Chapter 5, Problem 5.63E

(a)

Interpretation Introduction

Interpretation:

The equilibrium concentration of all species if the initial concentration of ${\text{SO}}_{\text{3}}$ is $0.150\text{atm}$ is to be calculated.

Concept introduction:

The static equilibrium is defined as a process in which the rate of forward reaction or the rate of backward reaction is zero. On the other hand, in dynamic equilibrium, the rate of forward and backward reaction is equal.

Answer to Problem 5.63E

The concentration of ${\text{SO}}_{\text{3}}$, ${\text{S}}_2$ and ${\text{O}}_{\text{2}}$ is $0.056\text{atm}$, $0.047\text{atm}$ and $0.141\text{atm}$, respectively.

Explanation of Solution

The equilibrium constant of the reaction is $4.33\times {10}^{-2}$.

The given balanced chemical reaction is,

$2{\text{SO}}_{\text{3}}\left(\text{g}\right)\to {\text{S}}_{\text{2}}\left(\text{g}\right)+3{\text{O}}_2\left(\text{g}\right)$

The equilibrium constant for the above reaction is expressed as,

$K_p=\frac{p_{{\text{S}}_2}\times p_{{\text{O}}_2}^3}{p_{{\text{SO}}_3}^2}$ (1)

Where,

• $p_{{\text{s}}_2}$ is the partial pressure of ${\text{S}}_{\text{2}}$ at equilibrium.

• $p_{{\text{O}}_2}$ is the partial pressure of ${\text{O}}_{\text{2}}$ at equilibrium.

• $p_{{\text{SO}}_3}$ is the partial pressure of ${\text{SO}}_{\text{3}}$ at equilibrium.

The ICE table for the given reaction is expressed as,

$\begin{array}{cccccc}& 2{\text{SO}}_3\left(\text{g}\right)& \rightleftharpoons & {\text{S}}_{\text{2}}\left(\text{g}\right)& +& {\text{3O}}_{\text{2}}\left(\text{g}\right)\\ \text{initial}& 0.150\text{atm}& & 0& & 0\\ \text{change}& -2x& & +x& & +3x\\ \text{equilibrium}& 0.150-2x& & x& & 3x\end{array}$ (2)

Substitute the equilibrium concentrations of ${\text{S}}_{\text{2}}$, ${\text{O}}_{\text{2}}$ and ${\text{SO}}_{\text{3}}$ and value of equilibrium constant in equation (1).

$\begin{array}{rll}4.33\times {10}^{-2}& =& \frac{x\times {\left(3x\right)}^3}{{\left(0.150-2x\right)}^2}\\ & =& \frac{27x^4}{{\left(0.150-2x\right)}^2}\end{array}$

Take square root on both sides.

$\begin{array}{rll}\sqrt{4.33\times {10}^{-2}}& =& \sqrt{\frac{27x^4}{{\left(0.150-2x\right)}^2}}\\ 0.208& =& \frac{5.20x^2}{0.150-2x}\\ 5.20x^2+0.416x-0.0312& =& 0\end{array}$

Solve this quadratic equation by the formula,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Where,

• $a$ is the coefficient of $x^2$.

• $b$ is the coefficient of $x$.

• $c$ is the coefficient of $1$.

Substitute the values of $a$, $b$ and $c$ in the above formula.

$\begin{array}{rll}x& =& \frac{-\left(0.416\right)\pm \sqrt{{\left(0.416\right)}^2-4\times 5.20\times \left(-0.0312\right)}}{2\times 5.20}\\ & =& \frac{-0.416\pm 0.9066}{10.4}\\ & =& \frac{0.4906}{10.4}\\ & =& 0.047\end{array}$

The value of $x$ is $0.047$.

Substitute the value of $x$ in equation (2).

$\begin{array}{ll}\text{The concentration of }{\text{SO}}_{\text{3}}\text{ is}& =0.150-2x\\ & =0.150-2\left(0.047\right)\\ & =0.056\text{atm}\end{array}$

$\begin{array}{ll}\text{The concentration of }{\text{S}}_2\text{ is}& =x\\ & =0.047\text{atm}\end{array}$

$\begin{array}{ll}\text{The concentration of }{\text{O}}_{\text{2}}\text{ is}& =3x\\ & =3\left(0.047\right)\\ & =0.141\text{atm}\end{array}$

Conclusion

The concentration of ${\text{SO}}_{\text{3}}$, ${\text{S}}_2$ and ${\text{O}}_{\text{2}}$ is $0.056\text{atm}$, $0.047\text{atm}$ and $0.141\text{atm}$, respectively.

(b)

Interpretation Introduction

Interpretation:

The equilibrium concentration of all species if the initial concentration of ${\text{SO}}_{\text{3}}$ is $0.100\text{atm}$ is to be calculated.

Concept introduction:

The static equilibrium is defined as a process in which the rate of forward reaction or the rate of backward reaction is zero. On the other hand in dynamic equilibrium, the rate of forward and backward reaction is equal.

Answer to Problem 5.63E

The concentration of ${\text{SO}}_{\text{3}}$, ${\text{S}}_2$ and ${\text{O}}_{\text{2}}$ is $0.0069\text{atm}$, $0.0348\text{atm}$ and $0.104\text{atm}$, respectively.

Explanation of Solution

The equilibrium constant of the reaction is $4.33\times {10}^{-2}$.

The given balanced chemical reaction is,

$2{\text{SO}}_{\text{3}}\left(\text{g}\right)\to {\text{S}}_{\text{2}}\left(\text{g}\right)+3{\text{O}}_2\left(\text{g}\right)$

The equilibrium constant for the above reaction is expressed as,

$K_p=\frac{p_{{\text{S}}_2}\times p_{{\text{O}}_2}^3}{p_{{\text{SO}}_3}^2}$ (1)

Where,

• $p_{{\text{S}}_2}$ is the partial pressure of ${\text{S}}_{\text{2}}$ at equilibrium.

• $p_{{\text{O}}_2}$ is the partial pressure of ${\text{O}}_{\text{2}}$ at equilibrium.

• $p_{{\text{SO}}_3}$ is the partial pressure of ${\text{SO}}_{\text{3}}$ at equilibrium.

The ICE table for the given reaction is expressed as,

$\begin{array}{cccccc}& 2{\text{SO}}_3\left(\text{g}\right)& \rightleftharpoons & {\text{S}}_{\text{2}}\left(\text{g}\right)& +& {\text{3O}}_{\text{2}}\left(\text{g}\right)\\ \text{initial}& 0.100\text{atm}& & 0& & 0\\ \text{change}& -2x& & +x& & +3x\\ \text{equilibrium}& 0.100-2x& & x& & 3x\end{array}$ (3)

Substitute the equilibrium concentrations of ${\text{S}}_{\text{2}}$, ${\text{O}}_{\text{2}}$ and ${\text{SO}}_{\text{3}}$ and value of equilibrium constant in equation (1).

$\begin{array}{rll}4.33\times {10}^{-2}& =& \frac{x\times {\left(3x\right)}^3}{{\left(0.100-2x\right)}^2}\\ & =& \frac{27x^4}{{\left(0.100-2x\right)}^2}\end{array}$

Take square root on both sides.

$\begin{array}{rll}\sqrt{4.33\times {10}^{-2}}& =& \sqrt{\frac{27x^4}{{\left(0.100-2x\right)}^2}}\\ 0.208& =& \frac{5.20x^2}{0.100-2x}\\ 5.20x^2+0.416x-0.0208& =& 0\end{array}$

Solve this quadratic equation by the formula,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Where,

• $a$ is the coefficient of $x^2$.

• $b$ is the coefficient of $x$.

• $c$ is the coefficient of $1$.

Substitute the values of $a$, $b$ and $c$ in the above formula.

$\begin{array}{rll}x& =& \frac{-\left(0.416\right)\pm \sqrt{{\left(0.416\right)}^2-4\times 5.20\times \left(-0.0208\right)}}{2\times 5.20}\\ & =& \frac{-0.416\pm 0.778}{10.4}\\ & =& \frac{0.362}{10.4}\\ & =& 0.0348\end{array}$

The value of $x$ is $0.0348$.

Substitute the value of $x$ in equation (3).

$\begin{array}{ll}\text{The concentration of }{\text{SO}}_{\text{3}}\text{ is}& =0.100-2x\\ & =0.100-2\left(0.0348\right)\\ & =0.0069\text{atm}\end{array}$

$\begin{array}{ll}\text{The concentration of }{\text{S}}_2\text{ is}& =x\\ & =0.0348\text{atm}\end{array}$

$\begin{array}{ll}\text{The concentration of }{\text{O}}_{\text{2}}\text{ is}& =3x\\ & =3\left(0.0348\right)\\ & =0.104\text{atm}\end{array}$

Conclusion

The concentration of ${\text{SO}}_{\text{3}}$, ${\text{S}}_2$ and ${\text{O}}_{\text{2}}$ is $0.0069\text{atm}$, $0.0348\text{atm}$ and $0.104\text{atm}$, respectively.

(c)

Interpretation Introduction

Interpretation:

The equilibrium concentration of all species if the initial concentration of ${\text{SO}}_{\text{3}}$ is $0.001\text{atm}$ is to be calculated.

Concept introduction:

The static equilibrium is defined as a process in which the rate of forward reaction or the rate of backward reaction is zero. On the other hand in dynamic equilibrium, the rate of forward and backward reaction is equal.

Answer to Problem 5.63E

The concentration of ${\text{SO}}_{\text{3}}$, ${\text{S}}_2$ and ${\text{O}}_{\text{2}}$ is $2\times {10}^{-5}\text{atm}$, $4.9\times {10}^{-4}\text{atm}$ and $1.47\times {10}^{-3}\text{atm}$, respectively.

Explanation of Solution

The equilibrium constant of the reaction is $4.33\times {10}^{-2}$.

The given balanced chemical reaction is,

$2{\text{SO}}_{\text{3}}\left(\text{g}\right)\to {\text{S}}_{\text{2}}\left(\text{g}\right)+3{\text{O}}_2\left(\text{g}\right)$

The equilibrium constant for the above reaction is expressed as,

$K_p=\frac{p_{{\text{S}}_2}\times p_{{\text{O}}_2}^3}{p_{{\text{SO}}_3}^2}$ (1)

Where,

• $p_{{\text{S}}_2}$ is the partial pressure of ${\text{S}}_{\text{2}}$ at equilibrium.

• $p_{{\text{O}}_2}$ is the partial pressure of ${\text{O}}_{\text{2}}$ at equilibrium.

• $p_{{\text{SO}}_3}$ is the partial pressure of ${\text{SO}}_{\text{3}}$ at equilibrium.

The ICE table for the given reaction is expressed as,

$\begin{array}{cccccc}& 2{\text{SO}}_3\left(\text{g}\right)& \rightleftharpoons & {\text{S}}_{\text{2}}\left(\text{g}\right)& +& {\text{3O}}_{\text{2}}\left(\text{g}\right)\\ \text{initial}& 0.001\text{atm}& & 0& & 0\\ \text{change}& -2x& & +x& & +3x\\ \text{equilibrium}& 0.001-2x& & x& & 3x\end{array}$ (4)

Substitute the equilibrium concentrations of ${\text{S}}_{\text{2}}$, ${\text{O}}_{\text{2}}$ and ${\text{SO}}_{\text{3}}$ and value of equilibrium constant in equation (1).

$\begin{array}{rll}4.33\times {10}^{-2}& =& \frac{x\times {\left(3x\right)}^3}{{\left(0.001-2x\right)}^2}\\ & =& \frac{27x^4}{{\left(0.001-2x\right)}^2}\end{array}$

Take square root on both sides.

$\begin{array}{rll}\sqrt{4.33\times {10}^{-2}}& =& \sqrt{\frac{27x^4}{{\left(0.001-2x\right)}^2}}\\ 0.208& =& \frac{5.20x^2}{0.001-2x}\\ 5.20x^2+0.416x-2.08\times {10}^{-4}& =& 0\end{array}$

Solve this quadratic equation by the formula,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Where,

• $a$ is the coefficient of $x^2$.

• $b$ is the coefficient of $x$.

• $c$ is the coefficient of $1$.

Substitute the values of $a$, $b$ and $c$ in the above formula.

$\begin{array}{rll}x& =& \frac{-\left(0.416\right)\pm \sqrt{{\left(0.416\right)}^2-4\times 5.20\times \left(-0.000208\right)}}{2\times 5.20}\\ & =& \frac{-0.416\pm 0.4211}{10.4}\\ & =& \frac{5.1\times {10}^{-3}}{10.4}\\ & =& 4.9\times {10}^{-4}\end{array}$

The value of $x$ is $4.9\times {10}^{-4}$.

Substitute the value of $x$ in equation (4).

$\begin{array}{ll}\text{The concentration of }{\text{SO}}_{\text{3}}\text{ is}& =0.001-2x\\ & =0.001-2\left(4.9\times {10}^{-4}\right)\\ & =2\times {10}^{-5}\text{atm}\end{array}$

$\begin{array}{ll}\text{The concentration of }{\text{S}}_2\text{ is}& =x\\ & =4.9\times {10}^{-4}\text{atm}\end{array}$

$\begin{array}{ll}\text{The concentration of }{\text{O}}_{\text{2}}\text{ is}& =3x\\ & =3\left(4.9\times {10}^{-4}\right)\\ & =1.47\times {10}^{-3}\text{atm}\end{array}$

Conclusion

The concentration of ${\text{SO}}_{\text{3}}$, ${\text{S}}_2$ and ${\text{O}}_{\text{2}}$ is $2\times {10}^{-5}\text{atm}$, $4.9\times {10}^{-4}\text{atm}$ and $1.47\times {10}^{-3}\text{atm}$, respectively.