Answer:
Energy released:
[tex]\Delta H=-71.03 kJ[/tex]
Explanation:
In Haber's process Ammonia formation takes place according to equation:
[tex]N_{2}+3H_{2}\leftrightharpoons 2NH_{3}[/tex]
[tex]\Delta H=-92 kJ/mol[/tex]
According to this equation ,when 2 mole of ammonia is formed from 3 mole of nitrogen and 1 mole of oxygen then 92 kJ of heat is released.
1 mole of [tex]NH_{3}[/tex] contains 17.03 g
mass of N=14.0067 g/mol
mass of H=1.007 g/mol
mass of [tex] NH_{3}[/tex] =14.0067+3.021
mass of [tex] NH_{3}[/tex] =17.03 g/mol
So, 2 mole of [tex] NH_{3}[/tex] contains[tex] 17.03\times 2[/tex]=34.06g
Since 34.06g (or 2mol) = -92 kJ
[tex]1 g=\frac{92}{34.06}[/tex]
[tex]26.3 g=\frac{92}{34.06}\times 26.3[/tex]
this gives [tex] \Delta H=-71.03 kJ[/tex]
Hence amount of energy released when 26.3g of ammonia is formed is -71.03 kJ
