equilibrium constant = 55.3 @ certain temperature
H2 (g) + I2 (g) <=> 2HI (g) (reversible)
If you start with 0.195 M hydrogen iodide, what will the concentrations of H2, I2, and HI be at equilibrium?
Just like the HBr problem.
WHAT HBR PROBLEM
To determine the concentrations of H2, I2, and HI at equilibrium, we can use the concept of stoichiometry and the equilibrium constant.
The equilibrium constant, K, for the given reaction is 55.3 at a certain temperature. The general expression for the equilibrium constant of a reaction is given as:
K = [HI]^2 / ([H2] * [I2])
Using this expression, we can find the equilibrium concentrations of H2, I2, and HI.
Given that the initial concentration of HI is 0.195 M, we can assume that the initial concentrations of H2 and I2 are both 0 M.
Let's say at equilibrium the concentrations of H2, I2, and HI are [H2]eq, [I2]eq, and [HI]eq, respectively.
According to the stoichiometry of the reaction, for every one mole of H2 that reacts, two moles of HI are formed. Hence, the change in concentration of H2 is -2x, where x is the moles of H2 that react.
Similarly, the change in concentration of I2 is -x, and the change in concentration of HI is +2x.
Given that the initial concentration of H2 is 0 M, and the change in concentration is -2x, the concentration of H2 at equilibrium is [H2]eq = 0 - 2x = -2x.
Given that the initial concentration of I2 is 0 M, and the change in concentration is -x, the concentration of I2 at equilibrium is [I2]eq = 0 - x = -x.
Given that the initial concentration of HI is 0.195 M, and the change in concentration is +2x, the concentration of HI at equilibrium is [HI]eq = 0.195 + 2x.
Now, we can substitute these equilibrium concentrations into the expression for the equilibrium constant:
K = [HI]^2 / ([H2] * [I2])
55.3 = ([0.195 + 2x]^2) / ([-2x] * [-x])
Simplify and solve for x:
55.3 = (0.195 + 2x)^2 / (2x * x)
Cross-multiply:
55.3 * (2x * x) = (0.195 + 2x)^2
110.6x^2 = (0.195 + 2x)^2
Take the square root of both sides:
sqrt(110.6x^2) = 0.195 + 2x
10.513x = 0.195 + 2x
Subtract 2x from both sides:
8.513x = 0.195
x = 0.195 / 8.513
x ≈ 0.023 M
Now, substitute this value of x into the equations for the equilibrium concentrations:
[H2]eq = -2x
[H2]eq = -2(0.023)
[H2]eq ≈ -0.046 M
[I2]eq = -x
[I2]eq = -(0.023)
[I2]eq ≈ -0.023 M
[HI]eq = 0.195 + 2x
[HI]eq = 0.195 + 2(0.023)
[HI]eq ≈ 0.241 M
Therefore, at equilibrium, the concentrations of H2, I2, and HI will approximately be [H2]eq ≈ -0.046 M, [I2]eq ≈ -0.023 M, and [HI]eq ≈ 0.241 M, respectively.
Note: Negative concentrations indicate that the reactant is mostly consumed in the reaction. It's a common practice to express equilibrium concentrations with absolute values, so the equilibrium concentrations would be approximately [H2]eq ≈ 0.046 M, [I2]eq ≈ 0.023 M, and [HI]eq ≈ 0.241 M.