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Sketch a potential-energy-versus-reaction-progress plot for the following reactions: $$ \begin{array}{l} \text { (a) } \mathrm{S}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g) \\ \Delta H^{\circ}=-296.06 \mathrm{~kJ} / \mathrm{mol} \\ \text { (b) } \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{Cl}(g)+\mathrm{Cl}(g) \\\ \Delta H^{\circ}=242.7 \mathrm{~kJ} / \mathrm{mol} \end{array} $$

Short Answer

Expert verified
The potential energy plot for the exothermic reaction (a) will slope downwards, reflecting the release of energy (enthalpy change of -296.06 kJ/mol), while the plot for the endothermic reaction (b) will slope upwards, indicating the absorption of energy (enthalpy change of 242.7 kJ/mol).

Step by step solution

01

Understand the chemical reactions and enthalpy change

In part (a), the reaction is S(s) + O2(g) -> SO2(g). The enthalpy change for this reaction is -296.06kJ/mol, which means heat is released in the reaction, hence it is an exothermic reaction. In part (b), the reaction is Cl2(g) -> Cl(g) + Cl(g). The enthalpy change is 242.7kJ/mol, implying that heat is absorbed during the reaction. This is an endothermic reaction.
02

Sketch the potential energy plot for exothermic reaction

For the exothermic reaction (a), the energy level of reactants will be higher than the products. Thus, the potential energy plot would slant downwards. It starts high and then drops to signify the release of energy.
03

Sketch the potential energy plot for endothermic reaction

For the endothermic reaction (b), the energy level of reactants is lower than the products. Hence, the potential energy plot would slant upwards. The diagram starts low and rises to show that energy is absorbed.
04

Label the potential energy plots

In both cases, the horizontal axis represents the reaction progress from reactants to products, and the vertical axis represents potential energy. The difference in energy levels between reactants and products in both graphs corresponds to the absolute values of the enthalpy changes provided.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enthalpy Change
Enthalpy change, denoted as \( \Delta H \), is a measurement of the heat change in a chemical reaction at constant pressure. It is one of the most fundamental concepts in thermochemistry, the study of energy changes during chemical reactions. Understanding enthalpy change is crucial for predicting whether a reaction will release or absorb energy, which further determines if a reaction is favorable under certain conditions.

When a chemical reaction occurs, bonds between atoms are broken and new bonds are formed. Bond breaking requires energy, while bond formation releases energy. The enthalpy change of a reaction is the net result of these processes. A negative \( \Delta H \), as seen in the given textbook exercise part (a) with sulfur and oxygen forming sulfur dioxide (\(\mathrm{S}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)\)), indicates that the reaction releases more energy than it consumes; therefore, it is termed exothermic. Conversely, a positive \( \Delta H \), portrayed by the exercise's part (b) with the dissociation of chlorine gas (\(\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{Cl}(g)+\mathrm{Cl}(g)\)), signifies the reaction absorbs energy, making it endothermic.
Exothermic Reaction
An exothermic reaction is a chemical process in which energy is released into the surroundings, usually in the form of heat. This release can also come in the form of light or electricity. When it comes to understanding the energy flow, the potential-energy-versus-reaction-progress plot serves as a crucial visual aid. In these plots, an exothermic reaction is characterized by a downward slope from the reactants to the products.

For example, the combustion of sulfur in oxygen to form sulfur dioxide is exothermic, illustrated in our textbook problem. The plot would show a higher potential energy for the reactants relative to the products. The difference in height between these points is equal to the enthalpy change, which in this case is -296.06 kJ/mol. The negative sign indicates that heat is being released. It's important to note that exothermic reactions often occur spontaneously because they tend to lead to a state of lower energy and increased disorder, or entropy, in the universe.
Endothermic Reaction
Opposite to exothermic reactions, an endothermic reaction is one in which energy is absorbed from the surroundings. This absorption is typically in the form of thermal energy, as the system requires heat to proceed to the products. On a potential-energy-versus-reaction-progress plot, an endothermic reaction is represented by an upward slope from the reactants to the products.

Reflecting on our textbook example of chlorine gas dissociating into chlorine atoms, the plot for this reaction would start at a lower point and rise to a higher potential energy for the products. The enthalpy change value of 242.7 kJ/mol is positive, reinforcing the fact that energy is being consumed in the process. Endothermic reactions, such as photosynthesis or evaporating water, are critical for various biological processes and industrial applications, despite often requiring an input of energy to occur.

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Most popular questions from this chapter

The rate constant for the second-order reaction $$ 2 \mathrm{NOBr}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ is \(0.80 / M \cdot \mathrm{s}\) at \(10^{\circ} \mathrm{C}\). (a) Starting with a concentration of \(0.086 M,\) calculate the concentration of \(\mathrm{NOBr}\) after \(22 \mathrm{~s}\) (b) Calculate the half-lives when \([\mathrm{NOBr}]_{0}=\) \(0.072 M\) and \([\mathrm{NOBr}]_{0}=0.054 \mathrm{M}\).

The integrated rate law for the zero-order reaction \(\mathrm{A} \longrightarrow \mathrm{B}\) is \([\mathrm{A}]_{t}=[\mathrm{A}]_{0}-k t .\) (a) Sketch the follow- ing plots: (i) rate versus \([\mathrm{A}]_{t}\) and (ii) \([\mathrm{A}]_{t}\) versus \(t\). (b) Derive an expression for the half-life of the reaction. (c) Calculate the time in half-lives when the integrated rate law is no longer valid, that is, when \([\mathrm{A}]_{t}=0\)

Determine the overall orders of the reactions to which these rate laws apply: (a) rate \(=k\left[\mathrm{NO}_{2}\right]^{2} ;\) (b) rate \(=k\) (c) rate \(=k\left[\mathrm{H}_{2}\right]\left[\mathrm{Br}_{2}\right]^{\frac{1}{2}} ;\) (d) rate \(=k[\mathrm{NO}]^{2}\left[\mathrm{O}_{2}\right]\)

The rate constant of a first-order reaction is \(4.60 \times\) \(10^{-4} \mathrm{~s}^{-1}\) at \(350^{\circ} \mathrm{C}\). If the activation energy is 104 \(\mathrm{kJ} / \mathrm{mol}\), calculate the temperature at which its rate constant is \(8.80 \times 10^{-4} \mathrm{~s}^{-1}\)

For the reaction \(X_{2}+Y+Z \longrightarrow X Y+X Z\) it is found that doubling the concentration of \(\mathrm{X}_{2}\) doubles the reaction rate, tripling the concentration of \(Y\) triples the rate, and doubling the concentration of \(Z\) has no effect. (a) What is the rate law for this reaction? (b) Why is it that the change in the concentration of \(Z\) has no effect on the rate? (c) Suggest a mechanism for the reaction that is consistent with the rate law.

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