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 $$HF(a q) \rightleftharpoons H^{+}(a q)+F^{-}(a q)$$ 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. \([\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 M\) b. \([\mathrm{HF}]=0.98 M,\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=2.7 \times 10^{-2} M\) c. \([\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-5} M\) 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?

Short Answer

Expert verified
From the given data and calculations, we find the following ΔG values for each set of conditions: a. ΔG < 0, the reaction proceeds in the forward direction (to the right) b. ΔG < 0, the reaction proceeds in the forward direction (to the right) c. ΔG = 0, the reaction is in equilibrium d. ΔG > 0, the reaction proceeds in the backward direction (to the left) e. ΔG > 0, the reaction proceeds in the backward direction (to the left)

Step by step solution

01

Calculate ΔGº

Standard Gibbs Free Energy change (ΔGº) can be calculated using the relationship between dissociation constant (Ka) and ΔGº: \[\Delta G^{\circ}=-R T \ln K_{a}\] At 25°C, T = 298 K, R = 8.314 J/mol K, and Ka = 7.2 × 10⁻⁴. Plug these values into the equation to find ΔG°: \[\Delta G^{\circ}=-((8.314 \, \text{J/mol K})(298 \, \text{K})\ln(7.2 \times 10^{-4}))\]
02

Calculate ΔG for each condition

We can find ΔG for each of the five sets of conditions using the equation \(\Delta G = \Delta G^\circ + RT \ln(Q)\), where Q is the reaction quotient, given by \([H^{+}][F^{-}]/[HF]\). Calculate Q for each set of conditions: a. Q = (1.0)(1.0)/(1.0) b. Q = (2.7 × 10⁻²)(2.7 × 10⁻²)/(0.98) c. Q = (1.0 × 10⁻⁵)(1.0 × 10⁻⁵)/(1.0 × 10⁻⁵) d. Q = (7.2 × 10⁻⁴)(0.27)/(0.27) e. Q = (1.0 × 10⁻³)(0.67)/(0.52) Now, plug in the corresponding Q values into the equation ΔG = ΔGº + RTln(Q) to find ΔG for each condition.
03

Determine the direction of reaction

To find in which direction the reaction will shift under each set of conditions, we can compare the calculated ΔG values to zero: - If ΔG < 0, the reaction proceeds in the forward direction (to the right) - If ΔG > 0, the reaction proceeds in the backward direction (to the left) - If ΔG = 0, the reaction is in equilibrium Compare the calculated ΔG values for each set of conditions and determine the direction of the reaction.

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Most popular questions from this chapter

Impure nickel, refined by smelting sulfide ores in a blast furnace, can be converted into metal from \(99.90 \%\) to \(99.99 \%\) purity by the Mond process. The primary reaction involved in the Mond process is $$\mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g)$$ a. Without referring to Appendix \(4,\) predict the sign of \(\Delta S^{\circ}\) for the above reaction. Explain. b. The spontaneity of the above reaction is temperature-dependent. Predict the sign of \(\Delta S_{\text {surr}}\) for this reaction. Explain. c. For \(\mathrm{Ni}(\mathrm{CO})_{4}(g), \Delta H_{\mathrm{f}}^{\circ}=-607 \mathrm{kJ} / \mathrm{mol}\) and \(S^{\circ}=417 \mathrm{J} / \mathrm{K}\) mol at 298 K. Using these values and data in Appendix 4 calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the above reaction. d. Calculate the temperature at which \(\Delta G^{\circ}=0(K=1)\) for the above reaction, assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

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