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

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.

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

3H2​(g)+N2​(g)2NH3​(g)

Answer 1

3H₂​(g) + N₂ ​(g)<==> 2NH₃​(g)

0500.....0.900.............0...........Initial

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

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

Kp = (NH₃)² / (H₂)³(N₂)

4.30x10^-4 = (2x)2 / (0.5 - 3x)(0.9 - x)

4.30x10^-4 = 4x² / 3x² - 3.2x + 0.45

Use quadratic to solve...

x = 0.00680 atm

Equilibrium partial pressure of NH3 = 2x = 2 x 0.00680 atm = 0.0136 atm