At a certain temperature, the Equilibrium constant, Kc, for this reaction is 53.3. At this temperature, 0.500 mol of H2 and 0.500 mol of I2 were placed in a 1.00-L container to react. What concentration of HI is present at Equilibrium?
equation: H2 + I2 = 2HI
...........H2 + I2 ==> 2HI
I.........0.50.0.50....0
C.........-x....-x.....2x
E........0.50-x.0.50-x..2x
Substitute the E line into Kc expression and solve for x, then 0.50-x and 2x.
Those of you looking at this in the future don't forget the quadratic equation!!!
To find the concentration of HI at equilibrium, we need to use the equilibrium constant, Kc. The equilibrium constant expression for the given reaction is:
Kc = [HI]^2 / ([H2] * [I2])
Given that the equilibrium constant, Kc, is 53.3, and the initial concentrations of H2 and I2 are both 0.500 mol in a 1.00-L container, we can substitute these values into the equilibrium constant expression.
Kc = [HI]^2 / (0.500 * 0.500)
To solve for [HI], we can rearrange the equation to isolate [HI]:
[HI]^2 = Kc * (0.500 * 0.500)
[HI] = √(Kc * 0.250)
Substituting the given value of Kc, we get:
[HI] = √(53.3 * 0.250)
[HI] ≈ √13.325
[HI] ≈ 3.65 M
Therefore, the concentration of HI at equilibrium is approximately 3.65 M.