Question

Consider the reaction:

H2(g)+I2(g)⇌2HI(g)

A reaction mixture in a 3.65 L flask at a certain temperature initially contains 0.764 g H2 and 97.0 g I2. At equilibrium, the flask contains 90.5 g HI.

Calculate the equilibrium constant (Kc) for the reaction at this temperature.

Kc = ?

Answer 1

Answer – Given, mass of H₂ = 0.746 g , mass of I₂ = 97.0 g , volume = 3.65 L

At equilibrium, HI = 90.5 g

First we need to calculate the moles of each –

Moles of H₂ = 0.746 g / 2.016 g.mol^-1

                    = 0.370 moles

Moles of I₂ = 97.0 g / 253.8 g.mol^-1

                    = 0.382 moles

Moles of HI = 90.5 g / 127.91 g.mol^-1

                    = 0.707 moles

Now. [H₂] = 0.370 moles / 3.65 L

                  = 0.101 M

[I₂] = 0.382 moles / 3.65 L

           = 0.105 M

[HI] = 0.707 moles / 3.65 L

           = 0.194 M

Now we need to put ICE chart

    H₂ (g)   +    I₂ (g) <----> 2HI(g)

I   0.101    0.105             0

C -x        -x                +2x

E 0.101-x    0.105-x     0.194

So, at equilibrium [HI] = 0.194

2x = 0.194

So, x = 0.0969 M

So, at equilibrium,

[H₂] = 0.101-x

          = 0.101-0.0969

            = 0.00446 M

[H₂] = 0.105-x

          = 0.151-0.0969

            = 0.00779 M

So, Kc = [HI]² / [H₂] [I₂]

            = (0.194)² / (0.00446) (0.00779)

           = 1082