consider the reaction 2hi(g) h2(g) + i2(g).Starting with 0.100 mol of HI in a 1.0 L container at 800K, it is observed that the equilibrium concentration of H2 is 0.01 mol/L. Find out equilibrium concentration of I2 and HI

To solve this problem, we need to analyze the given chemical reaction and apply the principles of equilibrium. The reaction in question is:

Understanding the Reaction

2 HI(g) ⇌ H2(g) + I2(g)

This reaction indicates that two moles of hydrogen iodide (HI) decompose to form one mole of hydrogen (H2) and one mole of iodine (I2). We start with an initial concentration of HI and need to find the equilibrium concentrations of all species involved.

Initial Conditions

We begin with:

• Initial moles of HI = 0.100 mol
• Volume of the container = 1.0 L
• Initial concentration of HI = 0.100 mol/L
• Equilibrium concentration of H2 = 0.01 mol/L

Setting Up the ICE Table

To find the equilibrium concentrations, we can use an ICE (Initial, Change, Equilibrium) table:

Substituting Known Values

From the problem, we know that at equilibrium, the concentration of H2 is 0.01 mol/L. Thus, we can set:

x = 0.01 mol/L

Calculating Changes in Concentration

Now we can substitute x back into our ICE table:

• For H2: At equilibrium, H2 = x = 0.01 mol/L
• For I2: At equilibrium, I2 = x = 0.01 mol/L
• For HI: HI = 0.100 - 2x = 0.100 - 2(0.01) = 0.100 - 0.02 = 0.080 mol/L

Final Equilibrium Concentrations

Now we can summarize the equilibrium concentrations:

• Equilibrium concentration of HI = 0.080 mol/L
• Equilibrium concentration of H2 = 0.01 mol/L
• Equilibrium concentration of I2 = 0.01 mol/L

In summary, at equilibrium, the concentrations are:

• HI: 0.080 mol/L
• H2: 0.01 mol/L
• I2: 0.01 mol/L

This methodical approach allows us to determine the equilibrium concentrations based on the initial conditions and the stoichiometry of the reaction. If you have any further questions or need clarification on any part of this process, feel free to ask!