The key step in the manufacture of sulfuric acid is the oxidation of sulfur dioxide in the presence of a catalyst, such as ${\mathbf{\text{V}}}_{\mathbf{\text{2}}}{\mathbf{\text{O}}}_{\mathbf{\text{5}}}$. At ${727}^o$C, role="math" localid="1663318144563" $\mathbf{\text{0.010}}$mol of role="math" localid="1663318136951" ${\mathbf{\text{SO}}}_{\mathbf{\text{2}}}$is injected into an empty 2.00-L container$\left(\text{Kp = 3.18}\right)\mathbf{\text{.}}$

(a) What is the equilibrium pressure of ${\mathbf{\text{O}}}_{\mathbf{\text{2}}}$that is needed to maintain a 1/1 mole ratio of ${\mathbf{\text{SO}}}_{\mathbf{\text{3}}}$to${\mathbf{\text{SO}}}_{\mathbf{\text{2}}}$ ?

(b) What is the equilibrium pressure of ${\mathbf{\text{O}}}_{\mathbf{\text{2}}}$needed to maintain a mole ratio of${\text{SO}}_{\text{3}}$ to ${\mathbf{\text{SO}}}_{\mathbf{\text{2}}}$?

Chemically, elements are substances that cannot be broken down into simpler things. Hydrogen (H), with an atomic number of 1, is one of nature's elements. This element gave origin to all others and makes up 75% of the mass of the cosmos. Carbon (C) is a six-atomic element.

(a)

Write the reaction for the procedure first:

${\text{2SO}}_{\text{2}}{\text{(g) +O}}_{\text{2}}\text{(g)}\to {\text{2SO}}_{\text{3}}\text{(g)}$

The reaction's equilibrium constant can thus be written as:

$K_p=\frac{{\left[S{O}_3\right]}^2}{{\left[S{O}_2\right]}^2\times \left[O_2\right]}$

Because all of the compounds involved are gaseous, their concentrations are represented by partial pressures of comparable gases.

$K_p=\frac{p{\left[S{O}_3\right]}^2}{p{\left[S{O}_2\right]}^2\times p\left[O_2\right]}$

Because the container's whole volume is known,

$K_p=\frac{{\left(\frac{n\left(S{O}_3\right)}{V}\right)}^2}{{\left(\frac{n\left(S{O}_2\right)}{V}\right)}^2\times \frac{n\left(O_2\right)}{V}}$

Substituting the known values,

The reaction equation shows that at equilibrium, the amount of${\text{SO}}_{\text{3}}$and${\text{SO}}_2$are the same, thus we can assign

$$\begin{gathered} a=n\left(S{O}_2\right) \\ =n\left(S{O}_3\right) \end{gathered}$$

After that, applying it to the problem and solving it

$$\begin{gathered} 3.18mol/L=\frac{{\left(\frac{a}{2L}\right)}^2}{{\left(\frac{a}{2L}\right)}^2\times \frac{n\left(O_2\right)}{2L}} \\ 3.18mol/L=\frac{1}{\frac{n\left(O_2\right)}{2L}} \\ n{O}_2=\frac{2L}{3.18mol/L} \\ n\left(O_2\right)=0.6289mol \end{gathered}$$

Thus, the amount of ${\text{O}}_{\text{2}}$present is $\text{0.6289}$mol. The equilibrium pressure of oxygen is then,

$$\begin{gathered} p\left(O_2\right)=\frac{n\times R\times T}{V} \\ p\left(O_2\right)=\frac{0.6289mol\times 0.0821\frac{L\times atm}{mol\times K}\times (727+273)K}{2L} \\ p\left(O_2\right)=25.816atm \end{gathered}$$

As the result, at equilibrium, the partial pressure of oxygen is $25.816$atm.

(b) Similarly, the${\text{K}}_{\text{p}}$ is represented as

Since${\text{SO}}_{\text{2}}{\text{:SO}}_{\text{3}}\text{= 95:5}$mole ratio, if assign

$\text{a = n}\left({\text{SO}}_{\text{3}}\right)$

Then,

$n\left(S{O}_2\right)=19\times a$

Adding the values to the equation and allocating them$\text{n}\left({\text{O}}_{\text{2}}\right)\text{= b}$,

$3.18mol/L=\frac{{\left(\frac{19\times a}{2L}\right)}^2}{{\left(\frac{a}{2L}\right)}^2\times \frac{b}{2L}}$

Simplifying,

role="math" localid="1663319078080" $$\begin{gathered} 3.18mol/L\times {\left(\frac{a}{2L}\right)}^2\times \frac{b}{2L}={\left(\frac{19\times a}{2L}\right)}^2 \\ =\frac{3.18mol/L\times a^2\times b}{8L^3} \\ =\frac{361\times a^2}{4L^2} \end{gathered}$$

Crossing out the repeating units, and rearranging,

$$\begin{gathered} b=\frac{361\times 2}{3.18}mol \\ b=227.044mol \end{gathered}$$

Thus, the amount ofgas is. The related oxygen equilibrium pressure is then,

$$\begin{gathered} p\left(O_2\right)=\frac{n\times R\times T}{V} \\ p\left(O_2\right)=\frac{227.044mol\times 0.0821\frac{L\times atm}{mol\times K}\times (727+273)K}{2L} \\ p\left(O_2\right)=9320.156atm \end{gathered}$$

As the result, the oxygen partial pressure is $\text{9320.156 atm}$.