The reaction below has Kc = 280 at 150C:
πΌ2 (π)+ π΅π2 (π) β2πΌπ΅π (π)
A 0.500 mol sample of IBr was placed in a 1.00 L flask at this temperature and was allowed to reach equilibrium. What are the equilibrium concentrations for I2, Br2, and IBr?
To determine the equilibrium concentrations for I2, Br2, and IBr in the given reaction at the specified temperature, we can use the stoichiometry of the reaction and the given value of the equilibrium constant Kc.
First, let's assume that the initial concentration of IBr is x mol/L. Since two moles of IBr molecules are needed to produce one mole each of I2 and Br2, the initial concentrations of I2 and Br2 are both zero.
At equilibrium, the change in the concentration of IBr will be -2x mol/L (as two moles of IBr are consumed for every one mole of I2 and Br2 produced).
The equilibrium concentrations for I2 and Br2 will both be x mol/L, considering the stoichiometry of the reaction.
Now, we can set up an expression for the equilibrium constant Kc and solve for x:
Kc = [I2] * [Br2] / [IBr]^2
Substituting the equilibrium concentrations in terms of x:
Kc = (x) * (x) / (2x)^2
280 = x^2 / (4x^2)
280 * 4x^2 = x^2
1120x^2 = x^2
1120x^2 - x^2 = 0
1119x^2 = 0
x^2 = 0
x = 0
Hence, the equilibrium concentrations of I2, Br2, and IBr in the given reaction at the specified temperature are:
[I2] = [Br2] = 0 mol/L
[IBr] = 0 mol/L