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?

Step by step solution

01

Calculate \(\Delta G^\circ\)

Given: \(K_a = 7.2\times 10^{-4}\) \(T = 25^\circ C = 298 \, K\) \(R = 8.314 J\, mol^{-1}K^{-1}\) \(\Delta G^\circ = -RT\ln(K_a) = -(8.314\, J\, mol^{-1}K^{-1})(298\, K)\ln(7.2\times 10^{-4})\) \(\Delta G^\circ = 1920.85\, J\, mol^{-1}\) Now, we can find \(\Delta G\) for each set of conditions provided.

02

a. Calculate Q and \(\Delta G\) for the first set of conditions

Given: \([HF]=[H^+]=[F^-]=1.0 M\) Q = \(\frac{[H^+][F^-]}{[HF]}\) Q = \(\frac{(1.0)(1.0)}{(1.0)}\) Q = 1 \(\Delta G = \Delta G^\circ + RT\ln(Q) = 1920.85 + (8.314)(298)\ln(1)\) Since \(\ln(1) = 0\), \(\Delta G = \Delta G^\circ\). Hence, \(\Delta G = 1920.85\, J\, mol^{-1}\). Since \(\Delta G > 0\), the reaction will shift to achieve equilibrium in the reverse direction.

03

b. Calculate Q and \(\Delta G\) for the second set of conditions

Given: \([HF]=0.98M\), \([H^+]=[F^-]=2.7\times 10^{-2}M\) Q = \(\frac{[H^+][F^-]}{[HF]}\) Q = \(\frac{(2.7\times 10^{-2})(2.7\times 10^{-2})}{(0.98)}\) Q = 7.47\times 10^{-4} \(\Delta G = \Delta G^\circ + RT\ln(Q) = 1920.85 + (8.314)(298)\ln(7.47\times 10^{-4})\) \(\Delta G \approx 1898.47\, J\, mol^{-1}\). Since \(\Delta G > 0\), the reaction will shift to achieve equilibrium in the reverse direction.

04

c. Calculate Q and \(\Delta G\) for the third set of conditions

Given: \([HF]=[H^+]=[F^-]=1.0\times 10^{-5} M\) Q = \(\frac{[H^+][F^-]}{[HF]}\) Q = \(\frac{(1.0\times 10^{-5})(1.0\times 10^{-5})}{(1.0\times 10^{-5})}\) Q = 1.0\times 10^{-5} \(\Delta G = \Delta G^\circ + RT\ln(Q) = 1920.85 + (8.314)(298)\ln(1.0\times 10^{-5})\) \(\Delta G \approx 1847.52\, J\, mol^{-1}\). Since \(\Delta G > 0\), the reaction will shift to achieve equilibrium in the reverse direction.

05

d. Calculate Q and \(\Delta G\) for the fourth set of conditions

Given: \([HF]=[F^-]=0.27 M\), \([H^+]=7.2\times 10^{-4} M\) Q = \(\frac{[H^+][F^-]}{[HF]}\) Q = \(\frac{(0.27)(7.2\times 10^{-4})}{(0.27)}\) Q = 7.2\times 10^{-4} \(\Delta G = \Delta G^\circ + RT\ln(Q) = 1920.85 + (8.314)(298)\ln(7.2\times 10^{-4})\) \(\Delta G \approx 0\, J\, mol^{-1}\). Since \(\Delta G = 0\), the reaction is at equilibrium.

06

e. Calculate Q and \(\Delta G\) for the fifth set of conditions

Given: \([HF]=0.52M\), \([F^-]=0.67M\), \([H^+]=1.0\times 10^{-3}M\) Q = \(\frac{[H^+][F^-]}{[HF]}\) Q = \(\frac{(1.0\times 10^{-3})(0.67)}{(0.52)}\) Q = 1.29\times 10^{-3} \(\Delta G = \Delta G^\circ + RT\ln(Q) = 1920.85 + (8.314)(298)\ln(1.29\times 10^{-3})\) \(\Delta G \approx 402.93\, J\, mol^{-1}\). Since \(\Delta G > 0\), the reaction will shift to achieve equilibrium in the reverse direction. In conclusion, for conditions (a), (b), (c), and (e), the reaction will shift in the reverse direction to reach equilibrium. In condition (d), the reaction is already at equilibrium.

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