Answer to Question #140874 in General Chemistry for Aisha

Question

Calculate Delta H for the reaction 4 NH3 (g) —-> 4 NO (g) —-> + H2O (g)

Answer 1

4NH₃ (g) + 5O₂ (g) → 4NO (g) + 6H₂O (g)

N₂ (g) + O₂ (g) → 2NO (g) delta H = -180.5 kJ (1)

N₂ (g) + 3H₂ (g) → 2NH₃ (g) delta H = -91.8 kJ (2)

2H₂ + O₂ → 2H₂O (g) delta H = -483.6 kJ (3)

Multiplying equation (1) by 2 to obtain

2N₂ (g) + 2O₂ (g) → 4NO (g) delta H $= 2 \times (-180.5\;kJ) = -361.0 \;kJ$

Multiplying equation (2) by 2 and then reversing it to obtain

4NH₃ (g) → 2N₂ (g) + 6H₂ (g) delta H $= 2 \times 91.8\;kJ = 183.6 \;kJ$

Multiplying equation (3) by 3 to obtain

6H₂ + 3O₂ → 6H₂O (g) delta H $= 2 \times (-483.6 \;kJ) = -967.2 \;kJ$

Adding the last three equations to obtain

4NH₃ (g) + 5O₂ (g) → 4NO (g) + 6H₂O (g) delta H = -361.0 kJ + 183.6 kJ – 967.2 kJ = -1144.6 kJ

Answer: -1144.6 kJ