Problem 75 - Chemistry: An Atoms-Focused Approach

Question

Given the following thermochemical data: $$\begin{array}{ll}\frac{1}{2} \mathrm{N}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \frac{1}{2} \mathrm{NO}(g) & \Delta H_{\mathrm{rxn}}^{\circ}=+90.3 \mathrm{kJ} \\\ \mathrm{NO}(g)+\frac{1}{2} \mathrm{Cl}_{2}(g) \rightarrow \mathrm{NOCl}(g) & \Delta H_{\mathrm{rxn}}^{\circ}=-38.6 \mathrm{kJ} \end{array}$$ what is the value of \(\Delta H_{\mathrm{rxn}}^{\circ}\) for the decomposition of NOCl? $$2 \mathrm{NOCl}(g) \rightarrow \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{Cl}_{2}(g) \quad \Delta H_{\mathrm{rxn}}^{\circ}=?$$

Step-by-Step Solution

Answer

Answer: The enthalpy change for the decomposition of NOCl is +167.5 kJ.

Step 1: Manipulate the given reactions

To match the desired reaction, we need to reverse the second reaction:$$\mathrm{NOCl}(g) \rightarrow \mathrm{NO}(g) + \frac{1}{2} \mathrm{Cl}_{2}(g)$$The enthalpy change for this reversed reaction will be the opposite of the given one: $$\Delta H_{\mathrm{rxn}}^{\circ}=+38.6 \mathrm{kJ}$$Now, let's multiply the reversed reaction by 2 to match the desired reaction:$$2\mathrm{NOCl}(g) \rightarrow 2\mathrm{NO}(g) + \mathrm{Cl}_{2}(g)$$The enthalpy change will also be multiplied by 2: $$\Delta H_{\mathrm{rxn}}^{\circ}=+77.2 \mathrm{kJ}$$

Step 2: Sum the manipulated reactions and cancel out common species

We now sum the first (unaltered) reaction and our manipulated second reaction:$$\begin{array}{rl} &\frac{1}{2} \mathrm{N}_{2}(g) + \frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \frac{1}{2} \mathrm{NO}(g) \\ + &2\mathrm{NOCl}(g) \rightarrow 2\mathrm{NO}(g) + \mathrm{Cl}_{2}(g)\end{array}$$And cancel out any common species on both sides of the reactions:$$2\mathrm{NOCl}(g) \rightarrow \frac{1}{2} \mathrm{N}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g)+\mathrm{Cl}_{2}(g)$$This now matches the desired reaction.

Step 3: Calculate the enthalpy change for the overall reaction

To obtain the enthalpy change for the overall reaction, we add the enthalpy changes for the first and manipulated second reactions:$$\Delta H_{\mathrm{rxn}}^{\circ} = +90.3 \mathrm{kJ} + 77.2 \mathrm{kJ} = +167.5 \mathrm{kJ}$$So, the enthalpy change for the decomposition of NOCl is: $$\Delta H_{\mathrm{rxn}}^{\circ} = +167.5 \mathrm{kJ}$$

Key Concepts

Enthalpy Change

Enthalpy change is a fundamental concept in thermochemistry that measures the total heat content of a chemical reaction. Enthalpy, symbolized by \( H \), refers to the amount of heat absorbed or released at constant pressure. When we talk about the enthalpy change, denoted by \( \Delta H \), we are considering the difference between the enthalpy of the products and the enthalpy of the reactants in a chemical reaction.

If \( \Delta H \) is positive, the reaction is endothermic, meaning it absorbs heat from the surroundings. Conversely, if \( \Delta H \) is negative, the reaction is exothermic and releases heat. In the context of the given exercise, the enthalpy change for the decomposition of NOCl was calculated to be +167.5 kJ, indicating an endothermic process. This means that energy must be provided to break the bonds in NOCl, resulting in the formation of \( \mathrm{N}_2(g) \), \( \mathrm{O}_2(g) \), and \( \mathrm{Cl}_2(g) \).

Understanding enthalpy change allows chemists to predict the heat exchange involved in chemical processes, which is crucial for designing energy-efficient reactions in industrial applications.

Chemical Reactions

Chemical reactions involve the transformation of substances through the breaking and forming of chemical bonds. During this process, reactants are converted into products, leading to an energy exchange in the form of heat or light. The enthalpy change associated with a reaction, such as those given in the exercise, provides insight into whether the reaction absorbs or releases energy.

Chemical reactions can be classified based on their enthalpy changes as either endothermic or exothermic. An endothermic reaction, like the decomposition of NOCl, absorbs energy (\( \Delta H \) is positive), whereas an exothermic reaction releases energy (\( \Delta H \) is negative).

To understand these reactions better, consider the bond-breaking and bond-forming processes:

• Bond breaking - Requires energy input, thus is an endothermic process.
• Bond forming - Releases energy, making it exothermic.

By analyzing enthalpy changes, students can predict the energetic behavior of reactions, which is pivotal for applications ranging from energy production to biochemical pathways.

Hess's Law

Hess's Law is a powerful principle in thermochemistry that allows the calculation of enthalpy changes for reactions where direct measurement is challenging. It states that the total enthalpy change for a chemical reaction is the same, regardless of the pathway taken from reactants to products, provided the initial and final conditions are identical.

In essence, Hess's Law is based on the conservation of energy, whereby energy changes in chemical reactions are additive. This law is particularly useful when dealing with complex reactions that can be broken down into simpler steps, as seen in the provided exercise.

The solution to the exercise illustrates using Hess's Law by reversing and scaling given reactions to match the desired equation. By "manipulating" reactions and summing their respective enthalpy changes, we derive the overall enthalpy for the decomposition of NOCl as +167.5 kJ. This demonstrates how Hess's Law simplifies the determination of enthalpy changes, even when direct measurement isn't feasible.

Chemists routinely employ Hess's Law as a tool to navigate the thermodynamics of chemical reactions, aiding in the design and optimization of efficient processes and energy systems.