Question

8) 4.00 mol HI are placed in an evacuated 5.00 L flask and then heated to 800 K. The system is allowed to reach equilibrium. Determine the equilibrium concentration of all species. What kind of problem do you think this is

Answer 1

Answer: Concentration of $H_2$ at equilibrium = 0.08 M

Concentration of $I_2$ at equilibrium =0.08 M

Concentration of $Hl$ = 0.64M

Explanation:

Moles of  $HI$ = 4.00 mole

Volume of solution = 5.00 L

Initial concentration of $HI$ =$\frac{moles}{Volume}=\frac{4.00mol}{5.00L}=0.8M$

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_c$

For the given chemical reaction:

         $2HI(g)\rightarrow H_2(g)+I_2(g)$

Initial conc.      0.8 M           0 M    0 M

   At eqm. conc.  (0.8-2x) M   (x) M   (x) M    

The expression for $K_c$ is written as:

$K_c=\frac{[H_2]^1[I_2]^1}{[HI]^2}$

$0.016=\frac{(x)\times (x)}{(0.8-2x)}$

$x=0.08M$

Concentration of $H_2$ at equilibrium = x M = 0.08 M

Concentration of $I_2$ at equilibrium = x M = 0.08 M

Concentration of $Hl$ = (0.8-2x) = $(0.8-2\times 0.08)M= 0.64M$