Problem 122 - Chemistry: An Atoms-Focused Approach

Question

Calculate \(\Delta H_{\mathrm{rxn}}^{\circ}\) for the reaction $$2 \mathrm{Ni}(s)+\frac{1}{4} \mathrm{S}_{8}(s)+3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{NiSO}_{3}(s) \quad \Delta H_{\mathrm{rxn}}^{\circ}=?$$ from the following data: (1) \(\mathrm{NiSO}_{3}(s) \rightarrow \mathrm{NiO}(s)+\mathrm{SO}_{2}(g) \quad \Delta H_{\mathrm{rxn}}^{\circ}=156 \mathrm{kJ}\) (2) \(\frac{1}{8} S_{8}(s)+O_{2}(g) \rightarrow \operatorname{SO}_{2}(g) \quad \quad \Delta H_{\operatorname{rxn}}^{\circ}=-297 \mathrm{kJ}\) (3) \(\mathrm{Ni}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{NiO}_{2}(s) \quad \Delta H_{\mathrm{rxn}}^{\circ}=-241 \mathrm{kJ}\)

Step-by-Step Solution

Answer

\(2 \mathrm{Ni}(s)+\frac{1}{4}\mathrm{S}_{8}(s)+3 \mathrm{O}_{2}(g) \rightarrow 2\mathrm{NiSO}_{3}(s)\) Answer: The enthalpy change for this reaction is \(\Delta H_{\mathrm{rxn}}^{\circ} = - 635\mathrm{kJ}\).

Step 1: Align the given reactions to fit the main reaction

We want to manipulate the given reactions so that they can be combined to create the main reaction: (1) \(\mathrm{NiSO}_{3}(s) \rightarrow \mathrm{NiO}(s)+\mathrm{SO}_{2}(g) \quad \Delta H_{\mathrm{rxn}}^{\circ}=156 \mathrm{kJ}\) (2) \(\frac{1}{8} \mathrm{S}_{8}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{SO}_{2}(g) \quad \quad \Delta H_{\mathrm{rxn}}^{\circ}=-297 \mathrm{kJ}\) (3) \(\mathrm{Ni}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{NiO}(s) \quad \Delta H_{\mathrm{rxn}}^{\circ}=-241 \mathrm{kJ}\) First, we need two moles of NiSO3, so we'll multiply reaction (1) by 2: (1') \(2\mathrm{NiSO}_{3}(s) \rightarrow 2\mathrm{NiO}(s)+2\mathrm{SO}_{2}(g) \quad 2(\Delta H_{\mathrm{rxn}}^{\circ})=312 \mathrm{kJ}\) Next, we want \(\frac{1}{4} \mathrm{S}_{8}\) in the reaction, so we'll multiply reaction (2) by 4: (2') \(\mathrm{S}_{8}(s)+4\mathrm{O}_{2}(g) \rightarrow 4\mathrm{SO}_{2}(g) \quad 4(\Delta H_{\mathrm{rxn}}^{\circ})=-1188 \mathrm{kJ}\) Finally, we need to reverse reaction (3) because we want \(\mathrm{NiO}\) to be in the reactant side: (3') \(\mathrm{NiO}(s) \rightarrow \mathrm{Ni}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \quad -(\Delta H_{\mathrm{rxn}}^{\circ})=241 \mathrm{kJ}\) Now, we have the manipulated reactions: (1') \(2\mathrm{NiSO}_{3}(s) \rightarrow 2\mathrm{NiO}(s)+2\mathrm{SO}_{2}(g) \quad 2(\Delta H_{\mathrm{rxn}}^{\circ})=312 \mathrm{kJ}\) (2') \(\mathrm{S}_{8}(s)+4\mathrm{O}_{2}(g) \rightarrow 4\mathrm{SO}_{2}(g) \quad 4(\Delta H_{\mathrm{rxn}}^{\circ})=-1188 \mathrm{kJ}\) (3') \(\mathrm{NiO}(s) \rightarrow \mathrm{Ni}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \quad -(\Delta H_{\mathrm{rxn}}^{\circ})=241 \mathrm{kJ}\)

Step 2: Combine the manipulated reactions and calculate \(\Delta H_{\mathrm{rxn}}^{\circ}\) for the main reaction

Now, we combine the adjusted reactions. (1'+ 3') \(2\mathrm{NiSO}_{3}(s)-2\mathrm{NiO}(s)+2\mathrm{NiO}(s) \rightarrow 2\mathrm{Ni}(s)+2\mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)\) \(\Delta H_{\mathrm{rxn}}^{\circ}(1'+ 3') = 312 \mathrm{kJ} + 241 \mathrm{kJ} = 553 \mathrm{kJ}\) (1'+ 3'+ 2') \(2\mathrm{Ni}(s)+\frac{1}{4} \mathrm{S}_{8}(s)+3 \mathrm{O}_{2}(g) \rightarrow 2\mathrm{NiSO}_{3}(s)\) \(\Delta H_{\mathrm{rxn}}^{\circ} (1'+ 3'+ 2') = 553 \mathrm{kJ} - 1188 \mathrm{kJ} = - 635\mathrm{kJ}\) So, the enthalpy change for the reaction \(2 \mathrm{Ni}(s)+\frac{1}{4}\mathrm{S}_{8}(s)+3 \mathrm{O}_{2} \rightarrow 2\mathrm{NiSO}_{3}\) is \(\Delta H_{\mathrm{rxn}}^{\circ} = - 635\mathrm{kJ}\).

Key Concepts

Understanding Enthalpy Change

Enthalpy change, symbolized as \( \Delta H \), is crucial in understanding the energy dynamics in chemical reactions. It represents the heat absorbed or released during a reaction at constant pressure. For any chemical reaction:

• When \( \Delta H \) is negative, the reaction releases heat, making it exothermic.
• When \( \Delta H \) is positive, the reaction absorbs heat, making it endothermic.

For example, in the given problem, the overall enthalpy change \( \Delta H_{\mathrm{rxn}}^{\circ} \) was calculated to be \(-635 \text{kJ}\), indicating that the reaction is exothermic. This means that, as the reaction proceeds, it releases 635 kJ of energy for every two moles of NiSO3 formed. Understanding whether a reaction releases or absorbs energy can provide insight into the feasibility and spontaneity of the reactions in thermodynamics.

Exploring Thermochemistry

Thermochemistry is the branch of chemistry that deals with the study of energy changes, particularly heat, during chemical reactions. It focuses on understanding how and why energy transfers occur. Thermochemistry is grounded in the first law of thermodynamics, which states that energy cannot be created or destroyed, only converted from one form to another.

In the context of the exercise, thermochemistry helps in calculating the overall enthalpy change of a reaction using Hess's Law. Hess's Law states that the enthalpy change of a chemical reaction is the same, regardless of the steps taken, because enthalpy is a state function. You can add or subtract enthalpies of intermediate reactions to get the total \( \Delta H_{\mathrm{rxn}}^{\circ} \). Understanding this principle allowed us to manipulate the given reactions to determine the overall enthalpy change accurately. By connecting individual reaction steps, we maintain the focus on the total energy exchanged rather than the path taken.

Insights into Chemical Reactions

Chemical reactions involve the breaking of old bonds and the formation of new ones, which leads to a transformation of substances. Each reaction step involves specific reactants and products and is accompanied by an energy change, primarily in the form of heat (enthalpy).

In this exercise, three reactions were manipulated and combined. Each step involved specific transformations:

• Decomposition of NiSO3 into NiO and SO2.
• Oxidation of S\(_8\) to form SO2.
• Formation of NiO from Ni and O\(_2\).

By reversing and multiplying these reactions, we simulated the process of forming NiSO3 from Ni, S\(_8\), and O\(_2\). This step-by-step manipulation is a perfect example of how chemical reactions can be strategically combined and analyzed using Hess’s Law to predict the overall reaction enthalpy. By understanding these transformations, students can appreciate the dynamics of bond formation and breaking at a deeper level, which is fundamental to mastering chemical reactions and their energy requirements.