Question

a at equilibrium when N2 is 0.10 atm and H2 is 0.15 atm? The decomposition of ammonia is: 2 NH3(g) ? N2(g) + 3 H2(g). If Kp is 1.5 × 103 at 400°C, what is the partial pressure of ammonia at equilibrium when N2 is 0.10 atm and H2 is 0.15 atm? 4.7 × 10-4 atm 2.2 × 10-7 atm 2.1 × 103 atm 4.4 × 106 atm

Answer 1

Given:

Kp = 1.5*10³

partial pressure pN2 = 0.10 atm

partial pressure pH2 = 0.15 atm

To determine:

Partial pressure pNH3 at equilibrium

Explanation:

The decomposition reaction is:-

2NH3(g) ↔N2(g) + 3H2(g)

Kp = [pH2]³[pN2]/[pNH3]²

pNH3 =√ [(pH2)³(pN2)/Kp]

pNH3 = √(0.15)³(0.10)/1.5*10³ = 4.74*10⁻⁴ atm

Ans (a)