A mixture of 0.586 M H2, 0.477 M I2, and 0.943 M HI is enclosed in a vessel and heated to 430 °C.

H2(g)+I2(g)↽−−⇀2HI(g)𝐾c=54.3 at 430 ∘C

Calculate the equilibrium concentrations of each gas at 430 ∘C .

[H2]=
M
[I2]=
M
[HI]=
M

To calculate the equilibrium concentrations of each gas at 430 °C, we need to use the equation for the equilibrium constant (Kc) and the given initial concentrations of the reactants.

Step 1: Write out the balanced chemical equation:
H2(g) + I2(g) ⇌ 2HI(g)

Step 2: Write the expression for the equilibrium constant (Kc):
Kc = [HI]^2 / ([H2] * [I2])

Step 3: Calculate the equilibrium concentrations using the given initial concentrations and the stoichiometry of the reaction. Let's assume that the initial volume of the mixture is 1 L.

[H2] = 0.586 M
[I2] = 0.477 M
[HI] = 0.943 M

Using the stoichiometry of the reaction, we know that 1 mole of H2 reacts with 1 mole of I2 to form 2 moles of HI. Therefore, at equilibrium, the change in concentrations is given by:

[HI] = 2x # since 2 moles of HI is formed for every mole of H2 reacted
[H2] = -x # the concentration of H2 will decrease by x moles
[I2] = -x # the concentration of I2 will decrease by x moles

Substituting these expressions into the equilibrium constant expression:

Kc = (2x)^2 / ((0.586 - x) * (0.477 - x))

Step 4: Substitute the given equilibrium constant value into the equation and solve for x:

54.3 = (2x)^2 / ((0.586 - x) * (0.477 - x))

Solving this equation will give us the value of x, which represents the change in concentration of H2, I2, and HI at equilibrium.

Step 5: Once we have the value of x, we can calculate the equilibrium concentrations of each gas:

[H2] = 0.586 - x M
[I2] = 0.477 - x M
[HI] = 0.943 + 2x M

Plug in the value of x into these equations to find the equilibrium concentrations of each gas.

These last two are the same thing. Is this a home work dump? Let me know if you have any problems.