a mixture of 0.500 mol H2 and 0.500 mol I2 was placed in a 2.00-L stainless-steel flask at 430C. Calculate the concentrations of H2, I2, and HI at equilibrium
Don't we need a Kc or Kp?
To calculate the concentrations of H2, I2, and HI at equilibrium in this reaction, we need to use the concept of equilibrium constant (Kc) and the stoichiometry of the balanced equation.
The balanced equation for the reaction of H2 and I2 to form HI is:
H2(g) + I2(g) ⇌ 2HI(g)
At equilibrium, the concentration of each species can be expressed as follows:
[H2] = concentration of H2 (mol/L)
[I2] = concentration of I2 (mol/L)
[HI] = concentration of HI (mol/L)
The equilibrium constant expression for this reaction is given by:
Kc = ([HI]²) / ([H2] * [I2])
Since the stoichiometry of the reaction shows that two moles of HI are produced for each mole of H2 and I2, the concentration of HI at equilibrium will be 2 times the concentration of H2 or I2. Thus, we can express:
[HI] = 2x (where x is the concentration of H2 or I2)
Now, let's calculate the concentrations at equilibrium using the provided information:
Given:
[H2] = 0.500 mol
[I2] = 0.500 mol
V = 2.00 L (volume of the flask)
To find the concentration of H2, I2, and HI at equilibrium, we need to convert the moles into concentrations by dividing the number of moles by the volume:
[H2] = 0.500 mol / 2.00 L = 0.250 M
[I2] = 0.500 mol / 2.00 L = 0.250 M
Since [HI] = 2x, we substitute the concentrations of H2 and I2 to find [HI]:
[HI] = 2 * [H2] = 2 * 0.250 M = 0.500 M
Therefore, at equilibrium, the concentrations are as follows:
[H2] = 0.250 M
[I2] = 0.250 M
[HI] = 0.500 M