The following thermochemical equations represent combustion of ammonia

1.	The following thermochemical equations represent combustion of ammonia and hydrogen: 4NH3(g) + 3O2(g) → 6H2O(l) + 2N2(g); ΔH = -1516 kJ 2H2(g) + O2(g) → 2H2O(l); ΔH = -572kJ Calculate enthalpy of formation of ammonia.
Answer» Required equation is 1/2N_2(g) + 3/2H_2(g) → NH_3(g) ΔH = ? 4NH_3(g) + 3O_2(g) → 2N_2(g) + 6H₂O_2(l)ΔH = -1516 kJ …(1) 2H_2(g) + O_2(g) → 2H₂O_(l) ΔH = -572 kJ …(2) Reserving equation (1) and dividing by 4 we get: 1/2N_2(g) + 2H₂O_2(l) → NH_3(g)+ 3/4O_2(g) ΔH = +379 kJ …(3) Multiplying equation (2), by 3/4 we get, 3/2H_2(g) + 3/4O₂ → 3/2H₂O_(l) ΔH = -572 × 3/4 = -429kJ ....(4) Adding (3) and (4) we get , 1/2N_2(g) + 3/2H_2(g) → NH_3(g) ;ΔH = +379 – 429 = -50 kJ/mol

The following thermochemical equations represent combustion of ammonia and hydrogen: 4NH3(g) + 3O2(g) → 6H2O(l) + 2N2(g); ΔH = -1516 kJ 2H2(g) + O2(g) → 2H2O(l); ΔH = -572kJ Calculate enthalpy of formation of ammonia.

Answer»

Required equation is 1/2N_2(g) + 3/2H_2(g) → NH_3(g) ΔH = ?

4NH_3(g) + 3O_2(g) → 2N_2(g) + 6H₂O_2(l)ΔH = -1516 kJ …(1)

2H_2(g) + O_2(g) → 2H₂O_(l) ΔH = -572 kJ …(2)

Reserving equation (1) and dividing by 4 we get:

1/2N_2(g) + 2H₂O_2(l) → NH_3(g)+ 3/4O_2(g) ΔH = +379 kJ …(3)

Multiplying equation (2), by 3/4 we get,

3/2H_2(g) + 3/4O₂ → 3/2H₂O_(l) ΔH = -572 × 3/4 = -429kJ ....(4)

Adding (3) and (4) we get ,

1/2N_2(g) + 3/2H_2(g) → NH_3(g) ;ΔH = +379 – 429 = -50 kJ/mol

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The following thermochemical equations represent combustion of ammonia and hydrogen: 4NH3(g) + 3O2(g) → 6H2O(l) + 2N2(g); ΔH = -1516 kJ 2H2(g) + O2(g) → 2H2O(l); ΔH = -572kJ Calculate enthalpy of formation of ammonia.

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