Chemistry, asked by dhruvbadaya1, 11 months ago

A mixture containing 8.07 moles of hydrogen and 9.08 moles of iodine was heated at 448° C till equilibrium was attained when 13.38 moles of hydrogen iodide was obtained. Find the degree of disassociation.

Answers

Answered by devianand787
2

Answer:

3.2 moles of HI(g) were heated in a sealed bulb at 444°C till the equilibrium was reached. It's degree of dissociation was found to be 20%. Calculate the number of moles of hydrogen iodide, hydrogen and iodine present at the equilibrium point and determine the value of equilibrium constant for the reaction 

 2HI (g)  (g). Considering the volume of the container 1 L.2(g) + I2  

Answered by Anonymous
0

Answer:

Explanation:

At a given temperature, the product of concentration of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentration of the reactants raised to their individual stoichiometric coefficients has a constant value.

- wherein

aA+bB\rightleftharpoons cC+dD

K_{c}=\frac{[C]^{c\:[D]^{d}}}{[A]^{a}\:[B]^{b}}

[A],\:[B],\:[C]\:[D]

are equilibrium concentration

 

The reaction undergoes as follows:

                                         2HI\:\rightleftharpoons \:H_{2}+I_{2}

At  t=0                                3.2                0         0

At equilibrium               3.2(1-\propto )   3.2\propto  3.2\propto

\propto\:=\:0.22\:(given)

Similar questions