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 $1.69$

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:

$K_p=p_{N_2O}\times (p_{H_2O})^2$

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:

$K_p=(0.75)\times (2\times 0.75)^2\\\\K_p=1.69$

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