Chemical Thermodynamics Question 382
Question: The enthalpies of formation of $ Al_2O_3 $ and $ Cr_2O_3 $ are $ -1596kJ $ and $ -1134kJ $ respectively. $ \Delta H $ for the reaction $ 2Al+Cr_2O_3\to 2Cr+Al_2O_3 $ is
Options:
A) $ -2730kJ $
B) $ -462,kJ $
C) $ -1365,kJ $
D) $ +2730,kJ $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ 2Al+\frac{3}{2}O_2\to Al_2O_3,\Delta H=-1596kJ $ …..(i) $ 2Cr+\frac{3}{2}O_2\to Cr_2O_3,,\Delta H=-~1134kJ $ …….(ii) By (i)-(ii) $ 2Al+Cr_2O_3\to 2Cr+Al_2O_3,\Delta H=-462kJ. $
