Question

Determine AH for the following reaction, 2NH3(g) + 5/20260) - 2NO(g) + 3H2019) given the thermochemical equations below. N2(g) + O2(0) - 2NO(g) N2(g) + 3H2(g) — 2NH3(0) 2H269) + O26) - 2H206) AH = +180.8 kJ AH = -91.8 kJ AH = -483.6 kJ -1178.2 kJ -452.8 kJ -394.6 kJ -211.0 kJ +1178.2 kJ

Answer 1

From the given data.

Equation number (1)

N_{​​​​​​2}(g) + O_{​​​​​​2}(g) -------> 2NO (g) ∆H = 180.8kJ .........(1)

By reversing equation 2.

2NH₃ ​​​​​​(g) ----------> N_{​​​​​​2} (g) + 3H₂(g). ∆H = 91.8kJ ..........(2)

Multiplying 3/2 with equation number 3.

3H₂ ​​​​​​(g) + (3/2)O_{​​​​​​2} (g) -----> 3H₂O ∆H = -725.4kJ ......(3)

Adding the equations (1), (2) and (3).

N_{​​​​​​2} + O_{​​​​​​2} + 2NH₃ + 3H_{2 ⁺} (3/2)O_{​​​​​​2} -------> 2NO+N_{​​​​​​2} + 3H₂ + 3H₂O

∆H = 180.8kj + 91.8kJ - 725.6kJ

Or

Nitrogen and hydrogen are equal on both sides. Hence the cancellation of nitrogen and hydrogen from both sides.

2NH₃(g) + (5/2)O_{​​​​​​2} (g)-----> 2NO(g) + 3H₂O(g) ∆H = -452.8kJ

Hence.

Answer = -452.8kJ