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Chapter 18: Problem 49

For reactions carried out under standard-state conditions, Equation (18.10) takes the form \(\Delta G^{\circ}=\Delta H^{\circ}-\) \(T \Delta S^{\circ} .\) (a) Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are independent of temperature, derive the equation $$\ln \frac{K_{2}}{K_{1}}=\frac{\Delta H^{\circ}}{R}\left(\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right)$$ where \(K_{1}\) and \(K_{2}\) are the equilibrium constants at \(T_{1}\) and \(T_{2}\), respectively. (b) Given that at \(25^{\circ} \mathrm{C} K_{\mathrm{c}}\) is \(4.63 \times 10^{-3}\) for the reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \quad \Delta H^{\circ}=58.0 \mathrm{~kJ} / \mathrm{mol} $$ calculate the equilibrium constant at \(65^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The equilibrium constant at 65掳C is approximately 0.133

Step by step solution

01

Derive the equation

Start with the equation \(\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}\). We can equate this to \(-RT\ln K\), since 螖G掳 = -RTlnK from thermodynamics, where R is the gas constant, T is temperature in Kelvin, and K is equilibrium constant. Thus, \(-RT\ln K = \Delta H^{\circ}-T\Delta S^{\circ}\). Divide by -RT, we get \(\ln K = -\frac{\Delta H^{\circ}}{R} + \frac{\Delta S^{\circ}}{R}\). Now put this equation (let's call it equation 1) firstly at T1 and then at T2 and subtract one from another, we get \(\ln K_2 - \ln K_1 = -\frac{\Delta H^{\circ}}{R}(\frac{1}{T2}-\frac{1}{T1})\), which simplifies to \(\ln \frac{K_{2}}{K_{1}}=\frac{\Delta H^{\circ}}{R}\left(\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right)\).
02

Convert given parameters into appropriate units

We are given 螖H掳 = 58.0 kJ/mol and T1 is 25掳C (or 298K) and T2 is 65掳C (or 338K). We need to convert 螖H掳 to J/mol by multiplying by 1000. So, 螖H掳 = 58000 J/mol.
03

Calculate the equilibrium constant at 65掳C using the derived formula and given values

Substitute 螖H掳 = 58000 J/mol, R = 8.314 J/mol/K, T1 = 298 K, T2 = 338 K and K1 = 4.63x10^-3 into the derived formula. This will lead to a value for K2 = \(K_2 = K_1* e^{(\Delta H^{\circ} / R)*((T_{2}-T_{1})/(T_{1} T_{2}))}\). Calculating this gives K2 鈮� 0.133.

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Key Concepts

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

Thermodynamics and Chemical Reactions
Thermodynamics plays an integral role in understanding how chemical reactions occur and proceed. It's a branch of physics that deals with heat, work, and the forms of energy involved in chemical processes. One of the fundamental concepts of thermodynamics is that energy cannot be created or destroyed, only converted from one form to another. This principle is essential when looking at chemical reactions which involve energy changes, typically in the form of heat and work.

Within the realm of chemical reactions, thermodynamics helps explain why certain reactions proceed spontaneously while others do not. For a reaction to be spontaneous, it must result in a decrease of the system's free energy. The term 'free energy' refers to the energy that is available to do work. This concept is crucial when discussing the Gibbs free energy and chemical equilibrium, which are directly related to the thermodynamics of a reaction.
Gibbs Free Energy
Gibbs free energy (G) is a thermodynamic property that can help predict the direction of a chemical reaction. It integrates the effects of enthalpy (heat content) and entropy (disorder) within a system under constant temperature and pressure. The change in Gibbs free energy (鈭咷) during a reaction provides valuable information:

If 鈭咷 is negative, the reaction is spontaneous and will occur without outside intervention. If 鈭咷 is positive, the reaction is non-spontaneous and requires energy input to proceed. And if 鈭咷 is zero, the system is at equilibrium, and no net change occurs over time.

The relationship between Gibbs free energy and the equilibrium constant (K) is given by the equation 鈭咷 = -RTlnK, where R is the gas constant and T is the temperature in Kelvin. This equation shows how the position of equilibrium is related to the energy change involved in a reaction, providing a quantitative measure of a reaction鈥檚 spontaneity.
Chemical Equilibrium
Chemical equilibrium is the state in a chemical reaction when the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant over time, not because the reactions have stopped, but because they are occurring at the same rate. The equilibrium constant (K) is a numerical value that represents the ratio of the concentrations of products to reactants at equilibrium, each raised to the power of their respective coefficients in the balanced equation.

In the context of the exercise, understanding how K changes with temperature is crucial. The equilibrium constant is not constant; it can change with varying temperatures. The Van 't Hoff equation, derived from thermodynamic principles, describes this dependency and can be used to calculate the value of K at different temperatures, given the standard enthalpy change (鈭咹掳) for the reaction.

The exercise shows the practical application of thermodynamics in predicting how changes in temperature can shift the position of equilibrium, signifying the importance of both Gibbs free energy and the equilibrium constant in comprehending chemical reactions.

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Most popular questions from this chapter

As an approximation, we can assume that proteins exist either in the native (or physiologically functioning) state and the denatured state $$\text { native } \rightleftharpoons \text { denatured }$$ The standard molar enthalpy and entropy of the denaturation of a certain protein are \(512 \mathrm{~kJ} / \mathrm{mol}\) and \(1.60 \mathrm{~kJ} / \mathrm{K} \cdot \mathrm{mol}\), respectively. Comment on the signs and magnitudes of these quantities, and calculate the temperature at which the process favors the denatured state.

Find the temperatures at which reactions with the following \(\Delta H\) and \(\Delta S\) values would become spontaneous: (a) \(\Delta H=-126 \mathrm{~kJ} / \mathrm{mol}, \Delta S=84 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\) (b) \(\Delta H=-11.7 \mathrm{~kJ} / \mathrm{mol}, \Delta S=-105 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\).

Consider the reaction $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ Given that \(\Delta G^{\circ}\) for the reaction at \(25^{\circ} \mathrm{C}\) is 173.4 \(\mathrm{kJ} / \mathrm{mol}\), (a) calculate the standard free energy of formation of \(\mathrm{NO},\) and (b) calculate \(K_{P}\) of the reaction. (c) One of the starting substances in smog formation is NO. Assuming that the temperature in a running automobile engine is \(1100^{\circ} \mathrm{C},\) estimate \(K_{P}\) for the above reaction. (d) As farmers know, lightning helps to produce a better crop. Why?

For each pair of substances listed here, choose the one having the larger standard entropy value at \(25^{\circ} \mathrm{C}\). The same molar amount is used in the comparison. Explain the basis for your choice. (a) \(\operatorname{Li}(s)\) or \(\operatorname{Li}(l)\) (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) or \(\mathrm{CH}_{3} \mathrm{OCH}_{3}(l)\) (c) \(\operatorname{Ar}(g)\) or \(\operatorname{Xe}(g)\) (d) \(\mathrm{CO}(g)\) or \(\mathrm{CO}_{2}(g)\) (e) \(\mathrm{O}_{2}(g)\) or \(\mathrm{O}_{3}(g)\) (f) \(\mathrm{NO}_{2}(g)\) or \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\)

The equilibrium constant \(\left(K_{P}\right)\) for the reaction $$ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) $$ is 4.40 at \(2000 \mathrm{~K}\). (a) Calculate \(\Delta G^{\circ}\) for the reaction. (b) Calculate \(\Delta G\) for the reaction when the partial pressures are \(P_{\mathrm{H}_{2}}=0.25 \mathrm{~atm}, P_{\mathrm{CO}_{2}}=0.78 \mathrm{~atm}\) \(P_{\mathrm{H}_{2} \mathrm{O}}=0.66 \mathrm{~atm},\) and \(P_{\mathrm{CO}}=1.20 \mathrm{~atm}\)
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