frac{5}{2}\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{II}_2\text{O}\text{ and }\text{CII}_3\text{COOII} + 2\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{II}_2\text{O}\]In a chemical reaction, it's crucial to understand the stoichiometry, which refers to the quantitative relationship between reactants and products. This is essential to ensure that the reaction equations are balanced, meaning the number of atoms for each element is the same on both sides of the equation. Balancing equations helps in calculating the enthalpy change, which is a foundational aspect of thermodynamics in chemical reactions.

Hess's Law in Enthalpy Calculations

Hess's Law states that the total enthalpy change for a reaction is the same, no matter what pathway is taken, as long as the initial and final states are the same. This principle is used to determine the enthalpy change of complex reactions. In our exercise, the goal is to find the enthalpy change (\(\text{ΔH}\)) for the target reaction: \[\text{CII}_3\text{CIIO} + \frac{1}{2}\text{O}_2 \rightarrow \text{CII}_3\text{COOII}\]To achieve this, we manipulate given reactions according to Hess's Law: 1. Reverse Reaction 2 to align the product \(\text{CII}_3\text{COOII}\) with our target reaction: \[\text{CII}_3\text{COOII} + 2\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{II}_2\text{O}\]When reversed, the enthalpy change becomes negative: \[\text{ΔH} = -876 \text{kJ/mol}\] 2. Combine this with Reaction 1 to form the target reaction. Simplifying the combined reactions gives the desired equation: \[\text{CII}_3\text{CIIO} + \frac{1}{2}\text{O}_2 \rightarrow \text{CII}_3\text{COOII}\] 3. Subtract the reversed Reaction 2 enthalpy from that of Reaction 1: \[\text{ΔH} = 1168 \text{kJ/mol} - 876 \text{kJ/mol} = 292 \text{kJ/mol}\] Thus, using Hess's Law allows us to calculate the enthalpy change for reactions that are not straightforward by breaking them down into manageable steps.

Thermodynamics in Chemical Reactions

Thermodynamics is the branch of physical science that deals with the relations between heat and other forms of energy. In chemical reactions, thermodynamics helps us understand how energy is transferred and transformed. Enthalpy (\(\text{ΔH}\)) is a key thermodynamic quantity that represents the total heat content of a system. When calculating \(\text{ΔH}\) for a reaction using enthalpy changes of individual steps, it is critical to consider whether the reaction is exothermic or endothermic:

• Exothermic reactions: These release heat, resulting in a negative \(\text{ΔH}\).
• Endothermic reactions: These absorb heat, leading to a positive \(\text{ΔH}\).

The given reactions and their respective enthalpy changes (\text{\ΔH}) illustrate this concept. By understanding the enthalpy changes in these individual steps: \[\text{CII}_3\text{CIIO} + \frac{5}{2}\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{II}_2\text{O} ; \text{ΔH} = 1168 \text{kJ/mol}\] \[\text{CII}_3\text{COOII} + 2 \text{O}_2 \rightarrow 2 \text{CO}_2 + 2 \text{II}_2\text{O} ; \text{ΔH} = 876 \text{kJ/mol}\] We see how combining reaction steps, reversing equations, and simplifying them helps calculate the total enthalpy change. By keeping track of energy transfers and understanding how bonds break and form, we can predict whether a reaction will be endothermic or exothermic, thus deepening our understanding of thermodynamic principles in chemical reactions.