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 $

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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. $


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