Question

an adiabatic vessel contains 1000g of liquid aluminum at 700C. Calculate the mass of Cr2O3 at room temperature, which, when added to the liquid aluminum (with whichreacts to form Cr and Al3O2) raises the temperature of the resulting mixture of Al2O3, Cr2O3 and Cr to 1600k.

Answer 1

Reaction- 2Al Cr,0, 2rAl,0, **All aluminum reacts with Cr2O3 Thermodynamic Data;

2c.is, +y,,eGas) Δ",-1,134. 700., H10,700J c, 20.67+1238x0TJImole.K c,-31.76 JImole.K e,-106.6+17.78x10T-28.530 Jlmole.k c. 24,43 +9.87x10"Т-368x10,T" J/mole.K e.-ו19.37 + 9.30x10-, T-1 5.65x10, 7" J Imole.K Cİs) , Using the thermodynamic data above one can solve for the thermodynamic values for the reaction 2A'm +CrO.-2Cİs) + A40, as: ΔΗΝ.--1,676, 700 + 1,134,700 541,000 J c-=-5.25 + 3.46x10"T-2024x10?T-3 J1mole.K and 24lyt Cr,o, ass Solving fr Mmx for the reaction 2Aha-crt-2G0) + Alo, : 3.46x0 -541000-.25(934-298934 2-1(934:-298*) +20.24x10,넒-1)-2(10700) -27.43(973-934)+(2832 m)(973°«W)-20.24x10,( 1) 974 934

Heat required to raise temperature of 1 mol Cr,0) to 973 K: Mini 1'937(973-298)+(умгіо )(973:-298) +1565x102(m-298) 930xlo +15.65xlo 973 298 ΔΙ¡mx 80,920 J Therefore, total heat available at 973K is Aflmx-569, l 16 + 80, 920 =-488, 197 J Heat required to raise Cr, Al2O, and CrO to 1600K: 1mol Cr 24.43(1600-973)-937101600* (9.87x10" )(160,--973°)+ 368r10° 1600 973 1400A Imol Al:0 ionspen-enr(뷔ieu.wv).assar(i늚-司 17.78r10 600 -973+28.53x10 AH,x-106.6(1600-973)(16 M80.031 w, 1600 973

Imol CrQ A1,000K 119.37(1600-973)+19.30,10-016002-9732)-1565ro'(i.-1 1600 973 Therefore, the heat available at 1600K to increase the temperature of excess Cr O is: ΔΙ1wax. 488.1972.23, 130-80.031. 361.906 J Number of excess moles of Cr 0s heated: 361,906 excess moles of C1,716 4.43 543moles 151.9904 g/mol 825.3 gof Cr0 per 53.96g of Al0 15.3kg of Cr0 per k of Al O