A mixture of 0.439 M H 2 , 0.317 M I 2 , and 0.877 M HI is enclosed in a vessel and heated to 430 °C. H 2 ( g ) + I 2 ( g ) − ⇀ ↽ − 2 HI ( g ) K c = 54.3 at 430 ∘ C Calculate the equilibrium concentrations of each gas at 430 ∘ C .

Answer : The concentration of at equilibrium is, 0.244 M, 0.122 M and 1.267 M respectively.

Explanation :

The given chemical reaction is:

Initial conc.       0.439      0.317       0.877

At eqm.          (0.439-x)    (0.317-x)  (0.877+2x)

As we are given:

The expression for equilibrium constant is:

Now put all the given values in this expression, we get:

x = 0.195 and x = 0.690

We are neglecting the value of x = 0.690 because equilibrium concentration can not be more than initial concentration.

Thus, the value of x = 0.195 M

The concentration of at equilibrium = (0.439-x) = (0.439-0.195) = 0.244 M

The concentration of at equilibrium = (0.317-x) = (0.317-0.195) = 0.122 M

The concentration of at equilibrium = (0.877+2x) = (0.877+2\times 0.195) = 1.267 M

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

it would be d. henry moseley

hope this !

and sorry if i'm late. have a good rest of your day! : d