Question

Gaseous hydrogen iodide is placed in a closed container at 425°C, where it partially decomposes to hydrogen and iodine: 2HI(g)⇌H₂(g)+I₂(g) At equilibrium it is found that [HI]= 3.51×10⁻³ M, [H₂]= 4.76×10⁻⁴ M, and [I₂]= 4.76×10⁻⁴ M.
What is the value of $K_c$ at this temperature? Express the equilibrium constant to three significant digits.

Answer 1

Answer : The value of $K_c$ at this temperature is 0.0184

Explanation : Given,

Concentration of $HI$ at equilibrium = $3.51\times 10^{-3}M$

Concentration of $H_2$ at equilibrium = $4.76\times 10^{-4}M$

Concentration of $I_2$ at equilibrium = $4.76\times 10^{-4}M$

The given equilibrium reaction is,

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

The expression of $Kc$ will be,

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

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

$K_c=\frac{(4.76\times 10^{-4})\times (4.76\times 10^{-4})}{(3.51\times 10^{-3})^2}$

$K_c=0.0184$

Therefore, the value of $K_c$ at this temperature is 0.0184