Question

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.

Short answer

The sign of ΔS° is negative, as the reaction results in a decrease in the number of moles of gaseous species. The sign of ΔSsurr is positive, as heat is released from the system to the surroundings. ΔH° and ΔS° for the reaction are calculated to be -165 kJ/mol and -404 J/Kmol, respectively. The temperature at which ΔG°=0 for the given reaction is approximately 408 K.

Step-by-step solution

Predicting the sign of ΔS°

To predict the sign of ΔS° (change in entropy) for the given reaction, we need to look at the number of moles of gas in the reactants and products. The reaction is: \(Ni(s) + 4CO(g) \rightleftharpoons Ni(CO)_4(g)\) There are 4 moles of gas on the reactant side (from 4 moles of CO) and 1 mole of gas on the product side (from Ni(CO)_4). As the process moves from the reactants to the products, it results in a decrease in the number of moles of gaseous species. A decrease in the number of moles of gas corresponds to a decrease in entropy, thus the sign of ΔS° would be negative.

Predicting the sign of ΔSsurr

The spontaneity of the reaction is temperature dependent. We know that the sign of ΔS° is negative. For a reaction to be spontaneous, ΔG should be negative. ΔG = ΔH - TΔS Since ΔS° is negative, TΔS will be positive. To ensure the reaction is spontaneous, ΔH must be negative, and its magnitude should be greater than TΔS. This means that the heat energy released (exothermic ΔH) is greater than the increase in entropy of the system. If the system releases heat energy, it means that the surroundings will absorb energy, leading to an increase in the entropy of the surroundings, ΔSsurr. Thus, the sign of ΔSsurr would be positive.

Calculating ΔH° and ΔS° for the reaction

We are given ΔHf° and S° for Ni(CO)_4(g) at 298K: ΔHf° = -607 kJ/mol, S° = 417 J/Kmol. We also need to find the values of ΔHf° and S° for Ni(s) and CO(g) from the provided Appendix 4. From Appendix 4: ΔHf°(Ni(s)) = 0 kJ/mol (since it is an element in its standard state) S°(Ni(s)) = 29.9 J/Kmol ΔHf°(CO(g)) = -110.5 kJ/mol S°(CO(g)) = 197.7 J/Kmol Using these values, we can calculate ΔH° and ΔS° for the reaction: ΔH° = ΔH°(products) - ΔH°(reactants) = (ΔHf°(Ni(CO)_4(g))) - (ΔHf°(Ni(s)) + 4 * ΔHf°(CO(g))) = (-607 kJ/mol) - (0 kJ/mol + 4 * (-110.5 kJ/mol)) = -607 kJ/mol + 442 kJ/mol = -165 kJ/mol ΔS° = S°(products) - S°(reactants) = (S°(Ni(CO)_4(g))) - (S°(Ni(s)) + 4 * S°(CO(g))) = (417 J/Kmol) - (29.9 J/Kmol + 4 * 197.7 J/Kmol) = 417 J/Kmol - 821 J/Kmol = -404 J/Kmol

Calculating the temperature at which ΔG° = 0

We need to find the temperature at which ΔG° = 0, while assuming that ΔH° and ΔS° do not depend on temperature. We know that ΔG° = ΔH° - TΔS° Since ΔG° = 0, 0 = -165 kJ/mol - T(-404 J/Kmol) Convert kJ to J for the calculation, 0 = -165000 J/mol + T(404 J/Kmol) Now, we can solve for T: T = 165000 J/mol / 404 J/Kmol T ≈ 408 K So, the temperature at which ΔG°=0 for the given reaction is approximately 408 K.

Key concepts

Entropy in Chemical Reactions

When considering the entropy in chemical reactions, we look at the disorder or randomness of the particles involved. Entropy, denoted by the symbol S, is a fundamental concept in thermodynamics that helps us predict how a chemical reaction will proceed.

In a situation where a solid reacts with a gas to form another gas, as in the Mond process, there's a change in the number of moles of gas. Entropy tends to increase when the number of gaseous molecules increases on going from reactants to products, making the system more disordered. However, in the Mond process, which involves the reaction of nickel solid with carbon monoxide gas to form nickel tetracarbonyl gas, the entropy actually decreases because the overall number of gaseous particles decreases from four moles to one mole.

Thus, we can predict that the change in entropy () for the reaction would have a negative sign, indicating a decrease in disorder as the reaction proceeds to form nickel tetracarbonyl.

Thermodynamics of the Mond Process

The thermodynamics of the Mond process can be analyzed by looking at its key parameters: enthalpy () and entropy (), which together determine the free energy change (). The Mond process is an intriguing balance of these thermodynamic factors where nickel solid reacts with carbon monoxide to produce nickel tetracarbonyl.

The reaction's enthalpy is the measure of heat absorbed or released. The process is designed to be exothermic, where the formation of the product releases heat, resulting in a negative . This release of heat means that energy is being transferred to the surroundings, consequently increasing the entropy of the surroundings ().

The spontaneity of any reaction depends on the Gibbs free energy equation: . In the Mond process, the negative enthalpy helps drive the reaction forward. Because the reaction also has a negative , it becomes increasingly spontaneous at higher temperatures, where the Tterm outweighs the negative impact of the decreased system entropy.

Temperature Dependence of Spontaneity

The spontaneity of a chemical reaction is not just a factor of enthalpy and entropy but also of temperature. The temperature dependence of spontaneity can be understood through the Gibbs free energy equation ( = - T). This equation highlights that the temperature, T, acts as a lever, amplifying the effect of the entropy change on the reaction's spontaneity.

In the context of the Mond process, the negative indicates a decrease in entropy, generally signifying a non-spontaneous reaction. However, this changes when we consider the reaction at high temperatures. As temperature increases, the value of Tterm increases, possibly overcoming the negative value of . This transformation allows for the possibility of the reaction becoming spontaneous at high temperatures, as shown in the exercise when calculating the temperature at which = 0.

Therefore, a higher temperature compensates for the negative entropy change, allowing the Mond process of refining nickel to be spontaneous under sufficiently high-temperature conditions.