Chemical Thermodynamics Question 494
Question: The enthalpies of formation of $ Al_2O_3 $ and $ Cr_2O_3 $ 1596 kJ and - 1134 kJ respectively. $ \Delta H $ for the reaction $ 2Al+Cr_2O_3\to 2Cr+Al_2O_3 $
Options:
A) - 2730 kJ
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. $
