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Chapter 17: Problem 110

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

Short Answer

Expert verified
The standard Gibbs free energy change, \(\Delta G^{\circ}\), for the dissociation reaction of the weak acid HA in water at \(25^{\circ}C\) is approximately 20.66 kJ/mol.

Step by step solution

01

Convert temperature to Kelvin

To convert the given temperature from Celsius to Kelvin, use the formula below: $$T(K) = T(^{\circ}C) + 273.15$$ Plug in the given value \(T(^{\circ}C) = 25^{\circ}C\): $$T(K) = 25 + 273.15 = 298.15\,K$$
02

Calculate the standard Gibbs free energy change

Now we can use the formula to calculate \(\Delta G^{\circ}\): $$\Delta G^{\circ} = -RT\ln{K_a}$$ Plug in the values: R = 8.314 J/mol·K, T = 298.15 K, and \(K_a = 4.5 \times 10^{-3}\): $$\Delta G^{\circ} = - (8.314 \,\text{J/mol·K})(298.15 \,\text{K})\ln{(4.5 \times 10^{-3})}$$
03

Simplify and find the answer

Calculate the value of \(\Delta G^{\circ}\) by using the values from the previous step: $$\Delta G^{\circ} = -(8.314 \,\text{J/mol·K})(298.15 \,\text{K})\ln{(4.5 \times 10^{-3})} \approx 20662 \,\text{J/mol}$$ Since we usually express Gibbs free energy in kJ/mol, divide the result by 1000: $$\Delta G^{\circ} = \dfrac{20662 \,\text{J/mol}}{1000} \approx 20.66 \, \text{kJ/mol}$$ Therefore, \(\Delta G^{\circ}\) for the dissociation reaction of the weak acid HA in water at \(25^{\circ} \mathrm{C}\) is approximately 20.66 kJ/mol.

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Most popular questions from this chapter

In the text, the equation $$\Delta G=\Delta G^{\circ}+R T \ln (Q)$$ was derived for gaseous reactions where the quantities in \(Q\) were expressed in units of pressure. We also can use units of mol/L for the quantities in \(Q\)specifically for aqueous reactions. With this in mind, consider the reaction $$\mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q)$$ for which \(K_{\mathrm{a}}=7.2 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\) . Calculate \(\Delta G\) for the reaction under the following conditions at \(25^{\circ} \mathrm{C} .\) a. $[\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 \mathrm{M}$ b. $[\mathrm{HF}]=0.98 M,\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=2.7 \times 10^{-2} M$ c. $[\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-5} \mathrm{M}$ d. $[\mathrm{HF}]=\left[\mathrm{F}^{-}\right]=0.27 M,\left[\mathrm{H}^{+}\right]=7.2 \times 10^{-4} M$ e. $[\mathrm{HF}]=0.52 M,\left[\mathrm{F}^{-}\right]=0.67 M,\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-3} \mathrm{M}$ Based on the calculated DG values, in what direction will the reaction shift to reach equilibrium for each of the five sets of conditions?

Given the values of \(\Delta H\) and \(\Delta S,\) which of the following changes will be spontaneous at constant \(T\) and \(P ?\) a. $\Delta H=+25 \mathrm{kJ}, \Delta S=+5.0 \mathrm{J} / \mathrm{K}, T=300 . \mathrm{K}$ b. $\Delta H=+25 \mathrm{kJ}, \Delta S=+100 . \mathrm{J} / \mathrm{K}, T=300 . \mathrm{K}$ c. $\Delta H=-10 . \mathrm{kJ}, \Delta S=+5.0 \mathrm{J} / \mathrm{K}, T=298 \mathrm{K}$ d. $\Delta H=-10 . \mathrm{kJ}, \Delta S=-40 . \mathrm{J} / \mathrm{K}, T=200 . \mathrm{K}$

You remember that \(\Delta G^{\circ}\) is related to \(R T \ln (K)\) but cannot remember if it's \(R T \ln (K)\) or \(-R T \ln (K)\) . Realizing what $\Delta G^{\circ}$ and K mean, how can you figure out the correct sign?

Liquid water at \(25^{\circ} \mathrm{C}\) is introduced into an evacuated, insulated vessel. Identify the signs of the following thermodynamic functions for the process that occurs: $\Delta H, \Delta S, \Delta T_{\text { water }} \Delta S_{\text { surr }}\( \)\Delta S_{\text { univ }}$

When most biological enzymes are heated, they lose their catalytic activity. This process is called denaturing. The change original enzyme \(\rightarrow\) new form that occurs on heating is endothermic and spontaneous. Is the structure of the original enzyme or its new form more ordered (has the smaller positional probability)? Explain.
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