The following reaction was performed in a sealed vessel at 772 ∘C : H2(g)+I2(g)⇌2HI(g) Initially, only H2 and I2 were present at concentrations of [H2]=3.90M and [I2]=2.95M . The equilibrium concentration of I2 is 0.0500 M . What is the equilibrium constant, Kc, for the reaction at this temperature?

The value of for the given equation is 160.2

Explanation:

We are given:

Initial concentration of hydrogen gas = 3.90 M

Initial concentration of iodine gas = 2.95 M

Equilibrium concentration of iodine gas = 0.0500 M

The given chemical equation follows:

Initial:          3.90     2.95

At eqllm:   3.90-x    2.95-x      2x

Evaluating the value of 'x':

So, equilibrium concentration of hydrogen gas = (3.95 - x) = (3.95 - 2.9) = 1.05 M

Equilibrium concentration of HI gas = x = 2.9 M

The expression of for above equation follows:

Putting values in above equation, we get:

Hence, the value of for the given equation is 160.2

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