If 1.00 mole of BrCl is introduced into a 500.0 ml flask, what are the concentrations of Br2, Cl2 and BrCl at equailbrium
You need Kp or Kc to work this problem.
To determine the concentrations of Br2, Cl2, and BrCl at equilibrium, we need to know the equilibrium constant (K) for the reaction. The balanced equation for the reaction involved is:
2 BrCl ↔ Br2 + Cl2
Assuming the reaction has reached equilibrium, let's assign x moles of BrCl that reacts, which would produce x moles of Br2 and x moles of Cl2.
Initially, we have 1.00 mole of BrCl in a 500.0 ml flask, which is equivalent to 0.00200 moles/ml (by dividing 1.00 mole by the volume in liters).
Since we assumed x moles of BrCl reacts, the new concentrations of Br2 and Cl2 will be x moles / 0.500 L (as the total volume is 500.0 ml, which is equal to 0.500 L).
Using the stoichiometry of the balanced equation, we know that the number of moles of Br2 and Cl2 produced is equal to the number of moles of BrCl consumed.
Hence, the concentrations of Br2 and Cl2 at equilibrium will be both x moles / 0.500 L.
Now, to find the value of x, we need to use the equilibrium constant, K. The expression for K can be written as:
K = [Br2][Cl2]/[BrCl]^2
We'll substitute the concentrations we just obtained into this expression:
K = (x / 0.500 L) * (x / 0.500 L) / (0.00200 moles/ml)^2
Now, you'll need to know the value of the equilibrium constant (K) for this reaction in order to calculate the value of x and determine the concentrations of Br2, Cl2, and BrCl at equilibrium.