H2(g) + I2(g) ⇌ 2HI(g)

The concentration of H2, I2 and HI are measured as 0.04 M, 0.08 M and 0.12 M at equilibrium and 200 K.

What is the value of equilibrium constant?

To find the value of the equilibrium constant (K), you need to use the equilibrium concentrations of the reactants and products.

Given:
[H2] = 0.04 M
[I2] = 0.08 M
[HI] = 0.12 M

The balanced equation is:
H2(g) + I2(g) ⇌ 2HI(g)

The equilibrium constant expression for this reaction is:
K = ([HI]^2) / ([H2] * [I2])

Substituting the given concentrations:
K = (0.12^2) / (0.04 * 0.08)

Now, calculate the value of K:
K = 0.0144 / 0.0032

Simplifying,
K = 4.5

Therefore, the value of the equilibrium constant (K) for the given reaction at 200 K is 4.5.

Write the expression for Keq. Substitute the values given in the problem and solve for Keq

Post your work if you get stuck but this really is nothing more than plug and chug..