Question

A sample of S₈(g) is placed in an otherwise empty, rigid container at 1325 K at an initial pressure of 1.00 atm, where it decomposes to S₂(g) by the reaction

${\mathbf{\text{S}}}_{\mathbf{\text{8}}}\mathbf{\text{(g)}}\stackrel{}{\mathbf{\rightleftharpoons }}{\mathbf{\text{4S}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}$

At equilibrium, the partial pressure of S₈ is 0.25 atm.Calculate K_p for this reaction at 1325 K.

Short answer

The K_p for this reaction at 1325 K is ${\text{3.2×10}}^{\text{2}}$.

Step-by-step solution

Step 1:

The expression for the equilibrium constant, K_p is written below:

${\text{K}}_{\text{p}}\text{=}\frac{{{\text{P}}_{{\text{S}}_{\text{2}}}}^{\text{4}}}{{\text{P}}_{{\text{S}}_{\text{8}}}}$ …(1)

Let x be the change in partial pressure of${\text{S}}_{\text{8}}$ to reach the equilibrium.

Setup the ICE table:

$\begin{array}{cccc}& {\text{S}}_{\text{8}}\left(\text{g}\right)& \stackrel{}{\rightleftharpoons }& {\text{4S}}_{\text{2}}\left(\text{g}\right)\\ \text{Iinitial}\left(\text{atm}\right)& \text{1.00}& & \text{0}\\ \text{Change}\left(\text{atm}\right)& \text{\hspace{0.17em}–\hspace{0.17em}x}& & \text{+4x}\\ \text{Equilibrium}\left(\text{atm}\right)& \text{1.00\hspace{0.17em}–\hspace{0.17em}x}& & \text{4x}\end{array}$

Step 2:

At equilibrium, the partial pressure of S₈ is given as 0.25 atm.

The x value can be calculated by using the equilibrium partial pressure of S₈.

$\begin{array}{c}{\mathrm{P}}_{{\mathrm{S}}_{\mathrm{8}}}=1.00\hspace{0.17em}–\hspace{0.17em}\mathrm{x}\mathrm{at}\hspace{0.17em}\mathrm{equilibrium}\\ 0.25=\mathrm{1.00\hspace{0.17em}–\hspace{0.17em}}\mathrm{x}\\ \mathrm{x}=0.75\end{array}$

Step 3:

At equilibrium, the partial pressures are,

${\text{P}}_{{\text{S}}_{\text{8}}}\text{=0.25\hspace{0.17em}atm}$

$\begin{array}{c}{\mathrm{P}}_{{\mathrm{S}}_{\mathrm{2}}}=4\mathrm{x}\\ =4\times 0.75\\ =3.0\hspace{0.17em}\mathrm{atm}\end{array}$

Substitute these values into Equation (1) to calculate the K_p.

role="math" localid="1657301494436" $\begin{array}{c}{\mathrm{K}}_{\mathrm{p}}=\frac{{\left(3.0\right)}^{\mathrm{4}}}{0.25}\\ =3{.2\times 10}^{\mathrm{2}}\end{array}$

Hence, the K_p for this reaction at 1325 K is $3{.2\times 10}^{\mathrm{2}}$.