Question

The K for the reaction below is3.34×10⁻⁶ at equilibrium at 500 K: 2IBr(g)⇌I₂(g)+Br₂(g)

The enthalpy change (ΔH) for the reaction is −568kJ/mol. Furthermore, if the equilibrium concentration of IBr is 0.0947 M, what are the equilibrium concentrations of I2 and Br2?

Answer 1

The equilibrium concentrations of I2 and Br2 are approximately 4.44×10⁻⁸ M.

Step-by-step explanation:

Given that the equilibrium concentration of IBr is 0.0947 M and the equilibrium constant (K) is 3.34×10⁻⁶ at 500 K, we can use the stoichiometry of the reaction to determine the equilibrium concentrations of I2 and Br2.

The balanced equation for the reaction is 2IBr(g) ⇌ I₂(g) + Br₂(g).

Let's assume the equilibrium concentration of I2 is x M and the equilibrium concentration of Br2 is y M. Using the equilibrium constant expression, we have:

K = [I₂][Br₂] / [IBr]²

Plugging in the given values, we get:

3.34×10⁻⁶ = x*y / (0.0947)²

Solving for x and y, we find that the equilibrium concentrations of I2 and Br2 are approximately:

I2 = 4.44×10⁻⁸ M

Br2 = 4.44×10⁻⁸ M