For the following equilibrium reactions, calculate \(\Delta G^{\circ}\) at the indicated temperature. [Hint: How is each equilibrium constant related to a thermodynamic equilibrium constant, \(K ?]\) (a) \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) \quad K_{\mathrm{c}}=50.2\) at \(445^{\circ} \mathrm{C}\) (b) \(\mathrm{N}_{2} \mathrm{O}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\) \(K_{c}=1.7 \times 10^{-13} \mathrm{at} 25^{\circ} \mathrm{C}\) (c) \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g})\) \(K_{c}=4.61 \times 10^{-3}\) at \(25^{\circ} \mathrm{C}\) (d) \(2 \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Hg}_{2}^{2+}(\mathrm{aq}) \rightleftharpoons\) \(2 \mathrm{Fe}^{2+}(\mathrm{aq})+2 \mathrm{Hg}^{2+}(\mathrm{aq})\) \(K_{\mathrm{c}}=9.14 \times 10^{-6} \mathrm{at} 25^{\circ} \mathrm{C}\)

Step by step solution

01

Step 1. Convert Temperatures

Convert the temperatures from Celsius to Kelvin: For part (a), \(445^{\circ}C = 445 + 273.15 = 718.15K\). For part (b) and (c), \(25^{\circ}C = 25 + 273.15 = 298.15K\).

02

Step 2. Calculate \(\Delta G^{\circ}\) for Part (a)

Use the formula \(\Delta G^{\circ} = -RT \ln K\) with \(R = 8.3145 J/K.mol\), \(T = 718.15 K\), and \(K_{c} = 50.2\) to get: \(\Delta G^{\circ} = -8.3145 J/K.mol \times 718.15K \times \ln 50.2 = -45778.2 J/mol = -45.78 kJ/mol\).

03

Step 3. Calculate \(\Delta G^{\circ}\) for Part (b)

Use the formula \(\Delta G^{\circ} = -RT \ln K\) with \(R = 8.3145 J/K.mol\), \(T = 298.15 K\), and \(K_{c} = 1.7 \times 10^{-13}\) to get: \(\Delta G^{\circ} = -8.3145 J/K.mol \times 298.15K \times \ln (1.7 \times 10^{-13}) = 76443 J/mol = 76.44 kJ/mol \).

04

Step 4. Calculate \(\Delta G^{\circ}\) for Part (c)

Use the formula \(\Delta G^{\circ} = -RT \ln K\) with \(R = 8.3145 J/K.mol\), \(T = 298.15 K\), and \(K_{c} = 4.61 \times 10^{-3}\) to get: \(\Delta G^{\circ} = -8.3145 J/K.mol \times 298.15K \times \ln (4.61 \times 10^{-3}) = 11693.19 J/mol = 11.69 kJ/mol \).

05

Step 5. Calculate \(\Delta G^{\circ}\) for Part (d)

Use the formula \(\Delta G^{\circ} = -RT \ln K\) with \(R = 8.3145 J/K.mol\), \(T = 298.15 K\), and \(K_{c} = 9.14 \times 10^{-6}\) to get: \(\Delta G^{\circ} = -8.3145 J/K.mol \times 298.15K \times \ln (9.14 \times 10^{-6}) = 27108.3 J/mol = 27.11 kJ/mol \).

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