A) -2730 kJ
B) -462 kJ
C) -1365 kJ
D) +2730 kJ
Correct Answer: B
Solution :
According to question \[2Al+\frac{3}{2}{{O}_{2}}\xrightarrow{{}}A{{l}_{2}}{{O}_{3}}\] ... (i) \[\Delta H=-1596\,\,kJ\] \[2Cr+\frac{3}{2}{{O}_{2}}\xrightarrow{{}}C{{r}_{2}}{{O}_{3}}\] ... (ii) \[\Delta H=-1134\,\,kJ\] Reversing eq. (ii) and adding both reactions, we get \[2Al+C{{r}_{2}}{{O}_{3}}\xrightarrow{{}}2Cr+A{{l}_{2}}{{O}_{3}}\] \[\Delta H=-462\,\,kJ\]You need to login to perform this action.
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