Chapter 13, Problem 54E

At 25°c, K = 0.090 for the reaction

${\text{H}}_2\text{O}\left(g\right)+{\text{Cl}}_2\text{O}\left(g\right)\rightleftharpoons 2\text{HOCI}\left(g\right)$

Calculate the concentrations of all species at equilibrium for each of the following cases.

a. 1.0 g H₂O and 2.0 g Cl₂O are mixed in a 1.0-L flask.

b. 1.0 mole of pure HOCI is placed in a 2.0-L flask.

Interpretation Introduction

Interpretation: The value of the equilibrium constant for the reaction between ${\text{H}}_2\text{O}$ and ${\text{Cl}}_{\text{2}}\text{O}$ and the initial mass of the species involved is given. The equilibrium concentrations of the three gases involved in the system are to be calculated.

Concept introduction: The state when the reactants involved in a chemical reaction and the products formed in the reaction exist in concentrations having no further tendency to change is known as an equilibrium state of the reaction. When the equilibrium constant is expressed in terms of concentration, it is represented as $K$.

To determine: The equilibrium concentrations of the three gases involved in the system

Answer to Problem 54E

The equilibrium concentrations of the three gases are calculated as follows.

Explanation of Solution

Given

The stated reaction is,

${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}{\text{+Cl}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}\rightleftharpoons {\text{2HOCl}}_{\left(\text{g}\right)}$

The initial mass of ${\text{H}}_2{\text{O}}_{\left(g\right)}$ is $1.0\text{g}$.

The initial mass of ${\text{Cl}}_2{\text{O}}_{\left(g\right)}$ is $2.0\text{g}$.

The equilibrium constant $\left(K\right)$ value is $0.090$.

The volume of the container is $1.00\text{L}$.

The molar mass of $\begin{array}{c}{\text{H}}_{\text{2}}\text{O}=2\text{H}+\text{O}\\ =\text{2}\left(1\right)+16\\ =18\text{g}/\text{mol}\end{array}$

The molar mass of $\begin{array}{c}{\text{Cl}}_{\text{2}}\text{O}=2\text{Cl}+\text{O}\\ =\text{2}\left(35.5\right)+16\\ =87\text{g}/\text{mol}\end{array}$

Formula

The number of moles of a substance is calculated by the formula,

$\text{Number}\text{of}\text{moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$

Substitute the values of the given mass and molar mass of ${\text{H}}_{\text{2}}\text{O}$ and ${\text{Cl}}_{\text{2}}\text{O}$ in the above expression.

For ${\text{H}}_{\text{2}}\text{O}$,

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}{\text{H}}_{\text{2}}\text{O}=\frac{\text{1}\text{.0}\text{g}}{\text{18}\text{.0}\text{g}/\text{mol}}\\ =0.056\text{mol}\end{array}$

For ${\text{Cl}}_{\text{2}}\text{O}$,

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}{\text{Cl}}_{\text{2}}\text{O}=\frac{\text{2}\text{.0}\text{g}}{\text{87}\text{.0}\text{g}/\text{mol}}\\ =0.023\text{mol}\end{array}$

The concentration of a reactant is calculated by the formula,

$\text{Concentration}=\frac{\text{Moles}}{\text{Volume}\left(\text{L}\right)}$

The initial concentration of ${\text{H}}_{\text{2}}\text{O}$ is calculated by the formula,

$\text{Concentration}\text{of}{\text{H}}_{\text{2}}\text{O}=\frac{\text{Moles}\text{of}{\text{H}}_{\text{2}}\text{O}}{\text{Volume}\text{of}\text{the}\text{flask}\left(\text{L}\right)}$

Substitute the given values of the number of moles of ${\text{H}}_{\text{2}}\text{O}$ and the volume of the flask in the above expression.

$\begin{array}{c}\text{Concentration}\text{of}{\text{H}}_{\text{2}}\text{O}=\frac{\text{0}\text{.056}\text{moles}}{\text{1}\text{.00}\text{L}}\\ =0.056M\end{array}$

The initial concentration of ${\text{Cl}}_{\text{2}}\text{O}$ is calculated by the formula,

$\text{Concentration}\text{of}{\text{Cl}}_{\text{2}}\text{O}=\frac{\text{Moles}\text{of}{\text{Cl}}_{\text{2}}\text{O}}{\text{Volume}\text{of}\text{the}\text{flask}\left(\text{L}\right)}$

Substitute the given values of the number of moles of ${\text{Cl}}_{\text{2}}\text{O}$ and the volume of the flask in the above expression.

$\begin{array}{c}\text{Concentration}\text{of}{\text{Cl}}_{\text{2}}\text{O}=\frac{\text{0}\text{.023}\text{moles}}{\text{1}\text{.00}\text{L}}\\ =0.023M\end{array}$

To determine: The equilibrium concentrations of the three gases involved in the system.

Explanation:

The concentration of ${\text{H}}_{\text{2}}\text{O}$ and ${\text{Cl}}_{\text{2}}\text{O}$ consumed is assumed to be $x$.

The equilibrium concentrations are represented as,

$\begin{array}{cccccc}& {\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}& \text{+}& {\text{Cl}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}& \rightleftharpoons & {\text{2HOCl}}_{\left(\text{g}\right)}\\ \text{Initial}\text{concentration}& \text{0}\text{.056}& & \text{0}\text{.023}& & \text{0}\\ \text{Change}& \text{-x}& & \text{-x}& & \text{+2x}\\ \text{Equilibrium}\text{concentration}& \text{0}\text{.056-x}& & \text{0}\text{.023-x}& & \text{2x}\end{array}$

At equilibrium, the equilibrium ratio is expressed by the formula,

$K=\frac{\text{Concentration}\text{of}\text{products}}{\text{Concentration}\text{of}\text{reactants}}$

Where,

• $K$ is the equilibrium constant.

The equilibrium ratio for the given reaction is,

$K=\frac{{\left[\text{HOCl}\right]}^2}{\left[{\text{H}}_2\text{O}\right]\left[{\text{Cl}}_2\text{O}\right]}$ (1)

According to the formulated ICE table,

The equilibrium concentration of ${\text{H}}_2{\text{O}}_{\left(g\right)}$ is $\left(0.056-x\right)\text{M}$.

The equilibrium initial concentration of ${\text{Cl}}_2{\text{O}}_{\left(g\right)}$ is $\left(0.023-x\right)\text{M}$.

The equilibrium initial concentration of ${\text{HOCl}}_{\left(g\right)}$ is $\left(2x\right)\text{M}$.

Substitute these values in equation (1).

$\begin{array}{l}K=\frac{{\left[\text{HOCl}\right]}^2}{\left[{\text{H}}_2\text{O}\right]\left[{\text{Cl}}_2\text{O}\right]}\\ K=\frac{{\left[2x\right]}^2}{\left[\text{0}\text{.056}-x\right]\left[\text{0}\text{.023}-x\right]}\end{array}$

The given value of $K$ is $0.090$.

Substitute the value of $K$ in the above expression.

$0.090=\frac{{\left[2x\right]}^2}{\left[\text{0}\text{.056}-x\right]\left[\text{0}\text{.023}-x\right]}$

Simplify the above expression.

$\begin{array}{c}0.090=\frac{{\left[2x\right]}^2}{\left[\text{0}\text{.056}-x\right]\left[\text{0}\text{.023}-x\right]}\\ \left(0.09x^2-0.00711x+0.00011592\right)=4x^2\\ \left(-3.91x^2-0.007x+0.00011592\right)=0\end{array}$

Comparing this with the general quadratic equation, $a{x}^2+bx+c=0$.

In the obtained equation,

• $a$ is $-3.91$.
• $b$ is $-0.007$.
• $c$ is $0.00011592$.

The formula to calculate the value of $x$ is,

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

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

$\begin{array}{l}x=\frac{-\left(-0.007\right)\pm \sqrt{{\left(-0.007\right)}^2-4\left(-3.91\right)\left(0.00011592\right)}}{2\left(-3.91\right)}\\ x=0.0046M\end{array}$

The equilibrium concentration of ${\text{H}}_2{\text{O}}_{\left(g\right)}$ is calculated by the formula,

$\text{Equilibrium}\text{concentration}\text{of}{\text{H}}_2\text{O}\left(g\right)=\left(0.056-x\right)$

Substitute the calculated value of $x$ in the above expression.

$\begin{array}{c}\text{Equilibrium}\text{concentration}\text{of}{\text{H}}_2\text{O}\left(g\right)=\left(0.056-x\right)\\ =\left(0.056-00.0046\right)\text{M}\\ =\underset{\_}{0.0514M}\end{array}$

The equilibrium concentration of ${\text{Cl}}_2{\text{O}}_{\left(g\right)}$ is calculated by the formula,

$\text{Equilibrium}\text{concentration}\text{of}{\text{Cl}}_2{\text{O}}_{\left(g\right)}=\left(0.023-x\right)$

Substitute the calculated value of $x$ in the above expression.

$\begin{array}{c}\text{Equilibrium}\text{concentration}\text{of}{\text{Cl}}_2{\text{O}}_{\left(g\right)}=\left(0.023-x\right)\\ =\left(0.023-0.0046\right)\text{M}\\ =\underset{\_}{0.0184M}\end{array}$

The equilibrium concentration of ${\text{HOCl}}_{\left(g\right)}$ is calculated by the formula,

$\text{Equilibrium}\text{concentration}\text{of}{\text{HOCl}}_{\left(g\right)}=\left(2x\right)$

Substitute the calculated value of $x$ in the above expression.

$\begin{array}{c}\text{Equilibrium}\text{concentration}\text{of}{\text{HOCl}}_{\left(g\right)}=\left(2x\right)\\ =\left(2\left(0.0046\right)\right)\text{M}\\ =\underset{\_}{0.0092M}\end{array}$

Conclusion

The equilibrium concentration of ${\text{H}}_2{\text{O}}_{\left(g\right)}$ $\underset{\_}{0.0514M}$ and ${\text{Cl}}_2{\text{O}}_{\left(g\right)}$ is $\underset{\_}{0.0184M}$ and of ${\text{HOCl}}_{\left(g\right)}$ is $\underset{\_}{0.0092M}$.

Interpretation Introduction

Interpretation: The value of the equilibrium constant for the reaction between ${\text{H}}_2\text{O}$ and ${\text{Cl}}_{\text{2}}\text{O}$ and the initial mass of the species involved is given. The equilibrium concentrations of the three gases involved in the system are to be calculated.

Concept introduction: The state when the reactants involved in a chemical reaction and the products formed in the reaction exist in concentrations having no further tendency to change is known as an equilibrium state of the reaction. When the equilibrium constant is expressed in terms of concentration, it is represented as $K$.

To determine: The equilibrium concentrations of the three gases involved in the system.

Answer to Problem 54E

The equilibrium concentrations of the three gases are calculated as follows.

Explanation of Solution

Given

The stated reaction is,

$2{\text{HOCl}}_{\begin{array}{l}\left(g\right)\\ \end{array}}\rightleftharpoons {\text{H}}_2{\text{O}}_{\left(g\right)}+{\text{Cl}}_2{\text{O}}_{\left(g\right)}$

The initial moles of ${\text{HOCl}}_{\left(g\right)}$ are $1.0\text{mole}$.

The value of $K$ for this reaction will be $\frac{1}{0.090}$.

The volume of the container is $2.0\text{L}$.

The concentration of a reactant is calculated by the formula,

$\text{Concentration}=\frac{\text{Moles}}{\text{Volume}\left(\text{L}\right)}$

The initial concentration of $\text{HOCl}$ is calculated by the formula,

$\text{Concentration}\text{of}\text{HOCl}=\frac{\text{Moles}\text{of}\text{HOCl}}{\text{Volume}\text{of}\text{the}\text{flask}\left(\text{L}\right)}$

Substitute the given values of the number of moles of $\text{HOCl}$ and the volume of the flask in the above expression.

$\begin{array}{c}\text{Concentration}\text{of}\text{HOCl}=\frac{\text{1}\text{.0}\text{mole}}{\text{2}\text{.00}\text{L}}\\ =0.5M\end{array}$

The concentration of $\text{HOCl}$ consumed is assumed to be $2x$.

The equilibrium concentrations are represented as,

$\begin{array}{cccccc}& {\text{2HOCl}}_{\left(\text{g}\right)}& \rightleftharpoons & {\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}& \text{+}& {\text{Cl}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}\\ \text{Initial}\text{concentration}& \text{0}\text{.5}& & \text{0}& & \text{0}\\ \text{Change}& \text{-2x}& & \text{+x}& & \text{+x}\\ \text{Equilibrium}\text{concentration}& \text{0}\text{.5-2x}& & \text{x}& & \text{x}\end{array}$

At equilibrium, the equilibrium ratio is expressed by the formula,

$K=\frac{\text{Concentration}\text{of}\text{products}}{\text{Concentration}\text{of}\text{reactants}}$

Where,

• $K$ is the equilibrium constant.

The equilibrium ratio for the given reaction is,

$K=\frac{\left[{\text{H}}_2\text{O}\right]\left[{\text{Cl}}_2\text{O}\right]}{{\left[\text{HOCl}\right]}^2}$ (1)

According to the formulated ICE table,

The equilibrium concentration of ${\text{H}}_2{\text{O}}_{\left(g\right)}$ is $\left(x\right)\text{M}$.

The equilibrium initial concentration of ${\text{Cl}}_2{\text{O}}_{\left(g\right)}$ is $\left(x\right)\text{M}$.

The equilibrium initial concentration of ${\text{HOCl}}_{\left(g\right)}$ is $\left(0.5-2x\right)\text{M}$.

Substitute these values in equation (1).

$\begin{array}{l}K=\frac{\left[{\text{H}}_2\text{O}\right]\left[{\text{Cl}}_2\text{O}\right]}{{\left[\text{HOCl}\right]}^2}\\ K=\frac{\left[x\right]\left[x\right]}{{\left[0.5-2x\right]}^2}\end{array}$

The value of $K$ for this reaction will be $\frac{1}{0.090}$.

Substitute the value of $K$ in the above expression.

$\frac{1}{0.090}=\frac{{\left[x\right]}^2}{{\left[0.5-2x\right]}^2}$

Taking square root on both sides,

$\frac{1}{0.3}=\frac{\left[x\right]}{\left[0.5-2x\right]}$

Simplify the above expression.

$\begin{array}{c}0.3x=0.5-2x\\ x=0.217\end{array}$

The equilibrium concentration of ${\text{H}}_2{\text{O}}_{\left(g\right)}$ is equal to the equilibrium concentration of ${\text{Cl}}_2\text{O}\left(g\right)$. These are equal to $x$.

The equilibrium concentration of ${\text{H}}_2{\text{O}}_{\left(g\right)}$ is equal to the equilibrium concentration of ${\text{Cl}}_2\text{O}\left(g\right)$, that is, $\underset{\_}{0.217M}$.

The equilibrium concentration of ${\text{HOCl}}_{\left(g\right)}$ is calculated by the formula,

$\text{Equilibrium}\text{concentration}\text{of}{\text{HOCl}}_{\left(g\right)}=\left(0.5-2x\right)$

Substitute the calculated value of $x$ in the above expression.

$\begin{array}{c}\text{Equilibrium}\text{concentration}\text{of}{\text{HOCl}}_{\left(g\right)}=\left(0.5-2x\right)\\ =\left(0.5-2\left(0.217\right)\right)\text{M}\\ =\underset{\_}{0.066M}\end{array}$

Conclusion

The equilibrium concentration of ${\text{H}}_2{\text{O}}_{\left(g\right)}$ and ${\text{Cl}}_2{\text{O}}_{\left(g\right)}$ is $\underset{\_}{0.217M}$ and of ${\text{HOCl}}_{\left(g\right)}$ is $\underset{\_}{0.066M}$