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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 \(-176.6\) \(\mathrm{J} / \mathrm{K} \cdot \mathrm{mol}\), respectively. Calculate the value of \(\mathrm{K}\) 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°C is approximately 4.47, and at 100°C it is approximately 705.82.

Step by step solution

01

Understand the van 't Hoff equation

The van 't Hoff equation relates the change in the equilibrium constant, K, with the change in temperature, T. It is given by: \[ \ln(K_2/K_1) = -\frac{\Delta H^\circ}{R} \cdot (\frac{1}{T_2} - \frac{1}{T_1}) \] Where: \(K_1\) and \(K_2\) are the equilibrium constants at temperatures \(T_1\) and \(T_2\), respectively ∆H° is the standard enthalpy change of the reaction R is the universal gas constant (8.314 J/mol·K) \(T_1\) and \(T_2\) are the initial and final temperatures in Kelvin
02

Find the equilibrium constant K at 25°C

First, we need to convert the temperature from Celsius to Kelvin: \(T_1 = 25.0°C + 273.15 = 298.15K \) Now we need to find the standard Gibbs free energy change, ∆G°, using the equation: \[\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\] Plugging in the given values, we get: \[\Delta G^\circ = -58.03 kJ/mol - (298.15K)(-176.6 J/(mol \cdot K))\] Converting the ∆S° term to kJ: \[\Delta G^\circ = -58.03 kJ/mol - (298.15K)(-0.1766 kJ/(mol \cdot K))\] \[\Delta G^\circ = -58.03 kJ/mol + 52.65 kJ/mol\] \[\Delta G^\circ = -5.38 kJ/mol\] Now, we can find the equilibrium constant K using the equation: \[K = e^{-\frac{\Delta G^\circ}{RT}}\] Plugging in the values, we get: \[K = e^{-\frac{-5380 J/mol}{(8.314 J/(mol \cdot K))(298.15K)}}\] \[K \approx 4.47\] So, the equilibrium constant K at 25°C is approximately 4.47.
03

Estimate the value of K at 100°C

First, convert the temperature from Celsius to Kelvin: \(T_2 = 100.0°C + 273.15 = 373.15K \) Now, we will use the van 't Hoff equation: \[\ln(K_2/4.47) = -\frac{-58.03 \times 10^3 J/mol}{8.314 J/mol \cdot K} \cdot (\frac{1}{373.15K} - \frac{1}{298.15K})\] Solving for \(K_2\): \[K_2 = 4.47 \times e^{\frac{58.03 \times 10^3 J/mol}{8.314 J/mol \cdot K} \cdot (\frac{1}{373.15K} - \frac{1}{298.15K})}\] \[K_2 \approx 705.82\] The equilibrium constant K at 100°C is approximately 705.82.

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Key Concepts

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

Equilibrium Constant
The equilibrium constant (K) is a vital concept in chemistry that quantifies the ratio of the concentrations of products to reactants at equilibrium for a reversible chemical reaction. It is a measure of the extent to which a reaction proceeds and is determined by the specific reaction's conditions, such as temperature. In our exercise, we are asked to calculate the value of K for the dimerization of nitrogen dioxide (NO2) to form dinitrogen tetroxide (N2O4) at different temperatures.

It's important when performing calculations involving K to pay attention to the units and to use Kelvin for temperature. Starting with the standard Gibbs free energy change (∆G°), we can determine the equilibrium constant at 25°C. As the temperature changes, the value of K also shifts, reflecting the temperature's effect on the reaction's equilibrium position. This is demonstrated in the calculation of K at 100°C using the van 't Hoff equation, which shows an increase in K, indicating that the reaction is more product-favored at higher temperatures.
Standard Enthalpy Change
The standard enthalpy change (∆H°) is the energy change associated with a chemical reaction taking place under standard conditions (1 atm pressure and a specified temperature, usually 25°C). It's a key factor for understanding heat transfer during a reaction—whether it's exothermic (releasing heat) or endothermic (absorbing heat). In the given exercise, ∆H° for the reaction is negative (-58.03 kJ/mol), indicating that the formation of N2O4 from NO2 is exothermic and releases heat into the surroundings.

Knowing the standard enthalpy change is crucial for predicting how a reaction's equilibrium will shift with temperature changes. As the van 't Hoff equation demonstrates, an exothermic reaction's K value will decrease with increasing temperature. Conversely, endothermic reactions will have their K value increase with temperature. This relationship is derived from the Le Chatelier's principle, which states that a system at equilibrium will adjust to counteract changes imposed upon it.
Gibbs Free Energy
Gibbs free energy (G) is a thermodynamic quantity that represents the maximum amount of work that can be performed by a thermodynamic system at constant temperature and pressure. In the realm of chemical reactions, the change in Gibbs free energy (∆G) determines the spontaneity of a process. A negative ∆G indicates that a reaction is spontaneous and can proceed without additional energy input, whereas a positive ∆G suggests that the reaction is non-spontaneous.

In our exercise, we calculate the standard Gibbs free energy change (∆G°) at 25°C which allows us to find the equilibrium constant K. The link between ∆G° and K is given by the equation K = e-(∆G°/RT), meaning that as ∆G° becomes less negative (or more positive), K decreases, and the reaction is less product-favored. This highlights the delicate interplay between thermodynamics and reaction equilibria—a foundational concept for students studying chemical thermodynamics.
Chemical Thermodynamics
Chemical thermodynamics is the study of the interrelation of heat and work with chemical reactions or with physical changes of state within the confines of the laws of thermodynamics. It encompasses all of the core concepts mentioned earlier: equilibrium constant, standard enthalpy change, and Gibbs free energy. These principles help us understand and predict how chemical reactions occur and how they are influenced by external conditions.

The van 't Hoff equation, applied in our exercise, is an application of thermodynamic principles that describes how the equilibrium constant varies with temperature. The exercise perfectly illustrates how thermodynamics can provide insights into the direction and extent of chemical reactions, aiding the students in grasping the larger picture of how energy and matter interact in chemical processes. Emphasizing connections between these thermodynamic quantities fosters a comprehensive understanding of chemical reactions and helps students develop the ability to predict the behavior of systems under various conditions.

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Most popular questions from this chapter

Consider the dissociation of a weak acid HA \(\left(K_{\mathrm{a}}=4.5 \times 10^{-3}\right)\) in water: $$ \mathrm{HA}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{A}^{-}(a q) $$ Calculate \(\Delta G^{\circ}\) for this reaction at \(25^{\circ} \mathrm{C}\).

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 temperaturedependent. Predict the sign of \(\Delta S_{\text {surr }}\) for this reaction. Explain. c. For \(\mathrm{Ni}(\mathrm{CO})_{4}(g), \Delta H_{\mathrm{f}}^{\circ}=-607 \mathrm{~kJ} / \mathrm{mol}\) and \(S^{\circ}=417 \mathrm{~J} / \mathrm{K}\). 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 \(\mathrm{Ni}(\mathrm{CO})_{4}\) 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 $$ \mathrm{Ni}(\mathrm{CO})_{4}(l) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) \quad K=P_{\mathrm{Ni}(\mathrm{CO})} $$ \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) will liquefy when the pressure of \(\mathrm{Ni}(\mathrm{CO})_{4}\) is greater than the \(K\) value.]

Given the following data: $$ \begin{aligned} 2 \mathrm{C}_{6} \mathrm{H}_{6}(l)+15 \mathrm{O}_{2}(g) \longrightarrow & 12 \mathrm{CO}_{2}(g)+& 6 \mathrm{H}_{2} \mathrm{O}(l) \\ & \Delta G^{\circ}=-6399 \mathrm{~kJ} \\ \mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) & \Delta G^{\circ}=-394 \mathrm{~kJ} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & \Delta G^{\circ}=-237 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta G^{\circ}\) for the reaction $$ 6 \mathrm{C}(s)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l) $$

Consider the reaction $$ 2 \mathrm{O}(g) \longrightarrow \mathrm{O}_{2}(g) $$ a. Predict the signs of \(\Delta H\) and \(\Delta S\). b. Would the reaction be more spontaneous at high or low temperatures?

Some water is placed in a coffee-cup calorimeter. When \(1.0 \mathrm{~g}\) of an ionic solid is added, the temperature of the solution increases from \(21.5^{\circ} \mathrm{C}\) to \(24.2^{\circ} \mathrm{C}\) as the solid dissolves. For the dissolving process, what are the signs for \(\Delta S_{\text {sys }}, \Delta S_{\text {surr }}\), and \(\Delta S_{\text {univ }}\) ?

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