# Question b584a

Jun 5, 2017
• ${\text{H}}_{2}$: $0.322$ $M$

• ${\text{I}}_{2}$: $0.00749$ $M$

• $\text{HI}$: $0.347$ $M$

#### Explanation:

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

From the chemical equation

$\text{H"_2 (g) + "I"_2 (g) rightleftharpoons 2"HI} \left(g\right)$

the equilibrium-constant expression is

${K}_{c} = \left(\left[{\text{HI"]^2)/(["H"_2]["I}}_{2}\right]\right)$

The given volume is $1$ ${\text{dm}}^{3}$, which is equivalent to $1$ $\text{L}$, 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 ($M$):

• ${\text{H}}_{2}$: $0.496$

• ${\text{I}}_{2}$: $0.181$

• $\text{HI}$: $0$

Change in Concentration ($M$):

• ${\text{H}}_{2}$: $- x$

• ${\text{I}}_{2}$: $- x$

• $\text{HI}$: $+ 2 x$

Final Concentration ($M$):

• ${\text{H}}_{2}$: $0.496 - x$

• ${\text{I}}_{2}$: $0.00749$

• $\text{HI}$: $2 x$

Since we're given both the initial and final concentrations for ${\text{I}}_{2}$, 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 ($M$):

• ${\text{H}}_{2}$: $0.496 - \textcolor{red}{0.17351} = 0.322$

• ${\text{I}}_{2}$: $0.00749$

• $\text{HI}$: $2 \left(\textcolor{red}{0.17351}\right) = 0.347$

The equilibrium concentrations are thus

• ${\text{H}}_{2}$: $0.322$ $M$

• ${\text{I}}_{2}$: $0.00749$ $M$

• $\text{HI}$: $0.347$ $M$