What information can be determined from \(\Delta G\) for a reaction? Does one get the same information from \(\Delta G^{\circ}\), the standard free energy change? \(\Delta G^{\circ}\) allows determination of the equilibrium constant \(K\) for a reaction. How? How can one estimate the value of \(K\) at temperatures other than \(25^{\circ} \mathrm{C}\) for a reaction? How can one estimate the temperature where \(K=1\) for a reaction? Do all reactions have a specific temperature where \(K=1\) ?

Short Answer

Expert verified
\(\Delta G\) determines the spontaneity of a reaction, while \(\Delta G^{\circ}\) provides information about the reaction's thermodynamics under standard conditions. The equilibrium constant, \(K\), is calculated using \(K = e^{-\frac{\Delta G^{\circ}}{RT}}\). To estimate the value of \(K\) at different temperatures, the Van't Hoff equation, \(\ln{\frac{K_2}{K_1}} = \frac{-\Delta H^{\circ}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1}\right)\), can be employed. To find the temperature where \(K=1\), solve for the temperature in the equation \(\Delta G^{\circ} = -RT \ln{K}\) when \(\Delta G^{\circ} = 0\). However, not all reactions have a specific temperature where \(K=1\), as it depends on the initial conditions and the properties of the reaction.

Step by step solution

01

Meaning of \(\Delta G\) and \(\Delta G^{\circ}\)

\(\Delta G\), the Gibbs free energy change for a reaction, represents the maximum amount of reversible work that can be done by the system. It determines if the reaction will proceed spontaneously or not. If \(\Delta G < 0\), the reaction is spontaneous in the forward direction, and if \(\Delta G > 0\), the reaction is non-spontaneous and will proceed in the reverse direction. If \(\Delta G = 0\), then the reaction is at equilibrium. \(\Delta G^{\circ}\), the standard free energy change, is the Gibbs free energy change calculated under standard conditions (1 atm pressure, 1 M concentration, and a specified temperature, usually \(25^{\circ} \mathrm{C}\)). It provides information about the thermodynamics of the reaction in standard conditions.
02

Determining the Equilibrium Constant \(K\)

The equilibrium constant, \(K\), can be found using the relationship: \[ \Delta G^{\circ} = -RT \ln{K} \] Where \(R\) is the gas constant (8.314 J/mol K), \(T\) is the temperature in Kelvin, and \(K\) is the equilibrium constant. By rearranging the equation, we can find the equilibrium constant, \(K\): \[ K = e^{-\frac{\Delta G^{\circ}}{RT}} \]
03

Estimating the Value of \(K\) at a Different Temperature

To estimate the value of \(K\) at a temperature other than \(25^{\circ} \mathrm{C}\), we can use the Van't Hoff equation: \[ \ln{\frac{K_2}{K_1}} = \frac{-\Delta H^{\circ}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1}\right) \] where \(K_1\) and \(K_2\) are the equilibrium constants at temperatures \(T_1\) and \(T_2\), and \(\Delta H^{\circ}\) is the standard enthalpy change. If we are provided with \(\Delta H^{\circ}\) and \(K_1\) at the initial temperature \(T_1\), we can estimate the value of \(K_2\) at the new temperature \(T_2\).
04

Estimating the Temperature Where \(K=1\)

To estimate the temperature where \(K=1\) for a reaction, we can use the relationship between \(\Delta G^{\circ}\) and \(K\): \[ \Delta G^{\circ} = -RT \ln{K} \] When \(K=1\), the logarithm term \(\ln{K}\) becomes zero. Hence, at this temperature, \(\Delta G^{\circ} = 0\). We can use this relationship and solve for the temperature where this condition is met.
05

Determining If All Reactions Have a Temperature Where \(K=1\)

For some reactions, there may be a temperature where \(K=1\). However, not all reactions have such a specific temperature. It depends on the initial conditions of the reactants and products, as well as the properties of the reaction itself. As temperature changes, the enthalpy and entropy changes can affect the value of \(K\) in various ways, and there might not be a temperature where \(K=1\) can be attained for certain reactions.

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Most popular questions from this chapter

At \(100 .{ }^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}, \Delta H^{\circ}=40.6 \mathrm{~kJ} / \mathrm{mol}\) for the vaporiza- tion of water. Estimate \(\Delta G^{\circ}\) for the vaporization of water at \(90 .{ }^{\circ} \mathrm{C}\) and \(110 .{ }^{\circ} \mathrm{C}\). Assume \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) at \(100 .{ }^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) do not depend on temperature.

The equilibrium constant for a certain reaction increases by a factor of \(6.67\) when the temperature is increased from \(300.0 \mathrm{~K}\) to \(350.0 \mathrm{~K} .\) Calculate the standard change in enthalpy \(\left(\Delta H^{\circ}\right)\) for this reaction (assuming \(\Delta H^{\circ}\) is temperature-independent).

Consider the system $$ \mathrm{A}(g) \longrightarrow \mathrm{B}(g) $$ at \(25^{\circ} \mathrm{C}\). a. Assuming that \(G_{\mathrm{A}}^{\circ}=8996 \mathrm{~J} / \mathrm{mol}\) and \(G_{\mathrm{B}}^{\circ}=11,718 \mathrm{~J} / \mathrm{mol}\), calculate the value of the equilibrium constant for this reaction. b. Calculate the equilibrium pressures that result if \(1.00 \mathrm{~mole}\) of \(\mathrm{A}(g)\) at \(1.00\) atm and \(1.00\) mole of \(\mathrm{B}(g)\) at \(1.00 \mathrm{~atm}\) are mixed at \(25^{\circ} \mathrm{C}\). c. Show by calculations that \(\Delta G=0\) at equilibrium.

For the reaction at \(298 \mathrm{~K}\), $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g) $$ the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03 \mathrm{~kJ}\) and \(-176.6 \mathrm{~J} / \mathrm{K}\), respectively. What is the value of \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\) ? Assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature, at what temperature is \(\Delta G^{\circ}=0 ?\) Is \(\Delta G^{\circ}\) negative above or below this temperature?

Consider the reactions $$ \begin{aligned} \mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) & \longrightarrow \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q) \\ \mathrm{Ni}^{2+}(a q)+3 \mathrm{en}(a q) & \longrightarrow \mathrm{Ni}(\mathrm{en})_{3}^{2+}(a q) \end{aligned} $$ where $$ \text { en }=\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2} $$ The \(\Delta H\) values for the two reactions are quite similar, yet \(\mathrm{K}_{\text {reaction } 2}>K_{\text {reaction } 1} .\) Explain.

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