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 $



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