Chapter 17: Problem 22
The deciding factor on why HF is a weak acid and not a strong acid like the other hydrogen halides is entropy. What occurs when HF dissociates in water as compared to the other hydrogen halides?
Chapter 17: Problem 22
The deciding factor on why HF is a weak acid and not a strong acid like the other hydrogen halides is entropy. What occurs when HF dissociates in water as compared to the other hydrogen halides?
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Get started for freeYou remember that \(\Delta G^{\circ}\) is related to \(R T \ln (K)\) but cannot remember if it's \(R T \ln (K)\) or \(-R T \ln (K)\) . Realizing what \(\Delta G^{\circ}\) and K mean, how can you figure out the correct sign?
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 \(\Delta S_{\text { sum }}\) for this reaction. Explain. c. For \(\mathrm{Ni}(\mathrm{CO})_{4}(g), \Delta H_{\mathrm{f}}^{\circ}=-607 \mathrm{kJ} / \mathrm{mol}\) and \(S^{\circ}=417 \mathrm{J} / \mathrm{K} \cdot \mathrm{mol}\) at 298 \(\mathrm{K}\) . 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. e. The first step of the Mond process involves equilibrating impure nickel with \(\mathrm{CO}(g)\) and \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) at about \(50^{\circ} \mathrm{C} .\) The purpose of this step is to convert as much nickel as possible into the gas phase. Calculate the equilibrium constant for the above reaction at \(50 .^{\circ} \mathrm{C}\) f. In the second step of the Mond process, the gaseous \(\mathrm{Ni}(\mathrm{CO})_{4}\) is isolated and heated to \(227^{\circ} \mathrm{C}\) . The purpose of this step is to deposit as much nickel as possible as pure solid (the reverse of the preceding reaction). Calculate the equilibrium constant for the preceding reaction at \(227^{\circ} \mathrm{C}\) . g. Why is temperature increased for the second step of the Mond process? h. The Mond process relies on the volatility of \(\mathrm{Ni}(\mathrm{CO})_{4}\) for its success. Only pressures and temperatures at which \(\mathrm{Ni}(\mathrm{CO})_{4}\) is a gas are useful. A recently developed variation of the Mond process carries out the first step at higher pressures and a temperature of \(152^{\circ} \mathrm{C}\) . Estimate the maximum pressure of \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) that can be attained before the gas will liquefy at \(152^{\circ} \mathrm{C}\) . The boiling point for Nic CO) is \(42^{\circ} \mathrm{C}\) and the enthalpy of vaporization is 29.0 \(\mathrm{kJ} / \mathrm{mol} .\) [Hint: The phase change reaction and the corresponding equilibrium expression are \(\mathrm{Ni}(\mathrm{CO})_{4}(l) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) \quad K=P_{\mathrm{NiCO} 4}\) greater than the \(K\) value. \(]\)
At \(100 .^{\circ} \mathrm{C}\) and \(1.00 \mathrm{atm}, \Delta H^{\circ}=40.6 \mathrm{kJ} / \mathrm{mol}\) for the vaporization of water. Estimate \(\Delta G^{\circ}\) for the vaporization of water at \(90 .^{\circ} \mathrm{C}\) and \(110 .^{\circ} \mathrm{C}\) . Assume \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) at \(100 .^{\circ} \mathrm{C}\) and 1.00 \(\mathrm{atm}\) do not depend on temperature.
Using Appendix 4 and the following data, determine \(S^{\circ}\) for \(\mathrm{Fe}(\mathrm{CO})_{5}(g) . \)\mathrm{Fe}(s)+5 \mathrm{CO}(g) \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(g) \quad \Delta S^{\circ}=?\( \)\mathrm{Fe}(\mathrm{CO})_{5}(l) \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(g) \quad \Delta S^{\circ}=107 \mathrm{J} / \mathrm{K}\( \)\mathrm{Fe}(s)+5 \mathrm{CO}(g) \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(l) \quad \Delta S^{\circ}=-677 \mathrm{J} / \mathrm{K}$
For the reaction $$\mathrm{CS}_{2}(g)+3 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{SO}_{2}(g)$$ \(\Delta S^{\circ}\) is equal to \(-143 \mathrm{JK}\) . Use this value and data from Appendix 4 to calculate the value of \(S^{\circ}\) for \(\mathrm{CS}_{2}(g) .\)
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