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Chapter 13: Problem 35

Sketch a potential energy versus reaction progress plot for the following reactions: $$ \begin{array}{l} \text { (a) } \mathrm{S}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g) \quad \Delta H^{\circ}= \\ \quad-296 \mathrm{~kJ} / \mathrm{mol} \\ \text { (b) } \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{Cl}(g)+\mathrm{Cl}(g) \Delta H^{\circ}=243 \mathrm{~kJ} / \mathrm{mol} \end{array} $$

Short Answer

Expert verified
The plots for both reactions have been sketched. Reaction 1: S(s) + O2(g) -> SO2(g) is exothermic because \( \Delta H^{\circ} = -296 \) kJ/mol. It starts from a high energy level and goes to a lower one. Reaction 2: Cl2(g) -> 2Cl(g) is endothermic because \( \Delta H^{\circ} = 243 \) kJ/mol. It starts from a low energy level and goes to a higher one.

Step by step solution

01

Analyze the first reaction

The first reaction is \( S(s)+O_{2}(g) \rightarrow SO_{2}(g) \), with \( \Delta H^{\circ}= -296 \) kJ/mol. Since \( \Delta H^{\circ} \) is negative, this reaction is exothermic - the products have less energy than the reactants. The potential energy of reactant is higher than that of the product in the reaction progress. Therefore, in the plot, the starting point for this reaction should be higher than the ending point.
02

Sketch the first reaction

Draw vertical axis representing potential energy and horizontal axis representing reaction progress. The reaction progress starts with reactants (S and O2) and ends with product (SO2). Mark the vertical axis as 'Potential energy' and horizontal axis as 'Reaction progress'. Start at a high point for the reactants on the y-axis, then draw a line sloping downwards towards the products, showing that potential energy is decreasing due to the exothermic nature of the reaction.
03

Analyze the second reaction

The second reaction is \( Cl_{2}(g) \rightarrow Cl(g)+Cl(g) \), with \( \Delta H^{\circ}= 243 \) kJ/mol. Since \( \Delta H^{\circ} \) is positive, this reaction is endothermic - the products have more energy than the reactants. The potential energy of reactant is lower than that of the product in the reaction progress. Therefore, in the plot, the starting point for this reaction should be lower than the ending point.
04

Sketch the second reaction

Draw vertical axis representing potential energy and horizontal axis representing reaction progress. The reaction progress starts with reactant (Cl2) and ends with product (2Cl). Mark the vertical axis as 'Potential energy' and horizontal axis as 'Reaction progress'. Start at a relatively low point for the reactant on the y-axis, then draw a line sloping upwards towards the products, showing that potential energy is increasing due to the endothermic nature of the reaction.

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Key Concepts

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

Exothermic Reaction
An exothermic reaction is a chemical process that releases heat into its surroundings. It's like lighting a match; the chemical energy stored in the match is released as light and heat. In terms of the potential energy versus reaction progress plot, exothermic reactions begin with high potential energy and end with lower potential energy.
  • The products have less potential energy compared to the reactants, indicating that the energy difference has been released to the surrounding environment.
  • On a graph, this is represented by a downward slope from the reactants to the products, showing a decrease in potential energy.
  • Exothermic reactions typically involve bond formation, which is an energy-releasing process.
Common examples include combustion reactions like burning wood in a fireplace or the rusting of iron. The fact that exothermic reactions release energy makes them self-sustaining once they are initiated, as they keep producing the energy required to maintain the reaction.
Endothermic Reaction
Conversely, an endothermic reaction absorbs heat from the surroundings. Think of it as ice melting on a hot day; it requires heat from the air to change from solid to liquid. In a potential energy versus reaction progress plot, these reactions start with lower potential energy and end with higher.
  • The products are higher in potential energy when compared to the reactants because they have absorbed energy from the external environment.
  • This absorption is depicted by an upward slope on the graph, indicating an increase in potential energy as the reaction proceeds.
  • Endothermic reactions commonly involve bond breaking, which requires energy input.
Melting ice cubes, evaporating water, and photosynthesis in plants are all examples of endothermic reactions. Since these reactions consume energy, they require a continuous supply of heat to proceed.
Enthalpy Change
Enthalpy change (ΔH) is the term used to describe the heat exchange in a reaction at constant pressure. It measures the total heat content, or 'enthalpy', of a system. This concept is a pivotal part of learning about thermodynamics in chemistry.
  • If ΔH is negative, the reaction is exothermic, releasing heat to the surroundings and resulting in less enthalpy in the products than in the reactants.
  • If ΔH is positive, the reaction is endothermic, absorbing heat from the surroundings and resulting in more enthalpy in the products than in the reactants.
  • Understanding ΔH helps predict whether a reaction will generate or require heat and consequently whether it will occur spontaneously under given conditions.
The values of ΔH, like -296 kJ/mol for the formation of SO2 from sulfur and oxygen, or 243 kJ/mol for the dissociation of Cl2 into Cl atoms, tell us how much energy is involved in these processes.
Considering the sketching exercise provided, the enthalpy changes illustrate why potential energy diagrams slope down for exothermic and up for endothermic reactions, highlighting the heat flow aspect of chemical reactions.

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Most popular questions from this chapter

The rate law for the reaction \(2 \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) is rate \(=k\left[\mathrm{H}_{2}\right][\mathrm{NO}]^{2} .\) Which of the following mechanisms can be ruled out on the basis of the observed rate expression? Mechanism I $$ \begin{array}{c} \mathrm{H}_{2}+\mathrm{NO} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{N} \\ \mathrm{N}+\mathrm{NO} \longrightarrow \mathrm{N}_{2}+\mathrm{O} \\ \mathrm{O}+\mathrm{H}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O} \end{array} $$ $$ \begin{aligned} &\text { Mechanism II }\\\ &\begin{array}{l} \mathrm{H}_{2}+2 \mathrm{NO} \longrightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} \\ \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \longrightarrow \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O} \end{array} \end{aligned} $$ Mechanism III $$ \begin{aligned} 2 \mathrm{NO} & \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{2} \\ \mathrm{~N}_{2} \mathrm{O}_{2}+\mathrm{H}_{2} & \longrightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} \\ \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \longrightarrow & \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O} \end{aligned} $$

Write the equations relating the half-life of a secondorder reaction to the rate constant. How does it differ from the equation for a first-order reaction?

Some reactions are described as parallel in that the reactant simultaneously forms different products with different rate constants. An example is and $$ \begin{array}{l} \mathrm{A} \stackrel{k_{1}}{k_{2}} \mathrm{~B} \\ \mathrm{~A} \stackrel{\longrightarrow}{\longrightarrow} \mathrm{C} \end{array} $$ The activation energies are \(45.3 \mathrm{~kJ} / \mathrm{mol}\) for \(k_{1}\) and \(69.8 \mathrm{~kJ} / \mathrm{mol}\) for \(k_{2}\). If the rate constants are equal at \(320 \mathrm{~K},\) at what temperature will \(k_{1} / k_{2}=2.00 ?\)

Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right),\) commonly called table sugar, undergoes hydrolysis (reaction with water) to produce fructose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) and glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right):\) $$ \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} $$ fructose \(\quad\) glucose This reaction is of considerable importance in the candy industry. First, fructose is sweeter than sucrose. Second, a mixture of fructose and glucose, called invert sugar, does not crystallize, so the candy containing this sugar would be chewy rather than brittle as candy containing sucrose crystals would be. (a) From the following data determine the order of the reaction. (b) How long does it take to hydrolyze 95 percent of sucrose? (c) Explain why the rate law does not include \(\left[\mathrm{H}_{2} \mathrm{O}\right]\) even though water is a reactant. $$ \begin{array}{cc} \hline \text { Time (min) } & {\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right]} \\ \hline 0 & 0.500 \\ 60.0 & 0.400 \\ 96.4 & 0.350 \\ 157.5 & 0.280 \\ \hline \end{array} $$

At \(25^{\circ} \mathrm{C},\) the rate constant for the ozone-depleting reaction \(\mathrm{O}(g)+\mathrm{O}_{3}(g) \longrightarrow 2 \mathrm{O}_{2}(g)\) is \(7.9 \times 10^{-15} \mathrm{~cm}^{3} /\) molecule \(\cdot\) s. Express the rate constant in units of \(1 / M \cdot \mathrm{s}\)
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