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.

The equilibrium constant for the given equation is

Explanation:

We are given:

Total pressure in the container = 2.25 atm

The given chemical equation follows:

Initial:              -

At eqllm:         -                   x              2x

Evaluating the value of 'x'

The expression of for above equation follows:

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:

Hence, the equilibrium constant for the given equation is

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole