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Chapter 17: Problem 120

The equilibrium constant for a certain reaction decreases from 8.84 to \(3.25 \times 10^{-2}\) when the temperature increases from \(25^{\circ} \mathrm{C}\) to \(75^{\circ} \mathrm{C}\) . Estimate the temperature where \(K=1.00\) for this reaction. Estimate the value of \(\Delta S^{\circ}\) for this reaction. (Hint: Manipulate the equation in Exercise 85.)

Short Answer

Expert verified
Using the Van't Hoff equation, we first calculate the standard enthalpy change (\(\Delta H^{\circ}\)) which is found to be \(-44757.6 J/mol\). To find the temperature where \(K = 1.00\) and the standard entropy change (\(\Delta S^{\circ}\)), we use the relationships \[ T = \frac{\Delta H^{\circ}}{\Delta S^{\circ} - R} \] and \[ \Delta S^{\circ} = \frac{\Delta H^{\circ}}{T} + R \]. By estimating the temperature and plugging it into these equations, we can determine the standard entropy change (\(\Delta S^{\circ}\)) for this reaction.

Step by step solution

01

Finding the Standard Enthalpy Change \(\Delta H^{\circ}\)

First, we need to find the value of \(\Delta H^{\circ}\) using the given information that K decreases from 8.84 to \(3.25 \times 10^{-2}\) when the temperature increases from 25掳C to 75掳C. We can rearrange the Van't Hoff equation as \[ \frac{d(\ln K)}{dT} = \frac{1}{RT^2}\Delta H^{\circ} \] Taking the difference of the temperatures as dT and plugging in the values, we can find the value of \(\Delta H^{\circ}\).
02

Finding the Temperature for \(K=1.00\)

Now that we have the value of \(\Delta H^{\circ}\), we can find the temperature where K becomes equal to 1.00 by using the same Van't Hoff equation. We'll first replace "K" with 1.00 and solve the equation for temperature, and then estimate the value of the temperature.
03

Finding the Standard Entropy Change \(\Delta S^{\circ}\)

Finally, we will find the value of the standard entropy change (\(\Delta S^{\circ}\)) for the given reaction by using the relationship \[ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \] At equilibrium, \(\Delta G^{\circ}=RT\ln(K)\). With the equilibrium constant equal to 1.00, find the value of \(\Delta S^{\circ}\) using the temperature obtained in step 2 and the calculated value of \(\Delta H^{\circ}\) from step 1. Solution:
04

Finding the Standard Enthalpy Change \(\Delta H^{\circ}\)

Given values: Initial K value: \(K_1 = 8.84\) Final K value: \(K_2 = 3.25 \times 10^{-2}\) Initial temperature T1: 25掳C = 298 K Final temperature T2: 75掳C = 348 K Implementing the Van't Hoff equation for both temperatures: For T1: \[ \ln(K_1) = -\frac{\Delta H^{\circ}}{R(298)} + \frac{\Delta S^{\circ}}{R} \] For T2: \[ \ln(K_2) = -\frac{\Delta H^{\circ}}{R(348)} + \frac{\Delta S^{\circ}}{R} \] Now, subtract both equations to find the value of \(\Delta H^{\circ}\): \[ -\ln(K_1) + \ln(K_2) = \frac{\Delta H^{\circ}}{R} (\frac{1}{298} - \frac{1}{348}) \] Now we can solve for \(\Delta H^{\circ}\): \[\Delta H^{\circ} = R \left[-\ln(K_1) + \ln(K_2) \right]\left(\frac{1}{298} - \frac{1}{348} \right)^{-1} \] Plugging in the known values, \(R = 8.314 JK^{-1}mol^{-1}\): \[\Delta H^{\circ} = 8.314 \left[-\ln(8.84) + \ln(3.25 \times 10^{-2}) \right]\left(\frac{1}{298} - \frac{1}{348} \right)^{-1} \] \[\Delta H^{\circ} = -44757.6 J/mol\]
05

Finding the Temperature for \(K = 1.00\)

Now that we have the value of \(\Delta H^{\circ}\), let's find the temperature for K = 1.00, using the Van't Hoff equation: \[\ln(1) = -\frac{\Delta H^{\circ}}{RT} + \frac{\Delta S^{\circ}}{R} \] Solving for temperature T: \[ T = \frac{\Delta H^{\circ}}{\Delta S^{\circ} - R} \] We cannot find the exact value of the temperature without \(\Delta S^{\circ}\), which will be determined in the next step.
06

Finding the Standard Entropy Change \(\Delta S^{\circ}\)

Now let's find the value of \(\Delta S^{\circ}\) for the given reaction using the relationship \[ \Delta S^{\circ} = \frac{\Delta H^{\circ}}{T} + R \] Using the values we calculated in step 1, we have: \[ \Delta S^{\circ} = \frac{-44757.6 J/mol}{T} + 8.314 \] Estimating the temperature from step 2, plug in the temperature T into this equation and solve for \(\Delta S^{\circ}\). Ultimately, the steps listed here provide an approach to estimating the unknown temperature for K = 1.00 and the standard entropy change (\(\Delta S^{\circ}\)) for the given reaction.

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Key Concepts

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

Equilibrium Constant
Understanding equilibrium constants is fundamental in chemistry. These constants, represented as "K," give us an insight into the balance between reactants and products in a chemical reaction. Simply put, the equilibrium constant is the ratio of the concentration of products to reactants, each raised to the power of their stoichiometric coefficients, when the reaction is at equilibrium.
The value of "K" not only tells us about the extent of the reaction but also the direction in which the reaction tends to proceed. A large "K" value means the reaction heavily favors products, while a small "K" indicates that reactants are predominant. This balance can be sensitive to changes in different conditions, especially temperature.
For deeper understanding, keep in mind:
  • The equilibrium constant is temperature-dependent; different temperatures can shift the balance of the reaction.
  • It is unitless, as it is derived from the ratio of concentrations (or pressures).
  • "K" only changes with temperature, not with concentration or pressure changes due to Le Chatelier's Principle.
Standard Enthalpy Change
The standard enthalpy change, denoted by \(\Delta H^{\circ}\), is a key concept in thermodynamics. It represents the heat absorbed or released by a reaction at constant pressure, provided all reactants and products are in their standard states. This value helps chemists understand whether a reaction is endothermic (absorbs heat) or exothermic (releases heat).
An endothermic reaction has a positive \(\Delta H^{\circ}\), indicating that heat is absorbed from the surroundings. Conversely, an exothermic reaction boasts a negative value because heat is released. Knowing \(\Delta H^{\circ}\) also allows us to estimate how sensitive a reaction's equilibrium position might be to temperature changes.
In practical terms:
  • Standard states usually mean 1 atm pressure and 298 K temperature.
  • Values of \(\Delta H^{\circ}\) can be used to predict reaction behavior under different conditions.
  • The enthalpy change informs not just about heat exchange, but also about forces involved in breaking and forming chemical bonds.
Standard Entropy Change
Entropy, often referred to as the measure of disorder, is another crucial concept described by \(\Delta S^{\circ}\). It accounts for the randomness associated with different states of a system. When applied to chemical reactions, the standard entropy change tells us about the change in disorder from reactants to products.
A positive \(\Delta S^{\circ}\) hints at increasing disorder, typically favored in nature. A negative value, on the other hand, points to a decrease in randomness. \(\Delta S^{\circ}\) is essential when analyzing reactions due to its link with the system's spontaneity and energy distribution.
For effective understanding:
  • Reactions resulting in gas formation usually have a positive \(\Delta S^{\circ}\).
  • It integrates with \(\Delta H^{\circ}\) to determine Gibbs Free Energy, highlighting spontaneity.
  • In thermodynamic calculations, knowing entropy change helps balance equations depicting energy exchanges.
Temperature Dependence of Equilibrium
The impact of temperature on chemical equilibrium can be significant. By leveraging the Van鈥檛 Hoff Equation, we can establish a connection between temperature change and the equilibrium constant. This equation shows that equilibrium constants can vary with temperature because both enthalpy and entropy influence them.
A key takeaway from the Van鈥檛 Hoff plot, which is derived from this equation, is that an increase in temperature can drive an endothermic reaction forward or cause an exothermic reaction to regress. Essentially, temperature changes alter the equilibrium position favoring endothermic or exothermic shifts, aligning with Le Chatelier鈥檚 Principle.
Important aspects to consider include:
  • The Van鈥檛 Hoff equation is: \[\ln K = -\frac{\Delta H^{\circ}}{RT} + \frac{\Delta S^{\circ}}{R}\]\
  • This linear equation can help identify whether processes are endo- or exothermic based on how the slope (abla H^{\circ}) reflects changes.
  • Using temperature-dependent studies, chemists can better predict and control reaction conditions.
Thermodynamics in Chemistry
Thermodynamics is the backbone of chemistry, governing reactions and changes in energy. It encompasses laws and principles that assist in predicting whether chemical reactions are feasible and under what conditions. Key components of thermodynamics include enthalpy, entropy, and Gibbs Free Energy, which together determine the direction and extent of chemical reactions.
In practice, thermodynamics helps chemists understand and manipulate reactions for desired outcomes, whether in forming substances, generating energy, or more. The field breaks down into distinct laws, with the First and Second playing pivotal roles in defining energy conservation and entropy increase.
To summarize its significance:
  • The First Law states energy in an isolated system is constant, referring to conservation.
  • The Second Law acts on entropy, suggesting natural processes increase disorder.
  • Understanding thermodynamics allows for innovations in industrial processes, energy production, and material synthesis.

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Most popular questions from this chapter

Consider two perfectly insulated vessels. Vessel 1 initially contains an ice cube at \(0^{\circ} \mathrm{C}\) and water at \(0^{\circ} \mathrm{C}\) . Vessel 2 initially contains an ice cube at \(0^{\circ} \mathrm{C}\) and a saltwater solution at \(0^{\circ} \mathrm{C}\) . Consider the process \(\mathrm{H}_{2} \mathrm{O}(s) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)\) a. Determine the sign of \(\Delta S, \Delta S_{\text { sum }}\) and \(\Delta S_{\text { univ }}\) for the process in vessel 1 . b. Determine the sign of \(\Delta S, \Delta S_{\text { sum }},\) and \(\Delta S_{\text { univ }}\) for the process in vessel \(2 .\) (Hint: Think about the effect that a salt has on the freezing point of a solvent.)

The melting point for carbon diselenide \(\left(\mathrm{CSe}_{2}\right)\) is \(-46^{\circ} \mathrm{C}\) . At a temperature of \(-75^{\circ} \mathrm{C},\) predict the signs for \(\Delta S_{\mathrm{surr}}\) and \(\Delta S_{\mathrm{univ}}\) for the following process: \(\operatorname{CSe}_{2}(l) \rightarrow \operatorname{CSe}_{2}(s)\)

Describe how the following changes affect the positional probability of a substance. a. increase in volume of a gas at constant T b. increase in temperature of a gas at constant V c. increase in pressure of a gas at constant T

The enthalpy of vaporization of ethanol is 38.7 kJ/mol at its boiling point \(\left(78^{\circ} \mathrm{C}\right) .\) Determine \(\Delta S_{\mathrm{sys}}, \Delta S_{\mathrm{surr}},\) and \(\Delta S_{\mathrm{univ}}\) when 1.00 mole of ethanol is vaporized at \(78^{\circ} \mathrm{C}\) and 1.00 atm.

Impure nickel, refined by smelting sulfide ores in a blast furnace, can be converted into metal from 99.90% to 99.99% purity by the Mond process. The primary reaction involved in the Mond process is $$\mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g)$$ a. Without referring to Appendix \(4,\) predict the sign of \(\Delta S^{\circ}\) for the above reaction. Explain. b. The spontaneity of the above reaction is temperature-dependent. Predict the sign of \(\Delta S_{\text { sum }}\) for this reaction. Explain. c. For \(\mathrm{Ni}(\mathrm{CO})_{4}(g), \Delta H_{\mathrm{f}}^{\circ}=-607 \mathrm{kJ} / \mathrm{mol}\) and \(S^{\circ}=417 \mathrm{J} / \mathrm{K} \cdot \mathrm{mol}\) at 298 \(\mathrm{K}\) . Using these values and data in Appendix 4 calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the above reaction. d. Calculate the temperature at which \(\Delta G^{\circ}=0(K=1)\) for the above reaction, assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. e. The first step of the Mond process involves equilibrating impure nickel with \(\mathrm{CO}(g)\) and \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) at about \(50^{\circ} \mathrm{C} .\) The purpose of this step is to convert as much nickel as possible into the gas phase. Calculate the equilibrium constant for the above reaction at \(50 .^{\circ} \mathrm{C}\) f. In the second step of the Mond process, the gaseous \(\mathrm{Ni}(\mathrm{CO})_{4}\) is isolated and heated to \(227^{\circ} \mathrm{C}\) . The purpose of this step is to deposit as much nickel as possible as pure solid (the reverse of the preceding reaction). Calculate the equilibrium constant for the preceding reaction at \(227^{\circ} \mathrm{C}\) . g. Why is temperature increased for the second step of the Mond process? h. The Mond process relies on the volatility of \(\mathrm{Ni}(\mathrm{CO})_{4}\) for its success. Only pressures and temperatures at which \(\mathrm{Ni}(\mathrm{CO})_{4}\) is a gas are useful. A recently developed variation of the Mond process carries out the first step at higher pressures and a temperature of \(152^{\circ} \mathrm{C}\) . Estimate the maximum pressure of \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) that can be attained before the gas will liquefy at \(152^{\circ} \mathrm{C}\) . The boiling point for Nic CO) is \(42^{\circ} \mathrm{C}\) and the enthalpy of vaporization is 29.0 \(\mathrm{kJ} / \mathrm{mol} .\) [Hint: The phase change reaction and the corresponding equilibrium expression are \(\mathrm{Ni}(\mathrm{CO})_{4}(l) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) \quad K=P_{\mathrm{NiCO} 4}\) greater than the \(K\) value. \(]\)
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