Question

At a particular temperature, ${\mathbf{K}}_p=\mathbf{1}.\mathbf{00}\times {\mathbf{10}}^2$ for the reaction

${\mathbf{H}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{I}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\stackrel{}{\mathbf{\rightleftharpoons }}\mathbf{2}\mathbf{HI}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

If 2.00 atm of ${\mathbf{H}}_2\left(g\right)$ and 2.00 atm of ${\mathbf{I}}_2\left(g\right)$ are introduced into a 1.00-L container, calculate the equilibrium partial pressures of all species.

Short answer

The equilibrium partial pressures of ${\text{H}}_{\text{2}}$, ${\text{I}}_{\text{2}}$ and $\text{HI}$ are 0.33 atm, 0.33 atm and 3.34 atm, respectively.

Step-by-step solution

Equilibrium constant value

For the given balanced equation, the expression for the equilibrium constant, K_p in terms of partial pressures is written below:

${\text{K}}_{\text{p}}\text{=}\frac{{{\text{P}}_{\text{HI}}}^{\text{2}}}{\left({\text{P}}_{{\text{H}}_{\text{2}}}\right)\left({\text{P}}_{{\text{I}}_{\text{2}}}\right)}$

The given initial partial pressures of hydrogen gas and iodine gas are 2.00 atm each.

The volume of the reaction container is 1.00 L.

The equilibrium constant value is given as, K_p = 1.00×10².

…(1)

Calculation of x

Let x be the change in partial pressure.

Setup the ICE table as:

$\begin{array}{cccccc}& {\text{H}}_{\text{2}}\left(\text{g}\right)& \text{+}& {\text{I}}_{\text{2}}\left(\text{g}\right)& \stackrel{}{\rightleftharpoons }& \text{2HI}\left(\text{g}\right)\\ \text{Iinitial}\left(\text{atm}\right)& \text{2.00}& & \text{2.00}& & \text{0}\\ \text{Change}\left(\text{atm}\right)& \text{\hspace{0.17em}–\hspace{0.17em}x}& & \text{\hspace{0.17em}–\hspace{0.17em}x}& & \text{+2x}\\ \text{Equilibrium}\left(\text{atm}\right)& \text{2.00\hspace{0.17em}–\hspace{0.17em}x}& & \text{2.00\hspace{0.17em}–\hspace{0.17em}x}& & \text{2x}\end{array}$

Now, substitute the values of K_p and the equilibrium partial pressures of all the gases from ICE table into equation (1) to determine the value of x.

$\begin{array}{l}{\text{1.00×10}}^{\text{2}}\text{=}\frac{{\left(\text{2x}\right)}^{\text{2}}}{\left(\text{2.00\hspace{0.17em}–\hspace{0.17em}x}\right)\left(\text{2.00\hspace{0.17em}–\hspace{0.17em}x}\right)}\\ \text{100=}\frac{{\left(\text{2x}\right)}^{\text{2}}}{{\left(\text{2.00\hspace{0.17em}–\hspace{0.17em}x}\right)}^{\text{2}}}\\ \sqrt{\text{100}}\text{=}\frac{\left(\text{2x}\right)}{\left(\text{2.00\hspace{0.17em}–\hspace{0.17em}x}\right)}\end{array}$

Solve for x.

$\text{x=1.67\hspace{0.17em}atm}$

Equilibrium partial pressures of the given gaseous species

The equilibrium partial pressures of all the gases can be calculated by using the x value as follows:

$\begin{array}{c}{\text{P}}_{{\text{H}}_{\text{2}}}\text{=2.00\hspace{0.17em}–\hspace{0.17em}x}\\ \text{=2.00\hspace{0.17em}–\hspace{0.17em}1.67}\\ \text{=0.33\hspace{0.17em}atm}\end{array}$

$\begin{array}{c}{\text{P}}_{{\text{I}}_{\text{2}}}\text{=2.00\hspace{0.17em}–\hspace{0.17em}x}\\ \text{=2.00\hspace{0.17em}–\hspace{0.17em}1.67}\\ \text{=0.33\hspace{0.17em}atm}\end{array}$

$\begin{array}{c}{\text{P}}_{\text{HI}}\text{=2x}\\ \text{=2×1.67}\\ \text{=3.34\hspace{0.17em}atm}\end{array}$

Hence, the equilibrium partial pressures of ${\text{H}}_{\text{2}}$, ${\text{I}}_{\text{2}}$ and $\text{HI}$ are 0.33 atm, 0.33 atm and 3.34 atm, respectively.