The equilibrium constant for HI decomposition at 500 °C is 5.8×10-3.
2HI(g) > H2(g) + I2(g)
At that temperature reactants and products are gaseous. (There is no change in the total number of moles, so Kc = Kp.)
A sample of 0.49 mol of HI is placed in a 1.00 L vessel which is then heated to 500 °C. When equilibrium is reached, what is the molar concentration of H2? (Give units).
[H2] = 0.032 M
Now I cant figure out how to fine the concentrations of I2 and HI.
I keep getting for I2=.035M and for HI=.46M
see above.
To find the molar concentrations of I2 and HI, we can use the stoichiometry of the reaction and the equilibrium constant.
Given:
Equilibrium constant (Kc) = 5.8 x 10^(-3)
Initial moles of HI = 0.49 mol
Volume of the vessel = 1.00 L
Let's denote the final molar concentrations of HI, H2, and I2 as [HI], [H2], and [I2], respectively.
First, we need to set up an ICE (Initial, Change, Equilibrium) table to solve for the concentrations at equilibrium.
Initial:
[HI] = 0.49 M
[H2] = 0 M (since it is not present initially)
[I2] = 0 M (since it is not present initially)
Change:
Since 2 moles of HI produce 1 mole of H2 and 1 mole of I2 in the balanced equation, the change in concentration for HI will be -2x, and the change in concentration for both H2 and I2 will be +x.
Equilibrium:
[HI] = 0.49 - 2x
[H2] = x
[I2] = x
Now, we need to substitute the equilibrium concentrations into the equilibrium expression for Kc and solve for x.
Kc = [H2] * [I2] / [HI]^2
5.8 x 10^(-3) = (x * x) / (0.49 - 2x)^2
Solving this equation may give a quadratic equation. However, we can simplify it by assuming that the value of x will be small compared to 0.49. This approximation is justified because the reaction is in the gas phase and the equilibrium constant is relatively small.
Assuming x << 0.49, we can ignore the 2x term in the denominator of the equation.
5.8 x 10^(-3) ≈ x^2 / 0.49^2
Simplifying further, we get:
x^2 ≈ 5.8 x 10^(-3) * 0.49^2
x^2 ≈ 1.438 x 10^(-5)
Taking the square root of both sides:
x ≈ 0.003794
Since x represents the molar concentration of H2, we can conclude that the molar concentration of H2 at equilibrium is approximately 0.003794 M (rounded to four decimal places).
To find the concentrations of HI and I2, we can substitute this value of x into the expressions we derived earlier:
[HI] = 0.49 - 2x ≈ 0.49 - 2 * 0.003794 ≈ 0.482412 M
[I2] = x ≈ 0.003794 M
Therefore, the molar concentration of HI is approximately 0.4824 M, and the molar concentration of I2 is approximately 0.0038 M.