/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 124 Suppose that a reaction has \(\D... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 124

Suppose that a reaction has \(\Delta H=-33 \mathrm{~kJ}\) and \(\Delta S=-58 \mathrm{~J} / \mathrm{K}\). At what temperature will it change from spontaneous to nonspontaneous?

Short Answer

Expert verified
The reaction becomes non-spontaneous at approximately 569 K.

Step by step solution

01

Understand Gibbs Free Energy

Gibbs Free Energy (\(G\)) change is given by the equation \(\Delta G = \Delta H - T \Delta S\), where \(\Delta G\) is the change in Gibbs free energy, \(\Delta H\) is the change in enthalpy, \(T\) is the temperature in Kelvin, and \(\Delta S\) is the change in entropy. For a reaction to be spontaneous, \(\Delta G\) must be less than zero.
02

Set \(\Delta G\) to Zero

To find the temperature at which the reaction changes from spontaneous to non-spontaneous, set \(\Delta G = 0\). The equation becomes \(0 = \Delta H - T \Delta S\).
03

Solve for Temperature \(T\)

Rearrange the equation \(0 = \Delta H - T \Delta S\) to solve for \(T\). This yields \(T = \frac{\Delta H}{\Delta S}\).
04

Convert Units

Since \(\Delta H\) is given in kJ and \(\Delta S\) in J/K, convert \(\Delta H\) to J by multiplying by 1000: \(\Delta H = -33,000 \, \text{J}\).
05

Calculate Temperature

Substitute the values into the formula: \(T = \frac{-33,000}{-58}\). Calculate \(T\) to find the temperature at which the reaction becomes non-spontaneous.
06

Obtain Result

Calculate \(T = 568.97\, \text{K}\). Since it's not practical to have a fraction of a Kelvin in this context, round to the nearest Kelvin: \(T \approx 569\, \text{K}\).

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Key Concepts

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

Spontaneity of Reaction
The spontaneity of a chemical reaction is determined by the change in Gibbs Free Energy (\(\Delta G\)). This concept helps us understand whether a reaction will occur on its own without any external input.
  • If \(\Delta G < 0\), the reaction is spontaneous and can proceed without any aid.
  • If \(\Delta G > 0\), the reaction is non-spontaneous and needs energy input to occur.
  • When \(\Delta G = 0\), the reaction is at equilibrium and can shift between reactants and products without any net change.
To determine if a reaction is spontaneous at a certain temperature, one crucial step involves finding the temperature at which \(\Delta G\) changes sign. In our exercise, we calculate the critical temperature where \(\Delta G\) becomes zero by setting the formula \(\Delta G = \Delta H - T\Delta S\) equal to zero and solving for \(T\). This shows at what point the reaction changes from being spontaneous to non-spontaneous.
Enthalpy Change
Enthalpy (\(\Delta H\)) represents the heat absorbed or released during a chemical reaction at constant pressure. It tells us if the reaction is endothermic or exothermic.
  • \(\Delta H < 0\) indicates an exothermic reaction, where heat is released.
  • \(\Delta H > 0\) indicates an endothermic reaction, where heat is absorbed.
In our example, the reaction has a \(\Delta H\) of \-33 kJ\, meaning it is exothermic and releases heat. An exothermic reaction generally favors spontaneity as it contributes to a negative \(\Delta G\) value. However, we also need to consider the entropy change and temperature to fully understand their combined effect on spontaneity.
Entropy Change
Entropy (\(\Delta S\)) measures the disorder or randomness in a system. A higher entropy generally means greater disorder and a tendency towards spontaneous change.
  • A positive \(\Delta S\) suggests increasing disorder in the products compared to the reactants, which can promote spontaneity.
  • A negative \(\Delta S\) suggests the products are more ordered than the reactants, which can inhibit spontaneity.
In the given reaction, \(\Delta S = -58 \text{ J/K}\), indicating that the system becomes more ordered upon reaction. This entropy change opposes the spontaneity driven by the exothermic nature of the reaction, as a more ordered outcome usually demands external energy, especially as temperature changes. Understanding how entropy interacts with enthalpy allows for precise predictions of reaction behaviors at different temperatures.

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Most popular questions from this chapter

Sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)\), the most widely produced chemical in the world, is made by a two-step oxidation of sulfur to sulfur trioxide, \(\mathrm{SO}_{3}\), followed by reaction with water. Calculate \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{SO}_{3}\) in \(\mathrm{kJ} / \mathrm{mol}\), given the following data: $$\begin{array}{ll}\mathrm{S}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g) & \Delta H^{\circ}=-296.8 \mathrm{~kJ} \\ \mathrm{SO}_{2}(g)+1 / 2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{3}(g) & \Delta H^{\circ}=-98.9 \mathrm{~kJ} \end{array}$$

A reaction inside a cylindrical container with a movable piston causes the volume to change from \(12.0 \mathrm{~L}\) to \(18.0 \mathrm{~L}\) while the pressure outside the container remains constant at \(0.975\) atm. (The volume of a cylinder is \(V=\pi r^{2} h\), where \(h\) is the height; \(1 \mathrm{~L} \cdot \mathrm{atm}=101.325 \mathrm{~J} .)\) (a) What is the value in joules of the work \(w\) done during the reaction? (b) The diameter of the piston is \(17.0 \mathrm{~cm}\). How far does the piston move?

Which of the following are state functions, and which are not? (a) The distance from your dorm room to your chemistry class. (b) The temperature in the room of your chemistry class. (c) The balance in your bank account.

How is it possible for a reaction to be spontaneous yet endothermic?

What are the two terms that make up the free-energy change for a reaction, \(\Delta G\), and which of the two is usually more important?
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