Thermodynamics Question 71
Question: The enthalpies of formation of $ Al _{2}O _{3} $ and $ Cr _{2}O _{3} $ are $ -1596kJ $ and $ -1134kJ $ respectively. $ \Delta H $ for the reaction $ 2Al+Cr _{2}O _{3}\to 2Cr+Al _{2}O _{3} $ is
[KCET 2003]
Options:
A) $ -2730kJ $
B) $ -462kJ $
C) $ -1365kJ $
D) $ +2730kJ $
Show Answer
Answer:
Correct Answer: B
Solution:
$ 2Al+\frac{3}{2}O _{2}\to Al _{2}O _{3},\Delta H=-1596kJ $ …(i)
$ 2Cr+\frac{3}{2}O _{2}\to Cr _{2}O _{3},\Delta H=-1134kJ $ …(ii)
By (i) . (ii) $ 2Al+Cr _{2}O _{3}\to 2Cr+Al _{2}O _{3},\Delta H=-462kJ $
