J.R. S. answered • 01/25/19

Ph.D. University Professor with 10+ years Tutoring Experience

Eq. 1: 4NH_{3} + 3O_{2} ==> 6H_{2}O + 2N_{2} ∆H = =1516 kJ

Eq. 2: 2H_{2} + O_{2} ==> 2H_{2}O ∆H = -572 kJ (note, the equation as you wrote it in the question is wrong)

Reverse Eq.1 to get 6H_{2}O + 2N_{2} ==> 4NH_{3} + 3O_{2} ∆H = +1516 kJ

Multiply Eq. 2 by 3 to get 6H_{2} + 3O_{2} ==> 6H_{2}O ∆H = 3 x -572 kJ = -1716 kJ

Add the 2 equations to get 6H_{2}O + 2N_{2} + 6H_{2} + 3O_{2} ==> 4NH_{3} + 3O_{2} + 6H_{2}O ∆H =-200 kJ

Cancel to get 2N_{2 }+ 6H_{2} ==> 4NH_{3} ∆H = -200 kJ

This is the ∆H for formation of FOUR moles NH_{3}. Thus, molar ∆H = -200/4 = **-50 kJ/mole** (Ans. B)