The following thermochemical equations represent combustion of ammonia and hydrogen.
`4NH_(3)(g) + 3O_(2)(g) rarr 6H_(2)O(l) + 2N_(2)(g), Delta H = -1516 kJ`
`2H_(2)(g) + O_(2)(g) rarr 2H_(2)O(l) , Delta H = -572 kJ`
Calculate enthelpy of formation of ammonia.

Explanation

Required equations is `(1)/(2)N_(2(g)) + (3)/(2)H_(2(g)) rarr NH_(3(g)) " "Delta H = ?`
`4NH_(3(g)) + 3O_(2(g)) rarr 2N_(2(g)) + 6H_(2)O_((l)) " "Delta H = -1516 kJ` …(1)
`2H_(2(g)) + O_(2(g)) rarr 2H_(2)O_((l)) " "Delta H = -572 kJ` ….(2)
Multiplying equation (2), by `(3)/(4)` we get,
`(3)/(2)H_(2(g)) + (3)/(4)O_(2(g)) rarr (3)/(2)H_(2)O_((l)) " "Delta H = -572 xx (3)/(4) = -429 kJ` ...(4)
Adding (3) and (4) we get,
`(1)/(2)N_(2(g)) + (3)/(2)H_(2(g)) rarr NH_(3(g)) " "Delta H = +379 - 429 = -50 kJ//mol`.