suitable at
At T=800?C=1073K, lnPO2 =-22.2, PO2 =2.28?10-10 atm. At 800?C, if the equilibrium is reached, nitrogen can be of high purity level. However, at this high temperature , particles of Cu will weld together to reduce effective work surface. So it is not suitable to use this high temperature in purification either.
(c ) The equilibrium oxygen pressure remains the same when the total pressure increases, which means a higher purity level of N2 . 5. The solubility of hydrogen(PH2 = 1 atm ) in liquid copper at 1200?C 7.34cm3(STP) per 100g of copper. Hydrogen in copper exists in monatomic form. (a) Write the chemical equation for the dissolution of H2 in copper; (b) What level of vacuum(atm) must be drown over a copper melt at 1200?C to reduce its hydrogen content to 0.1 cm3 (STP) per 100g? (c) A 100g melt of copper at 1200?C contains 0.5 cm3(STP) of H2. Argon is bubbled through the melt slowly so that each bubble equilibrates with the melt. How much argon must be bubbled through the melt to reduce the H2 content to 0.1 cm3(STP) per 100g ? Note: STP means standard temperature and pressure(298K and 1 atm). (5.5, p145) Solution: (a) H2(g) = 2H (b) ?H??KP
P?1atm,?H??7.34cm1a212H2H23/100gCu
Ka2
1is a constant,
1/2PH[H]'0.1[H][H]''1/2?'1/2?(PH2)?2??0.0136 1/2[H]7.34PH2(PH2)'(PH2)?0.00019atm?18.56?10130?18.8Pa (c ) The amount of H2 needed to be brought out by Ar is:
P?V10130?(0.5?0.1)?10?6n???1.6?10?6mol RT8.314?298This amount of H2 is in equilibrium with the melt in the bubble, ie. The partial pressure of H2 in the bubbles is 18.8Pa.
'?2?PH2Vbubble?nRT?4.05?10
3P'H2?4.05?10/18.8?0.00215m?2.15L?22.15L Ar is needed to be bubbled into the melt.
6. The following equilibrium data have been determined for the reaction: (a) (b) (c)
(d) Plot the data using appropriate axes and find ?Ho, K and ?Go at 1000K;
(e) Will an atmosphere of 15%CO2, 5%CO, and 80%N2 N2 oxidize
NiO(s)?CO(g)?Ni(s)?CO2(g)T(?C) K?10-3 663 4.535 716 3.323 754 2.554 793 2.037 852 1.577 nickel at 1000K? (5.6, p145) Solution: (a)
?Ho1 dlnKa??d()RT Plot
lnKa~1/T
8.68.48.28.0 Kduishu Linear Fit of Data1_KduishulnKa7.87.67.47.20.880.900.920.940.960.981.00-3lnKa =2.01+6003(1/T)1.021.041.061.081/T, 10.
dlnKa?Ho???6003??Ho??R?6003??49909JdTR
At T=1000K, lnKa =8.01, Ka = 3010
o?G1000??RTlnKa??8.314?1000?8.01?66600J?66.6kJ
(b)?G??Go?RTlnJ??RTlnKa?RTlnJ?RTln15%?3?Ka5%JKa
J?So the atmosphere will oxidize Ni.
7. At 1 atm pressure and 1750?C, 100 g of iron dissolve 35cm3 (STP) of nitrogen. Under the same conditions, 100 g of iron dissolves 35 cm3 of hydrogen. Argon is insoluble in molten iron. How much gas will 100 g of iron dissolve at 1750?C and 760 mm pressure under an atmosphere that consists of: (a) 50% nitrogen and 50% hydrogen? (b) 50% argon and 50% hydrogen? (c) 33% nitrogen, 33 hydrogen, and 34 argon? (5.7, p145)
Solution: N2 =2N, H2 = 2H
2, ?N??Ka,2NPN22 ， ?H??Ka,2HPH2
1111For N2 dissolving :
[N][N]'?'1/21/2PN(PN2)2
[H][H]'?'1/21/2For H2 dissolving：PH(PH2)2
（a）For dissolving N2, PN2 = 1 atm, [N]=35cm3/100g melt,
'1/2(PN[N]2)[N]??35?(0.5)1/2?24.75cm3/100gmelt1/2PN2‘ similarly: [H] =24.75cm/100g melt
’3
total gas : [H]?＋[N]? = 49.5 cm3/100g melt (b) [H]? =24.75 cm3/100g melt
(c ) [H]?＋[N]? = [N](0.33)1/2 /1+[H](0.33)1/2 /1=20.10+20.10 = 40.2cm3/100g melt
8. Solid silicon in contact with solid silicon dioxide is to be heated to a temperature of 1100 K in a vaccum furnace. The two solid phases are not soluble in each other, but is known that silicon and silicon dioxide can react to form gaseous silicon monoxide. For the reaction:
Si(s)?SiO2?2SiO(g)the Gibbs free
?Go?667000?25.0TlnT?510Tenergy change (J)
is . (a) Calculate the equilibrium pressure of SiO gas at 1100K; (b) For the reaction above, calculate ?Ho and ?So at 1100K; (c) Using the Ellingham chart (Figure 5.7), estimate the pressure of oxygen (O2) in equilibrium
