Determine the equilibrium partial pressure of NH3 in a reaction vessel that initially contained 0.900 atm N2 and 0.500 atm H2 at 648 K.

Question

The value of K_p for the reaction below is is 4.30 × 10^–4 at 648 K.

3H2​(g)+N2​(g)2NH3​(g)

Determine the equilibrium partial pressure of NH₃ in a reaction vessel that initially contained 0.900 atm N₂ and 0.500 atm H₂ at 648 K. 

Answer 1

3H₂​(g) + N₂​(g) <==> 2NH₃​(g)

0.500....0.900..............0.............Initial

-3x..........-x...............+2x...........Change

0.5-3x....0.9-x...........2x............Equilibrium

Kp = (NH₃)² / (H₂)³(N₂)

4.3x10^-4 = (2x)² / (0.5-3x)³(0.9-x)

Since Kp is small (10^-4) we can assume x is small and ignore it in the denominator, simplifying to...

4.3x10^-4 = (2x)² / (0.5)³(0.9)

4.3x10^-4 = 4x² / 0.1125

4x² = 4.84x10^-5

x² = 1.21x10^-5

x = 3.48x10^-3 (note: this is small compared to 0.5 and 0.9 so above assumption was valid)

Equilibrium partial pressure of NH₃ = 2x

Equilibrium partial pressure of NH₃ = 6.96x10^-3 atm