Question

Ammonium iodide dissociates reversibly to ammonia and hydrogen iodide. NH4I (s) ⇌ NH3 (g) + HI (g) At 400 ºC, Kp = 0.215. Calculate the partial pressure of ammonia at equilibrium when a sufficient quantity of ammonium iodide is heated to 400 ºC. Complete the ICE box below as part of your answer.

A. 0.103 atmB. 0.215 atmC. 0.232 atmD. 0.464 atmE. 2.00 atm

Answer 1

Answer: The partial pressure of ammonia is 0.464 atm

Step-by-step explanation:

For the given chemical equation:

`NH_4I(s)\rightleftharpoons NH_3(g)+HI(g)`

Initial: 1

At eqllm: 1-x x x

The expression for
`K_p` for the following equation is:

`K_p=p_(NH_3)* p_(HI)`

The partial pressures of solids and liquids are taken as 1 in the equilibrium expression.

We are given:

`K_p=0.215`

Putting values in above equation, we get:

`0.215=x* x\\\\x=0.464atm`

Hence, the partial pressure of ammonia is 0.464 atm