Question

Question #b584a

Answer 1

• `"H"_2`: `0.322` `M`

• `"I"_2`: `0.00749` `M`

• `"HI"`: `0.347` `M`

Explanation:

We're asked to find the molar concentrations of the substances at equilibrium from the given data.

From the chemical equation

`"H"_2 (g) + "I"_2 (g) rightleftharpoons 2"HI" (g)`

the equilibrium-constant expression is

`K_c = (["HI"]^2)/(["H"_2]["I"_2])`

The given volume is `1` `"dm"^3`, which is equivalent to `1` `"L"`, so the given mole values in the equation are the same as their molar concentrations.

Let's create a makeshift I.C.E. chart, using bullet points:

Initial Concentration (`M`):

• `"H"_2`: `0.496`

• `"I"_2`: `0.181`

• `"HI"`: `0`

Change in Concentration (`M`):

• `"H"_2`: `-x`

• `"I"_2`: `-x`

• `"HI"`: `+2x`

Final Concentration (`M`):

• `"H"_2`: `0.496-x`

• `"I"_2`: `0.00749`

• `"HI"`: `2x`

Since we're given both the initial and final concentrations for `"I"_2`, we can calculate the change in concentration, and thus `x`:

`x = overbrace(0.181)^("initial I"_2)-overbrace(0.00749)^("final I"_2) = color(red)(0.17351`

Now that we know `x`, we can use this to calculate the remaining equilibrium concentrations:

Final Concentrations (`M`):

• `"H"_2`: `0.496-color(red)(0.17351) = 0.322`

• `"I"_2`: `0.00749`

• `"HI"`: `2(color(red)(0.17351)) = 0.347`

The equilibrium concentrations are thus

• `"H"_2`: `0.322` `M`

• `"I"_2`: `0.00749` `M`

• `"HI"`: `0.347` `M`