Question

Suppose a 250.mL flask is filled with 0.70mol of H2 and 1.6mol of HCl. The following reaction becomes possible:

H2 (g) + Cl2 (g) <---> 2HCl (g)

The equilibrium constant K for this reaction is 0.262 at the temperature of the flask. Calculate the equilibrium molarity of HCl. Round your answer to two decimal places.

Answer 1

Answer:

6.97 M

Explanation:

The reaction that takes place is

H₂(g) + Cl₂(g) ↔ 2HCl (g)

With a equilibrium constant ke =  $\frac{[HCl]^2}{[H_{2}][Cl_{2}]}$ = 0.262

The original concentrations for each species are:

250 mL ⇒ 250 / 1000 = 0.25 L

• H₂ = 0.70 mol/0.25 L = 2.8 M

• Cl₂ = 0 M

• HCl = 1.6 mol/0.25 L = 6.4 M

The ICE table thus is:

                      H₂(g) + Cl₂(g) ↔ 2HCl (g)

Initial                 2.8      0           6.4

Change            -x          -x             +2x

Equilibium        2.8-x      0-x        6.4 + 2x

• 0.262 = $\frac{(6.4+2x)^2}{(2.8-x)(-x)}$

• x = 0.285  

So the equilibrium molarity of HCl is 6.4 + 2*0.285 = 6.97 M