An equilibrium was established after 0.100 mol of hydrogen gas and 0.100mol of iodine gas were added to an empty 1.00L reaction vessel and heated to 700K. The color intensity of the mixture changed from deep purple to a lighter purple color. At equilibrium, concentration of iodine was 0.0213 mol/L.

Question

An equilibrium was established after 0.100 mol of hydrogen gas and 0.100mol of iodine gas were added to an empty 1.00L reaction vessel and heated to 700K. The color intensity of the mixture changed from deep purple to a lighter purple color. At equilibrium, concentration of iodine was 0.0213 mol/L.

Calculate the equilibrium concentrations of hydrogen gas and hydrogen iodide gas as well as Kc.

Answer 1

Start with an equation

H2 + I2 -> 2HI

so 2 moles of HI formed for each mole of I2 used.

at start

0.100 M/L (H2) and 0.100 M/L (I2)

at equilibrium

0.0213 M/L (I2) so 0.0787 M/L I2 used

so 0.1574 M/L HI formed

and 0.0213 M/L equlibrium concentration of H2

Kc=[HI]^2/[H2][I2]

plug in the values and find Kc

Note that this Kc has no units.

Answer 2

To solve this question, we need to use the concept of equilibrium constant (Kc) and the stoichiometry of the chemical reaction. Let's first write the balanced chemical equation for the reaction:

H2(g) + I2(g) ↔ 2HI(g)

According to the balanced equation, 1 mole of hydrogen gas reacts with 1 mole of iodine gas to form 2 moles of hydrogen iodide gas.

Now, let's define the equilibrium concentrations of hydrogen gas (H2), iodine gas (I2), and hydrogen iodide gas (HI) as [H2], [I2], and [HI], respectively.

Given:
Initial moles of H2 = 0.100 mol
Initial moles of I2 = 0.100 mol
Equilibrium concentration of I2, [I2] = 0.0213 mol/L

Using the stoichiometry from the balanced equation, we can write the corresponding expressions for the equilibrium concentrations of H2 and HI:

[H2] = (Initial moles of H2 - change in moles of H2) / Volume of the reaction vessel
[HI] = (Change in moles of HI) / Volume of the reaction vessel

Since 1 mole of hydrogen gas reacts with 1 mole of iodine gas to form 2 moles of hydrogen iodide gas, the change in moles of HI is twice the change in moles of H2.

Change in moles of H2 = -0.5 × change in moles of HI
Change in moles of HI = 2 × change in moles of H2

Now, we can calculate the change in moles of HI by subtracting the moles at equilibrium from the initial moles:

Change in moles of HI = Initial moles of HI - Equilibrium moles of HI = 0 - 0.0213 mol = -0.0213 mol

Using the relationship between the changes in moles of HI and H2:

Change in moles of H2 = -0.5 × (-0.0213 mol) = 0.01065 mol

Finally, we can substitute these values into the expressions for the equilibrium concentrations:

[H2] = (0.100 mol - 0.01065 mol) / 1.00 L = 0.08935 mol/L
[HI] = (-0.0213 mol) / 1.00 L = -0.0213 mol/L (negative since HI is being consumed)

Since the concentration of a reactant or product cannot be negative, we have to take the magnitude of the negative value as the concentration:
[HI] = 0.0213 mol/L

Now, we have the equilibrium concentrations of H2 and HI. To calculate Kc, we use the formula:

Kc = [HI]^2 / ([H2] × [I2])

Substituting the values:

Kc = (0.0213 mol/L)^2 / (0.08935 mol/L × 0.0213 mol/L)
Kc = 0.000454 / 0.001902
Kc = 0.2387 (rounded to four significant figures)

Therefore, the equilibrium concentrations of hydrogen gas, hydrogen iodide gas, and the value of Kc are:
[H2] = 0.0894 mol/L
[HI] = 0.0213 mol/L
Kc = 0.2387