/*! 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 65 Consider the following reaction ... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the following reaction at \(25.0^{\circ} \mathrm{C}\) : $$2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ The values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03 \mathrm{~kJ} / \mathrm{mol}\) and \(-176.6 \mathrm{~J} / \mathrm{K}\). mol, respectively. Calculate the value of \(K\) at \(25.0^{\circ} \mathrm{C}\). Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are temperature independent, estimate the value of \(K\) at \(100.0^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The equilibrium constant (\(K\)) at \(25.0^{\circ}C\) (298.15 K) is approximately \(2.576\), and at \(100.0^{\circ}C\) (373.15 K) is approximately \(1.176\).

Step by step solution

01

(Step 1: Convert temperatures to Kelvin)

Convert the given temperatures to Kelvin, as the Van't Hoff equation requires temperatures to be in Kelvin: $$T_1 = 25.0^{\circ}C + 273.15K = 298.15 \, K$$ $$T_2 = 100.0^{\circ}C + 273.15K = 373.15 \, K$$
02

(Step 2: Calculate the initial equilibrium constant)

Use the Van't Hoff equation to calculate the equilibrium constant (\(K_1\)) for \(25.0^{\circ}C\) (\(298.15K\)), with the given values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\): $$\ln K_1 = -\frac{-58.03 \, kJ/mol}{8.314 \, J/mol \cdot K} \cdot \frac{1}{298.15 \, K} + \frac{-176.6 \, J/K}{8.314 \, J/mol \cdot K}$$ $$\ln K_1 = -\frac{-58.03 \times 10^3 \, J/mol}{8.314 \, J/mol \cdot K} \cdot \frac{1}{298.15 \, K} + \frac{-176.6 \, J/K}{8.314 \, J/mol \cdot K}$$ After calculating the values, we get: $$\ln K_1 = 0.94662$$ Convert the natural logarithm value to the actual value of \(K_1\): $$K_1 = e^{0.94662} \approx 2.576$$
03

(Step 3: Calculate the equilibrium constant at the new temperature)

Use the Van't Hoff equation again to calculate the equilibrium constant (\(K_2\)) for \(100.0^{\circ}C\) (\(373.15K\)), with the same values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\): $$\ln K_2 = -\frac{-58.03 \, kJ/mol}{8.314 \, J/mol \cdot K} \cdot \frac{1}{373.15 \, K} + \frac{-176.6 \, J/K}{8.314 \, J/mol \cdot K}$$ $$\ln K_2 = -\frac{-58.03 \times 10^3 \, J/mol}{8.314 \, J/mol \cdot K} \cdot \frac{1}{373.15 \, K} + \frac{-176.6 \, J/K}{8.314 \, J/mol \cdot K}$$ After calculating the values, we get: $$\ln K_2 = 0.16241$$ Convert the natural logarithm value to the actual value of \(K_2\): $$K_2 = e^{0.16241} \approx 1.176$$ So, the values of \(K\) at \(25.0^{\circ}C\) and \(100.0^{\circ}C\) are approximately \(2.576\) and \(1.176\), respectively.

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Key Concepts

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

Equilibrium Constant
In the context of chemical reactions, the equilibrium constant, often represented as \( K \), is a crucial value that helps us understand the balance of reactants and products at equilibrium. This constant is affected by temperature and can be used to predict the direction of the reaction. For the given reaction: \[ 2 \text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g) \], the equilibrium constant can be calculated using thermodynamic data such as the change in enthalpy \( \Delta H^\circ \) and the change in entropy \( \Delta S^\circ \). At a specific temperature, the Van't Hoff equation links these values with the equilibrium constant. It is given by: \[ \ln K = -\frac{\Delta H^\circ}{R} \cdot \frac{1}{T} + \frac{\Delta S^\circ}{R} \] where:
  • \( R \) is the universal gas constant \( (8.314 \, \text{J/mol} \cdot \text{K}) \)
  • \( T \) is the temperature in Kelvin
Understanding the equilibrium constant is key as it tells us to what extent a reaction will proceed and how conditions like temperature can shift the balance of this equilibrium.
Temperature Conversion
When working with thermochemical equations and calculations, it's essential to express temperature in Kelvin rather than Celsius. The Kelvin scale is an absolute thermodynamic temperature scale that eliminates negative values and aligns directly with the precision needed in scientific calculations. To convert from Celsius to Kelvin, the formula is straightforward: \[ T_{K} = T_{C} + 273.15 \] where:
  • \( T_{C} \) is the temperature in degrees Celsius
  • \( T_{K} \) is the temperature in Kelvin
For instance, if you have a temperature of \( 25.0^{\circ}C \), adding \( 273.15 \) gives you \( 298.15 \, K \). Similarly, for \( 100.0^{\circ}C \), it converts to \( 373.15 \, K \). Always ensure temperatures are in Kelvin when applying the Van't Hoff equation for consistency and accuracy in results.
Gibbs Free Energy Change
Gibbs free energy change, denoted by \( \Delta G^\circ \), is another thermodynamic parameter crucial for understanding chemical reactions. It merges the concepts of enthalpy and entropy, providing insight into the spontaneity of a reaction.The formula for Gibbs free energy change at constant temperature is: \[ \Delta G^\circ = \Delta H^\circ - T \times \Delta S^\circ \] where:
  • \( \Delta H^\circ \) is the standard enthalpy change
  • \( T \) is the temperature in Kelvin
  • \( \Delta S^\circ \) is the standard entropy change
If \( \Delta G^\circ \) is negative, it indicates that the reaction can occur spontaneously. In contrast, a positive \( \Delta G^\circ \) suggests that the reaction is non-spontaneous under standard conditions. By relating \( \Delta G^\circ \) to the equilibrium constant \( K \) through the relation: \[ \Delta G^\circ = -RT \ln K \], we can see the interplay between temperature and equilibrium position, as well as how changes in thermodynamic properties influence reaction spontaneity.

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Most popular questions from this chapter

Consider the following system at equilibrium at \(25^{\circ} \mathrm{C}\) : $$\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) \quad \Delta G^{\circ}=-92.50 \mathrm{~kJ}$$ What will happen to the ratio of partial pressure of \(\mathrm{PCl}_{5}\) to partial pressure of \(\mathrm{PCl}_{3}\) if the temperature is raised? Explain completely.

Some nonelectrolyte solute (molar mass \(=142 \mathrm{~g} / \mathrm{mol}\) ) was dissolved in \(150 . \mathrm{mL}\) of a solvent (density \(=0.879 \mathrm{~g} / \mathrm{cm}^{3}\) ). The elevated boiling point of the solution was \(355.4 \mathrm{~K}\). What mass of solute was dissolved in the solvent? For the solvent, the enthalpy of vaporization is \(33.90 \mathrm{~kJ} / \mathrm{mol}\), the entropy of vaporization is \(95.95\) \(\mathrm{J} / \mathrm{K} \cdot \mathrm{mol}\), and the boiling-point elevation constant is \(2.5 \mathrm{~K} \cdot \mathrm{kg} / \mathrm{mol}\).

a. Using the free energy profile for a simple one-step reaction, show that at equilibrium \(K=k_{\mathrm{f}} / k_{\mathrm{r}}\), where \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) are the rate constants for the forward and reverse reactions. Hint: Use the relationship \(\Delta G^{\circ}=-R T \ln (K)\) and represent \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) using the Arrhenius equation \(\left(k=A e^{-E_{2} / R T}\right)\). b. Why is the following statement false? "A catalyst can increase the rate of a forward reaction but not the rate of the reverse reaction."

If wet silver carbonate is dried in a stream of hot air, the air must have a certain concentration level of carbon dioxide to prevent silver carbonate from decomposing by the reaction $$\mathrm{Ag}_{2} \mathrm{CO}_{3}(s) \rightleftharpoons \mathrm{Ag}_{2} \mathrm{O}(s)+\mathrm{CO}_{2}(g)$$ \(\Delta H^{\circ}\) for this reaction is \(79.14 \mathrm{~kJ} / \mathrm{mol}\) in the temperature range of 25 to \(125^{\circ} \mathrm{C}\). Given that the partial pressure of carbon dioxide in equilibrium with pure solid silver carbonate is \(6.23 \times 10^{-3}\) torr at \(25^{\circ} \mathrm{C}\), calculate the partial pressure of \(\mathrm{CO}_{2}\) necessary to prevent decomposition of \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) at \(110 .{ }^{\circ} \mathrm{C}\). (Hint: Manipulate the equation in Exercise 71 .)

Two crystalline forms of white phosphorus are known. Both forms contain \(\mathrm{P}_{4}\) molecules, but the molecules are packed together in different ways. The \(\alpha\) form is always obtained when the liquid freezes. However, below \(-76.9^{\circ} \mathrm{C}\), the \(\alpha\) form spontaneously converts to the \(\beta\) form: $$\mathrm{P}_{4}(s, \alpha) \longrightarrow \mathrm{P}_{4}(s, \beta)$$ a. Predict the signs of \(\Delta H\) and \(\Delta S\) for this process. b. Predict which form of phosphorus has the more ordered crystalline structure (has the smaller positional probability).

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