Question

For the reaction I2 + Br2(g) 2 IBr(g), Kc= 280 at 150oC. Suppose that 0.500 molIBr in a 1.00- L flask is allowed to reach equilibrium at 150oC. What are the equilibrium concentrations of IBr, I2, and Br2?

Answer 1

To solve this problem, we'll need to set up an ICE (Initial, Change, Equilibrium) table and use the given equilibrium constant (Kc) to calculate the equilibrium concentrations.

Let's define the initial concentrations of IBr, I2, and Br2 as follows: [IBr]₀ = 0.500 mol/L [I2]₀ = 0 mol/L (since there is no initial I2) [Br2]₀ = 0 mol/L (since there is no initial Br2)

Now, let's assume that the change in concentration for IBr, I2, and Br2 is represented as x (in mol/L). After the reaction reaches equilibrium, we can calculate the equilibrium concentrations using the ICE table.

The ICE table will look like this:

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