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Chapter 17: Problem 24

Calculate \(\Delta G^{\circ}\) at \(45^{\circ} \mathrm{C}\) for reactions for which (a) \(\Delta H^{\circ}=293 \mathrm{~kJ} ; \Delta S^{\circ}=-695 \mathrm{~J} / \mathrm{K}\) (b) \(\Delta H^{\circ}=-1137 \mathrm{~kJ} ; \Delta S^{\circ}=0.496 \mathrm{~kJ} / \mathrm{K}\) (c) \(\Delta H^{\circ}=-86.6 \mathrm{~kJ} ; \Delta S^{\circ}=-382 \mathrm{~J} / \mathrm{K}\)

Short Answer

Expert verified
Question: Calculate the standard Gibbs free energy change (ΔG°) at 45°C for the following reactions: a) ΔH° = 293 kJ and ΔS° = -695 J/K b) ΔH° = -1137 kJ and ΔS° = 0.496 kJ/K c) ΔH° = -86.6 kJ and ΔS° = -382 J/K Answer: The standard Gibbs free energy changes for the given reactions are: a) ~515.13 kJ, b) ~-1178.38 kJ, and c) ~-5.97 kJ.

Step by step solution

01

Conversion of temperature to Kelvin

Convert 45°C to Kelvin by adding 273.15: \(T = 45\,^{\circ}\mathrm{C} + 273.15 = 318.15\, \mathrm{K}\)
02

Step 2a: Calculate \(\Delta G^{\circ}\) for given values in part (a)

Use the formula, \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\), and given values \(\Delta H^{\circ}=293 \mathrm{~kJ}\) and \(\Delta S^{\circ}=-695 \mathrm{~J} / \mathrm{K}\) for part (a). First, convert \(\Delta S^{\circ}\) to kJ/K, then substitute values and calculate: \(\Delta G^{\circ} = 293\,\mathrm{kJ} - 318.15\,\mathrm{K} \cdot (-695\,\mathrm{J/K} \cdot 10^{-3}\,\mathrm{kJ/J})\) \(\Delta G^{\circ} = 293\,\mathrm{kJ} - 318.15\,\mathrm{K} \cdot (-0.695\,\mathrm{kJ/K})\) \(\Delta G^{\circ} \approx 515.13\,\mathrm{kJ}\) (rounded to two decimal places)
03

Step 2b: Calculate \(\Delta G^{\circ}\) for given values in part (b)

Use the formula, \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\), and given values \(\Delta H^{\circ}=-1137 \mathrm{~kJ}\) and \(\Delta S^{\circ}=0.496 \mathrm{~kJ}/ \mathrm{K}\) for part (b). Substitute values and calculate: \(\Delta G^{\circ} = -1137\,\mathrm{kJ} - 318.15\,\mathrm{K} \cdot (0.496\,\mathrm{kJ/K})\) \(\Delta G^{\circ} \approx -1178.38\,\mathrm{kJ}\) (rounded to two decimal places)
04

Step 2c: Calculate \(\Delta G^{\circ}\) for given values in part (c)

Use the formula, \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\), and given values \(\Delta H^{\circ}=-86.6 \mathrm{~kJ}\) and \(\Delta S^{\circ}=-382 \mathrm{~J} / \mathrm{K}\) for part (c). First, convert \(\Delta S^{\circ}\) to kJ/K, then substitute values and calculate: \(\Delta G^{\circ} = -86.6\,\mathrm{kJ} - 318.15\,\mathrm{K} \cdot (-382\,\mathrm{J/K} \cdot 10^{-3}\,\mathrm{kJ/J})\) \(\Delta G^{\circ} = -86.6\,\mathrm{kJ} - 318.15\,\mathrm{K} \cdot (-0.382\,\mathrm{kJ/K})\) \(\Delta G^{\circ} \approx -5.97\,\mathrm{kJ}\) (rounded to two decimal places) Hence, the standard Gibbs free energy changes for the given reactions are: (a) ~515.13 kJ, (b) ~-1178.38 kJ, and (c) ~-5.97 kJ.

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Most popular questions from this chapter

\(K_{a}\) for acetic acid \(\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\) at \(25^{\circ} \mathrm{C}\) is \(1.754 \times 10^{-5} .\) At \(50^{\circ} \mathrm{C}, K_{\mathrm{a}}\) is \(1.633 \times 10^{-5}\). Assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are not affected by a change in temperature, calculate \(\Delta S^{\circ}\) for the ionization of acetic acid.

For the reaction $$ \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) $$ \(K=50.0\) at \(721 \mathrm{~K}\) (a) What is \(\Delta G^{\circ}\) at \(721 \mathrm{~K} ?\) (b) What is \(K\) at \(25^{\circ} \mathrm{C}\) ? \(\left(\Delta G_{f}^{\circ} \mathrm{I}_{2}(g)=+19.4 \mathrm{~kJ} / \mathrm{mol}\right.\) )

Consider the reaction $$ \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{NO}(g) \quad K=4.4 \times 10^{-19} $$ (a) Calculate \(\Delta G^{\circ}\) for the reaction at \(25^{\circ} \mathrm{C}\). (b) Calculate \(\Delta G_{i}^{\circ}\) for \(\mathrm{N}_{2} \mathrm{O}\) at \(25^{\circ} \mathrm{C}\).

The normal boiling point for ethyl alcohol is \(78.4^{\circ} \mathrm{C} .5^{\circ}\) for \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)\) is \(282.7 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}\). At what temperature is the vapor pressure of ethyl alcohol \(357 \mathrm{~mm} \mathrm{Hg} ?\)

I Sodium carbonate, also called "washing soda," can be made by heating sodium hydrogen carbonate: $$ \begin{gathered} 2 \mathrm{NaHCO}_{3}(s) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H^{\circ}=+135.6 \mathrm{~kJ} ; \Delta G^{\circ}=+34.6 \mathrm{~kJ} \text { at } 25^{\circ} \mathrm{C} \end{gathered} $$ (a) Calculate \(\Delta S^{\circ}\) for this reaction. Is the sign reasonable? (b) Calculate \(\Delta G^{\circ}\) at \(0 \mathrm{~K} ;\) at \(1000 \mathrm{~K}\).
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