The equilibrium constant, kc, for the following reaction is 9.49×10-4 at 533 k. cocl2(g) co(g) + cl2(g)when a sufficiently large sample of cocl2(g) is introduced into an evacuated vessel at 533 k, the equilibrium concentration of cl2(g) is found to be 0.190 m. calculate the concentration of cocl2 in the equilibrium mixture. m

38.0 M

Explanation:

Let's consider the following reaction.

COCl₂(g) ⇄ CO(g) + Cl₂(g)

To find the concentrations at equilibrium we will use an ICE chart. We recognize 3 stages (Initial, Change, and Equilibrium) and complete each row with concentrations or changes in concentrations. Let's call unknown concentrations x and y.

      COCl₂(g) ⇄ CO(g) + Cl₂(g)

I           y                0          0

C        -x                +x        +x

E        y-x               x           x

The equilibrium concentration of Cl₂(g) is 0.190 M, that is, x = 0.190 M

The equilibrium constant (Kc) is:

The concentration of COCl₂ in the equilibrium mixture is

y - x = 38.2 M - 0.190 M = 38.0 M

answer: -

reaction:

na + cl ⇌ nacl

balanced chemical equation:

na + 2cl ⇌ nacl2

sodium + clorine goes to form sodium chloride (table salt).

the product is nacl2.

and the reactants are na and cl.

(note: if you don’t know what ⇌ means, read below: -

⇌(double head arrow) means reversible reaction).

i'm pretty sure it is letter c.) car