What is the equilibrium constant, K_c, for the reaction at this temperature?

The following reaction was performed in a sealed vessel at 753 ∘C : H2(g)+I2(g)⇌2HI(g) Initially, only H2 and I2 were present at concentrations of [H2]=3.10M and [I2]=2.55M. The equilibrium concentration of I2 is 0.0200 M .

Mar 4, 2018

${K}_{c} = 2.25 \cdot {10}^{3}$

Explanation:

You know that when the following reaction takes place at ${753}^{\circ} \text{C}$

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

the initial concentrations of the two reactants are

["H"_ 2]_ 0 = "3.10 M" " " and $\text{ "["I"_ 2]_ 0 = "2.55 M}$

and the equilibrium concentration of iodine gas is

["I"_2] = "0.0200 M"

This tells you that the concentration of iodine gas decreased by

$| \text{0.0200 M " - " 2.55 M"| = "2.53 M}$

In other words, the reaction consumed $\text{2.53 M}$ of iodine gas. Since iodine gas and hydrogen gas react in a $1 : 1$ mole ratio, you can say that the reaction also consumed $\text{2.53 M}$ of hydrogen gas.

This means that the equilibrium concentration of hydrogen gas will be

["H"_ 2] = ["H"_ 2]_ 0 - "2.53 M"

["H"_ 2] = "3.10 M" - "2.53 M"

["H"_ 2] = "0.57 M"

Now, notice that hydrogen iodide, the product of the reaction, is produced in a $\textcolor{red}{2} : 1$ mole ratio with iodine gas and hydrogen gas.

This means that for every $1$ mole of hydrogen gas and $1$ mole of iodine gas that the reaction consumes, you get $\textcolor{red}{2}$ moles of hydrogen iodide.

Therefore, the equilibrium concentration of hydrogen iodide will be

["HI"] = color(red)(2) * "2.53 M"

["HI"] = "5.06 M"

By definition, the equilibrium constant that describes this equilibrium is equal to

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

Plug in your values to find--I'll leave the calculation without added units!

${K}_{c} = {\left(5.06\right)}^{\textcolor{red}{2}} / \left(0.57 \cdot 0.0200\right) = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{2.25 \cdot {10}^{3}}}}$

The answer is rounded to three sig figs.

So, does this value make sense?

Notice that the concentrations of the two reactants decrease significantly as the reaction proceeds. This tells you that at this temperature, the equilibrium lies to the right, i.e. the forward reaction is favored.

Therefore, you should expect to find

${K}_{c} > 1$

which is exactly what happens here.