Question

N2(g) + 3H2(g) (eqilibrium symbol) 2NH3(g), Kp = 1.47 x 10–5 at 508 oC.

At equilibrium at 508 oC, PH2 = 0.588 atm and PN2 = 0.645 atm. What is the partial pressure of ammonia at equilibrium?

Answer 1

The partial pressure of ammonia at the equilibrium is 0.00139 atm

Step-by-step explanation:

Step 1: Data given

Kp = 1.47 * 10^-5

Temperature = 508 °C = 781 K

Partial pressure of H2 = 0.588 atm

Partial pressure of N2 = 0.645 atm

Step 2: The balanced equation

N2(g) + 3H2(g) ⇆ 2NH3(g)

Step 3: Calculate the partial pressure of ammonia

Kp = (pNH3)² / (pH2)³*(pN2)

⇒with Kp = 1.47 * 10^-5

⇒with pNH3 = the partial pressure of ammonia = TO BE DETERMINED

⇒with pH2 = the partial pressure of H2 = 0.588 atm

⇒with pN2 = the partial pressure of N2 = 0.645 atm

1.47 *10^-5 = (pNH3)² / ((0.588³)*(0.645))

(pNH3)² = 1.93 * 10^-6

pNH3 = 0.00139 atm

The partial pressure of ammonia at the equilibrium is 0.00139 atm