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? 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

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)

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm