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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 Ni(s) + 4CO(g) ⇌ Ni(CO)4(g) 

a. Without referring to Appendix 4, predict the sign of ΔSo for the preceding reaction. Explain. 
b. The spontaneity of the preceding reaction is temperature dependent. Predict the sign of ΔSsurr for this reaction. Explain. 
c. For Ni(CO)4(g), ΔHof = –607 kJ/ mol and So = 417 JK-1mol-1 at 298 K. Using these values and data in Appendix 4, calculate ΔHo and ΔSo for the preceding reaction. 
d. Calculate the temperature at which DGo = 0 (K = 1) for the preceding reaction, assuming that ΔHo and ΔSo do not depend on temperature. 
e. The first step of the Mond process involves equilibrating impure nickel with CO(g) and Ni(CO)4(g) at about 50oC. The purpose of this step is to convert as much nickel as possible into the gas phase. Calculate the equilibrium constant for the preceding reaction at 50oC. 
f. In the second step of the Mond process, the gaseous Ni(CO)4 is isolated and heated at 227oC. 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 227oC. 
g. Why is temperature increased for the second step of the Mond process? 
h. The Mond process relies on the volatility of Ni(CO)4 for its success. Only pressures and temperatures at which Ni(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 152oC. Estimate the maximum pressure of Ni(CO)4(g) that can be attained before the gas will liquefy at 152oC. The boiling point for Ni(CO)4 is 42oC, and the enthalpy of vaporization is 29.0 kJ/mol. [Hint: The phase- change reaction and the corresponding equilibrium expression are Ni(CO)4(l) ⇌ Ni(CO) 4(g) K = PNi(CO)4 Ni(CO)4(g) will liquefy when the pressure of Ni(CO)4 is greater than the K value.]

asked in Chemistry by anonymous

1 Answer

Degree's high and low.
answered by Shed (5,220 points)