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 \(\mathrm{mol} / \mathrm{L}\) 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. \([\mathrm{HF}]=0.52 M,\left[\mathrm{~F}^{-}\right]=0.67 M,\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-3} M\) Based on the calculated \(\Delta G\) 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 \(\Delta G^\circ\) from \(K_a\).

For any reaction, the relationship between \(\Delta G^\circ\), \(K_a\), \(R\), and \(T\) is given as: $$ \Delta G^\circ = -RT \ln(K_a) $$ Here, \(K_a = 7.2 \times 10^{-4}\), \(R = 8.314 \, \mathrm{J/(mol \cdot K)}\), and \(T = 25 + 273.15 = 298.15 \, \mathrm{K}\). Plugging in the values, we get: $$ \Delta G^\circ = - (8.314 \, \mathrm{J/(mol \cdot K)}) (298.15 \, \mathrm{K}) \ln(7.2 \times 10^{-4}) $$ Calculating this, we find that \(\Delta G^\circ\): $$ \Delta G^\circ ≈ 3320.5 \, \mathrm{J/mol} $$ #Step 2: Calculate the reaction quotient Q for different conditions#

02

Calculate the reaction quotient \(Q\) for different conditions.

For the given reaction, the reaction quotient \(Q\) is defined as the product of the concentrations of products divided by the concentration of the reactants: $$ Q = \frac{[H^+][F^-]}{[HF]} $$ Now, calculate \(Q\) for each set of conditions (a-e): a. \(Q_a = \frac{1.0 \times 1.0}{1.0} = 1\) b. \(Q_b = \frac{(7.2 \times 10^{-4}) \times (7.2 \times 10^{-4})}{0.98} ≈ 5.3 \times 10^{-6}\) c. \(Q_c = \frac{1.0 \times 10^{-5} \times 1.0 \times 10^{-5}}{1.0 \times 10^{-5}} = 1.0 \times 10^{-5}\) d. \(Q_d = \frac{(7.2 \times 10^{-4}) \times 0.27}{0.27} = 7.2 \times 10^{-4}\) e. \(Q_e = \frac{(1 \times 10^{-3}) \times 0.67}{0.52} ≈ 1.3 \times 10^{-3}\) #Step 3: Calculate \(\Delta G\) for different conditions#

03

Calculate \(\Delta G\) for different conditions.

Using the equation \(\Delta G = \Delta G^\circ + RT \ln(Q)\), we can calculate \(\Delta G\) for different conditions (a-e): a. \(\Delta G_a = 3320.5 \, \mathrm{J/mol} + (8.314 \, \mathrm{J/(mol \cdot K)})(298.15 \, \mathrm{K}) \ln(1) = 3320.5 \, \mathrm{J/mol}\) b. \(\Delta G_b = 3320.5 \, \mathrm{J/mol} + (8.314 \, \mathrm{J/(mol \cdot K)})(298.15 \, \mathrm{K}) \ln(5.3 \times 10^{-6}) ≈ 6681.2 \, \mathrm{J/mol}\) c. \(\Delta G_c = 3320.5 \, \mathrm{J/mol} + (8.314 \, \mathrm{J/(mol \cdot K)})(298.15 \, \mathrm{K}) \ln(1.0 \times 10^{-5}) ≈ 5960.0 \, \mathrm{J/mol}\) d. \(\Delta G_d = 3320.5 \, \mathrm{J/mol} + (8.314 \, \mathrm{J/(mol \cdot K)})(298.15 \, \mathrm{K}) \ln(7.2 \times 10^{-4}) = 0 \, \mathrm{J/mol}\) e. \(\Delta G_e = 3320.5 \, \mathrm{J/mol} + (8.314 \, \mathrm{J/(mol \cdot K)})(298.15 \, \mathrm{K}) \ln(1.3 \times 10^{-3}) ≈ 1151.6 \, \mathrm{J/mol}\) #Step 4: Determine the direction of the reaction for different conditions#

04

Determine the direction of the reaction for different conditions.

The direction of the reaction can be determined based on the sign of \(\Delta G\): a. \(\Delta G_a > 0\): The reaction will proceed in the reverse direction to reach equilibrium. b. \(\Delta G_b > 0\): The reaction will proceed in the reverse direction to reach equilibrium. c. \(\Delta G_c > 0\): The reaction will proceed in the reverse direction to reach equilibrium. d. \(\Delta G_d = 0\): The reaction is at equilibrium. e. \(\Delta G_e > 0\): The reaction will proceed in the reverse direction to reach equilibrium.

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