Answer:
The Kc is 1.36 (but this is not an option, may be the options are wrong, or may be I was .. Thanks!)
Explanation:
Let's think all the situation.
2 ICl(g) ⇄ I₂(g) + Cl₂(g)
Initially 0.20 - -
Initially I have only 0.20 moles of reactant, and nothing of products. In the reaction, an x amount of compound has reacted.
React x x/2 x/2
Because the ratio is 2:1, in the reaction I have the half of moles.
So in equilibrium I will have
(0.20 - x) x/2 x/2
Notice that I have the concentration in equilibrium so:
0.20 - x = 0.060
x = 0.14
So in equilibrium I have formed 0.14/2 moles of I₂ and H₂ (0.07 moles)
Finally, we have to make, the expression for Kc and remember that must to be with concentration in M (mol/L).
As we have a volume of 2L, the values must be /2
Kc = ([I₂]/2 . [H₂]/2) / ([ICl]/2)²
Kc = (0.07/2 . 0.07/2) / (0.060/2)²
Kc = 1.225x10⁻³ / 9x10⁻⁴
Kc = 1.36
