Question

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 0.20 atm and H2 is 0.15 atm?

Answer 1

"
`6.7* 10^(-4) \ atm`" is the right answer.

Step-by-step explanation:

Given:

Partial pressure of
`N_2`,

= 0.20 atm

Partial pressure of
`H_2`,

= 0.15 atm

`K_p = 1.5* 10^3` at
`400^(\circ) C`

As we know,

⇒
`K_p = (pN_2* pH_2^3)/(pNH_3^2)`

By putting the values, we get

`1.5* 10^3=(0.20* (0.15)^3)/(pNH_3^2)`

`pNH_3^2 = (0.000675)/(1.5* 10^3)`

`=6.7* 10^(-4) \ atm`