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O3Al(s) + 2C(s) + 302(g) → Al3(CO3)2(S) 2A1(s) + 2C(s) + 202(g) → Al2(CO2)2(s) O2Al(s) + 3C(s) + 9/2 O2(g) → Al2(CO3)3(S) O4Al(s) + 6C(s) +9 02(g) → 2 Al2(CO3)3(s) 12. Calculate AG (in kJ/mol) at 298 K when the equilibrium constant equals 4.07 x 105. 32000 kJ/mol 32.0 kJ/mol -32.0 kJ/mol 316 kJ/mol -316 kJ/mol K 86°= RTIN K eq 86°= 8.3145 (298) In 4.0 06°

O3Al(s) + 2C(s) + 302(g) → Al3(CO3)2(S) 2A1(s) + 2C(s) + 202(g) → Al2(CO2)2(s) O2Al(s) + 3C(s) + 9/2 O2(g) → Al2(CO3)3(S) O4Al(s) + 6C(s) +9 02(g) → 2 Al2(CO3)3(s) 12. Calculate AG (in kJ/mol) at 298 K when the equilibrium constant equals 4.07 x 105. 32000 kJ/mol 32.0 kJ/mol -32.0 kJ/mol 316 kJ/mol -316 kJ/mol K 86°= RTIN K eq 86°= 8.3145 (298) In 4.0 06°

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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O3Al(s) + 2C(s) + 302(g) → Al3(CO3)2(S) D2Al(s) + 2C(s) +202(g) → Al2(CO2)2(s) O2Al(s) + 3C(s) + 9/2 O2(g) → Al2(CO3)3(S) O4Al(s) + 6C(s) +9 O2(g) → 2 Al2(CO3)3(S) 12. Calculate AG (in kJ/mol) at 298 K when the equilibrium constant equals 4.07 x 105. 32000 kJ/mol 32.0 kJ/mol -32.0 kJ/mol 316 kJ/mol -316 kJ/mol K All Pages 86°= RTIN K eq 86°= 8.3145 (298) In 4.0 46° Do
Transcribed Image Text:O3Al(s) + 2C(s) + 302(g) → Al3(CO3)2(S) D2Al(s) + 2C(s) +202(g) → Al2(CO2)2(s) O2Al(s) + 3C(s) + 9/2 O2(g) → Al2(CO3)3(S) O4Al(s) + 6C(s) +9 O2(g) → 2 Al2(CO3)3(S) 12. Calculate AG (in kJ/mol) at 298 K when the equilibrium constant equals 4.07 x 105. 32000 kJ/mol 32.0 kJ/mol -32.0 kJ/mol 316 kJ/mol -316 kJ/mol K All Pages 86°= RTIN K eq 86°= 8.3145 (298) In 4.0 46° Do
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