Question

The following reaction was performed in a sealed vessel at 760 ∘C : H2(g)+I2(g)⇌2HI(g) Initially, only H2 and I2 were present at concentrations of [H2]=3.85M and [I2]=2.35M . The equilibrium concentration of I2 is 0.0500 M . What is the equilibrium constant, Kc, for the reaction at this temperature?

Answer 1

Answer:

The equilibrium constant is 273.0322

Explanation:

For the given chemical reaction ,

ICE table can be written as -

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

initial moles         3.85             2.35                -

at equilibrium      3.85 - x       2.35 - x           2x

From question , at equilibrium the concentration of I₂ = 0.0500 M

The concentration of I₂ ( ICE table ) = concentration of I₂ (given in question )

2.35 - x = 0.0500

x = 2.3

Putting the value of x in ICE table , to obtain the concentration  terms as-

[H₂] = 3.85 - x

[H₂] = 3.85 - 2.3

[H₂] = 1.55 M

[HI] = 2x

[HI] = 2* 2.3

[HI] = 4.6 M

[I₂] = 0.0500M       (Given)

Equilibrium Constant ( Kc )

The value of equilibrium constant is written as the concentration of the products each raised to the power of their respective stoichiometry by the concentration of reactants each raised to the power of their respective stoichiometry.

Kc = [HI]² / [H₂][I₂]

Kc = ( 4.6 )² / (1.55)*(0.0500)

Kc = 273.0322 .