Question

A sample of solid NH4NO3 is placed in an empty container. It decomposes according to the following reaction: NH4NO3(s) LaTeX: \Leftrightarrow⇔ N2O(g) + 2H2O(g) At equilibrium, the total pressure in the container is 2.25 atm. Calculate Kp.

Answer 1

Answer: The equilibrium constant for the given equation is
`1.69`

Step-by-step explanation:

We are given:

Total pressure in the container = 2.25 atm

The given chemical equation follows:

`NH_4NO_3(s)\rightleftharpoons N_2O(g)+2H_2O(g)`

Initial: -

At eqllm: - x 2x

Evaluating the value of 'x'

`\Rightarrow (x+2x)=2.25\\\\x=0.75`

The expression of
`K_p` for above equation follows:

`K_p=p_(N_2O)* (p_(H_2O))^2`

The partial pressure of pure solids and pure liquids are taken as 1 in the equilibrium constant expression.

Putting values in above expression, we get:

`K_p=(0.75)* (2* 0.75)^2\\\\K_p=1.69`

Hence, the equilibrium constant for the given equation is
`1.69`