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Chapter 16: Problem 24

A reaction has \(\Delta H=98 \mathrm{kJ}\) and \(\Delta S=292 \mathrm{JK}\) . Investigate the spontaneity of the reaction at room temperature. Would increasing the temperature have any effect on the spontaneity of the reaction?

Short Answer

Expert verified
\(\Delta G\) is positive at room temperature, but increasing the temperature makes the reaction spontaneous.

Step by step solution

01

Understand the given values

Identify the given values in the problem. The enthalpy change (\(\Delta H\)) is \(98 \mathrm{kJ}\).The entropy change (\(\Delta S\)) is \(292 \mathrm{J/K}\).Note that we need to convert \(\Delta H\) to joules to match the units of \(\Delta S\).
02

Convert \(\Delta H\) to joules

Convert \(\Delta H\) from kilojoules to joules by multiplying by 1000:\(\Delta H = 98 \mathrm{kJ} \times 1000 \frac{\mathrm{J}}{\mathrm{kJ}} = 98000 \mathrm{J}\).
03

Use Gibbs free energy equation

Write the Gibbs free energy equation for spontaneity:\(\Delta G = \Delta H - T\Delta S\).Determine if the reaction is spontaneous by checking if \(\Delta G\) is negative. Room temperature is typically 298 K.
04

Calculate \(\Delta G\) at room temperature

Substitute the values into the Gibbs free energy equation:\[\Delta G = 98000 \mathrm{J} - 298 \mathrm{K} \times 292 \mathrm{J/K}\].Calculate the result:\[\Delta G = 98000 \mathrm{J} - 87016 \mathrm{J} = 10984 \mathrm{J}\].Since \(\Delta G\) is positive, the reaction is not spontaneous at room temperature.
05

Investigate the effect of increasing temperature

As temperature increases, the term \(T\Delta S\) becomes larger. Substitute a higher temperature value (e.g., 500 K) to check the effect on \(\Delta G\):\[\Delta G = 98000 \mathrm{J} - 500 \mathrm{K} \times 292 \mathrm{J/K}\].Calculate the result:\[\Delta G = 98000 \mathrm{J} - 146000 \mathrm{J} = -48000 \mathrm{J}\].Since \(\Delta G\) is now negative, the reaction becomes spontaneous at higher temperatures.

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Key Concepts

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

enthalpy change
The enthalpy change, denoted as \(\Delta H\), quantifies the heat involved in a chemical reaction occurring at constant pressure. It essentially indicates whether a reaction absorbs or releases heat. When \(\Delta H\) is positive, the reaction absorbs heat (endothermic), and when it is negative, the reaction releases heat (exothermic). In the given exercise, \(\Delta H\) is 98 kJ, meaning the reaction absorbs heat. It's important to convert this into joules to match units with entropy change. Thus, \(\Delta H\ = 98000 J\). Understanding this concept is crucial as it reveals whether the thermal energy aspect of the reaction favors or opposes spontaneity.
entropy change
Entropy change, represented as \(\Delta S\), measures the disorder or randomness in a system during a reaction. A positive \(\Delta S\) indicates an increase in randomness, while a negative \(\Delta S\) indicates a decrease. For the exercise, the entropy change is 292 J/K, showing an increase in disorder. Entropy is a key factor in determining spontaneity because it indicates the degree to which the energy of a system is dispersed. The term \(\Delta S\) is temperature-dependent, playing a significant role in the Gibbs free energy equation.
spontaneity of reaction
The spontaneity of a reaction is determined using the Gibbs free energy equation: \(\Delta G = \Delta H - T\Delta S\). A reaction is spontaneous if \(\Delta G\) is negative. For the given problem, let's put this into practice. At room temperature (298 K), \(\Delta G\) is calculated as: \[\Delta G = 98000 J - 298 K \times 292 J/K = 10984 J\]. Since \(\Delta G\) is positive, the reaction is not spontaneous at this temperature. However, increasing the temperature to 500 K shows: \[\Delta G = 98000 J - 500 K \times 292 J/K = -48000 J\], making \(\Delta G\) negative and the reaction spontaneous. Therefore, temperature significantly impacts the spontaneity of the reaction.

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Most popular questions from this chapter

How does the enthalpy of the products of a reaction system compare with the enthalpy of the reactants when the reaction is a. endothermic? b. exothermic?

What is the main characteristic of a calorimeter in a bomb calorimeter experiment, and why is this characteristic essential?

Rewrite each equation below with the \(\Delta H\) value included with either the reactants or the products, and identify the reaction as endothermic or exothermic. a. \(\mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) \(\Delta H^{0}=-285.83 \mathrm{kJ}\) b. \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s)\) \(\Delta H^{0}=-1200 \mathrm{kJ}\) c. \(\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(g) ; \Delta H^{0}=+62.4 \mathrm{kJ}\) d. \(3 \mathrm{CO}(g)+\mathrm{Fe}_{2} \mathrm{O}_{3}(s) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)$$\Delta H^{0}=-24.7 \mathrm{kJ}\)

Relating ldeas Given the entropy change for the first two reactions below, calculate the entropy change for the third reaction below. \(\mathrm{S}_{8}(s)+8 \mathrm{O}_{2}(s) \longrightarrow 8 \mathrm{SO}_{2}(g) \qquad \Delta S=89 \mathrm{J} / \mathrm{K}\) \(2 \mathrm{SO}_{2}(s)+\mathrm{O}_{2}(s) \longrightarrow 2 \mathrm{SO}_{3}(g) \qquad \Delta S=-188 \mathrm{J} / \mathrm{K}\) \(\mathrm{S}_{8}(s)+12 \mathrm{O}_{2}(s) \longrightarrow 8 \mathrm{SO}_{3}(g) \qquad \Delta S=?\)

Performance Design a simple calorimeter investigation to determine the molar enthalpy of fusion of water. Use the following materials: a large plastic-foam cup with cover, a thermometer, a balance, water at room temperature, and an ice cube. Allow your teacher to review your design. Then carry out the investigation, and write a laboratory report including your calculations and a comparison of your quantitative results with known values. Try to account for any disagreements between the experimental and actual values.
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