The following reaction was performed in a sealed vessel at 726 ∘C :H2(g)+I2(g)⇌2HI(g)Initially, only H2 and I2 were present at concentrations of [H2]=3.00M and [I2]=2.20M . The equilibrium concentration of I2 is 0.0800 M . What is the equilibrium constant, Kc, for the reaction at this temperature?

Kc = 255

Explanation:

Step 1: Write the balanced equation

H₂(g) + I₂(g) ⇌ 2 HI(g)

Step 2: Make an ICE chart

         H₂(g)   +   I₂(g)   ⇌   2 HI(g)

I         3.00        2.20             0

C         -x             -x              +2x

E      3.00-x      2.20-x           2x

Step 3: Find the value of x

The concentration at equilibrium of I₂ is 0.0800 M. Then,

2.20 - x = 0.0800

x = 2.12

Step 4: Calculate the concentrations at equilibrium

[H₂] = 3.00-x = 0.88 M

[I₂] = 0.0800 M

[HI] = 2x = 4.24 M

Step 5: Calculate the equilibrium constant (Kc)

Kc = [HI]²/[H₂]×[I₂]

Kc = 4.24²/0.88×0.0800

Kc = 255