Chemical Thermodynamics Question 492
Question: The standard heat of combustion of Al is - 837.8 kJ/mol at $ 25{}^\circ C $ . Which of the following releases 250 kcal of heat-
Options:
A) The reaction of 0.624 mol of Al
B) The formation of 0.624 mol of $ Al_2O_3 $
C) The reaction of 0.312 mol of $ Al $
D) The formation of 0.150 mol of $ Al_2O_3 $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ Al+\frac{3}{4}O_2\xrightarrow{{}}\frac{1}{2}Al_2O_3;\Delta H=-837.8kJ $
Realize energy $ =250kcal=250\times 4.2=1050kJ $
Since 837.8 kJ energy is realized in formation of 0.5 mol $ Al_2O_3, $ therefore 1050 kJ energy realized $ \frac{0.5}{837.8}\times 1050=0.624,mol $
