When 0.20 mol of hydrogen gas and 0.15 mol of iodine gas are heated at 723 K until equilibrium is established, the equilibrium mixture is found to contain 0.26 mol of hydrogen iodide.


The equation for the reaction is as follows.

H2(g) + I2(g) ↔ 2HI(g)

What is the equilibrium constant Kc of this reaction? Show the way with initial, react, and equilibrium mol.

To determine the equilibrium constant Kc for the given reaction, we need to set up an expression using the initial, reactant, and equilibrium molar amounts.

First, let's understand the molar changes that occur during the reaction.

H2(g) + I2(g) ↔ 2HI(g)
Initial: 0.20 mol 0.15 mol 0 mol
Change: -x -x +2x
Equilibrium: 0.20-x 0.15-x 2x

Based on the stoichiometry of the balanced equation, the change in moles for H2(g) and I2(g) will be -x, while the change for HI(g) will be +2x.

Given that the equilibrium mixture contains 0.26 mol of HI(g), we can write the equation as follows:

0.26 mol = 2x
x = 0.26 mol / 2
x = 0.13 mol

Now, let's substitute the values into the equilibrium expression:

Kc = [HI]² / ([H2] * [I2])

Kc = (0.26 mol)² / (0.20 - 0.13)(0.15 - 0.13)

Simplifying further:

Kc = 0.0676 mol² / (0.07)(0.02)

Kc = 0.0676 mol² / 0.0014 mol²

Kc = 48.28

Therefore, the equilibrium constant Kc for this reaction is 48.28.

To solve for the equilibrium constant Kc of the reaction, you need to use the balanced equation and the concentrations of the species at equilibrium.

Step 1: Write down the given information and the balanced equation:
Initial moles: H2 = 0.20 mol, I2 = 0.15 mol, HI = 0 mol
Equilibrium moles: H2 = 0.20 - x, I2 = 0.15 - x, HI = 0 + 2x

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

Step 3: Determine the concentrations at equilibrium:
[H2] = 0.20 - x
[I2] = 0.15 - x
[HI] = 0 + 2x

Step 4: Substitute the equilibrium concentrations into the Kc expression:
Kc = (0 + 2x)^2 / ((0.20 - x)*(0.15 - x))

Step 5: Since the equilibrium mixture is found to contain 0.26 mol of hydrogen iodide, we can set up an equation to solve for x:
0 + 2x = 0.26
Solve for x: x = 0.26 / 2 = 0.13 mol

Step 6: Substitute the value of x back into the expressions for the equilibrium concentrations:
[H2] = 0.20 - 0.13 = 0.07 mol
[I2] = 0.15 - 0.13 = 0.02 mol
[HI] = 0 + 2(0.13) = 0.26 mol

Step 7: Plug the equilibrium concentrations into the Kc expression to find the equilibrium constant:
Kc = (0.26)^2 / ((0.07)*(0.02))
Kc = 0.0674 / 0.0014
Kc ≈ 48.142

Therefore, the equilibrium constant Kc for the reaction is approximately 48.142.

Technically, one needs a volume since M is substituted into the Kc expression and not mols. You don't have a volume listed. However, the volume cancels as follows:

Kc = (mols HI/v)^2/(mols H2/v)(mols I2/v). That means we can assume any volume or ignore volume (or use 1 L)
(H2) initial = 0.2
(I2) initial = 0.15
(HI) equilibrium = 0.26

........H2 + I2 ==> 2HI
I......0.2..0.15......0
C......-x....-x......2x
E...0.2-x...0.15-x...2x
but you know 2x = 0.26 and that allows you to evaluate mols H2 and mols I2.

Substitute the E line into Kc expression and solve for Kc.

Post your work if you get stuck.