A closed system initially containing 1×10^-3 hydrogen 2×10^-3M iodine at 448 degree Celsius and is allowed to reach equilibrium. The concentration of the equilibrium shows concentration of hydrogen iodude as 1.87×10^-3M.Calculate kc at this temperature

Kc = 50.5

Explanation:

We determine the reaction:

H₂  +  I₂   ⇄   2HI

Initially we have 0.001 molesof H₂

and 0.002 moles of I₂

If we have produced 0.00187 moles of HI in the equilibrium we have to know, how many moles of I₂ and H₂, have reacted.

           H₂     +      I₂      ⇄   2HI

In:     0.001       0.002           -

R:       x                 x                2x

Eq:  0.001-x    0.002-x      0.00187  

x = 0.00187/2 = 9.35×10⁻⁴ moles that have reacted

So in the equilibrium we have:

0.001 - 9.35×10⁻⁴ = 6.5×10⁻⁵  moles of H₂

0.002 - 9.35×10⁻⁴ = 1.065×10⁻³ moles of I₂

Expression for Kc is =  (HI)² / (H₂) . (I₂)

0.00187 ² /  6.5×10⁻⁵ . 1.065×10⁻³ = 50.5