Consider the reaction: Al 2 O 3 (s) + 6HF(g) AlF 3 (g) + H 2 O(g) ! Given the fo
ID: 840994 • Letter: C
Question
Consider the reaction: Al 2 O 3 (s) + 6HF(g) AlF 3 (g) + H 2 O(g) ! Given the following table of thermodynamic data at 298K:
Substance Delta H S(J/K*Mol)
Al2 -1669.3 50.99
HF(g) -271.6 173.5
AlF -1209.3 276.7
H2 -241.8 188.7
What is the value of Kc for this reaction at 298K?
PLease include all steps
Explanation / Answer
Al 2 O 3 (s) + 6HF(g) ----> 2 AlF 3 (g) + 3 H 2 O(g)
Delta H for the reaction
= 3 * -241.8 + 2* -1209.3 - ( 6 * -271.6 - 1669.3)
= 154.9 kJ = 154900 J
Delta S = 3*188.7 + 2*276.7 - ( 6*173.5 + 50.99)
= 27.51 J
Delta G = Delta H - T*Delta S
= 154900 - 298*27.51
= 146702.02 J
Kc = e^ (-Delta G /( R T )
Kc = e^ (-146702.02 / (8.314 * 298))
Kc = 1.925 x 10^-26