A) - 2730 kJ
B) - 462 kJ
C) - 1365 kJ
D) + 2730 kJ
Correct Answer: B
Solution :
[b] \[2Al+\frac{3}{2}{{O}_{2}}\to A{{l}_{2}}{{O}_{3}},\Delta H=-1596\text{ }kJ\] | ...(i) |
\[2Cr+\frac{3}{2}{{O}_{2}}\to C{{r}_{2}}{{O}_{3}},\Delta H=-1134\text{ }kJ\] | ...(ii) |
By (i) - (ii) | |
\[2Al+C{{r}_{2}}{{O}_{3}}\to 2Cr+A{{l}_{2}}{{O}_{3}},\Delta H=-462kJ.\] |
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