Question

At a certain temperature, the equilibrium constant, K c , for this reaction is 53.3. H 2 ( g ) + I 2 ( g ) − ⇀ ↽ − 2 HI ( g ) K c = 53.3 At this temperature, 0.800 mol H 2 and 0.800 mol I 2 were placed in a 1.00 L container to react. What concentration of HI is present at equilibrium?

Answer 1

Answer: Concentration of HI is present at equilibrium is 1.26 M

Explanation:

Moles of  $H_2$ = 0.800 mole

Moles of  $I_2$ = 0.800 mole

Volume of solution = 1.00 L

Initial concentration of $H_2$ = 0.800 M

Initial concentration of $I_2$ = 0.800 M

The given balanced equilibrium reaction is,

                          $H_2(g)+I_2(g)\rightleftharpoons 2Hl(g)$

Initial conc.         0.800     0.800             0

At eqm. conc.     (0.800-x) M   (0.800-x) M   (2x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[HI]^2}{[H_2]\times [l_2]}$

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

$53.3=\frac{(2x)^2}{(0.800-x)^2}$

By solving the term 'x', we get :

x = 0.63

Concentration of $HI$ at equilibrium = 2 x = $2\times 0.63=1.26$

Concentration of HI is present at equilibrium is 1.26 M