Chapter 14, Problem 14.34QE

(a)

Interpretation Introduction

Interpretation:

If $0.50\text{ mol}$ of $\text{CO}$, $0.40\text{ mol}$ of ${\text{H}}_2\text{O}$, $0.80\text{ mol}$ of ${\text{CO}}_2$, and $0.90\text{ mol}$ of ${\text{H}}_2$ is place in $1.0\text{ L}$ container then the direction of the following reaction has to be determined.

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

Concept Introduction:

The condition of equilibrium is a state of balance of processes that runs in opposite directions. At equilibrium, the formation of a product from the reactant balances the formation of reactant from the product. Also, the change in concentration of reaction and product seems to be negligible at the equilibrium state.

A reaction quotient $Q$ is an algebraic form of equilibrium constant $\left(K_{\text{eq}}\right)$ for all concentrations including reaction concentrations at equilibrium. The reaction quotient $Q$ can also help in prediction of direction of reaction when compared with the equilibrium constant $\left(K_{\text{eq}}\right)$.

The general equilibrium reaction is as follows:

  $a\text{A}+b\text{B}\rightleftharpoons c\text{C}+d\text{D}$

Here,

$\text{A}$ and $\text{B}$ are the reactants.

$\text{C}$ and $\text{D}$ are products.

$a$ and $b$ are the stoichiometric coefficients of reactants.

$c$ and $d$ are the stoichiometric coefficients of products.

The expression of the reaction quotient for the above reaction is as follows:

  $Q=\frac{{\left[\text{C}\right]}^c{\left[\text{D}\right]}^d}{{\left[\text{A}\right]}^a{\left[\text{B}\right]}^b}$

Here,

$K_{\text{c}}$ is the equilibrium constant.

$\left[\text{C}\right]$ is the concentration of $\text{C}$.

$\left[\text{D}\right]$ is the concentration of $\text{D}$.

$\left[\text{A}\right]$ is the concentration of $\text{A}$.

$\left[\text{B}\right]$ is the concentration of $\text{B}$.

The concentration of reactants and products changes in order to bring reaction quotient and equilibrium constant closer. Therefore, the direction of reaction can be predicted as follows:

(1) If $Q$ is less than $K_{\text{eq}}$ then the reaction moves to increase the concentration of products, which means reaction moves towards right direction.

(2)If $Q$ is greater than $K_{\text{eq}}$ then the reaction moves to decrease the concentration of products, that means reaction moves towards the left direction.

(3)If $Q$ is equal to $K_{\text{eq}}$ then the reaction is in equilibrium.

Explanation of Solution

The given reaction occurs is as follows:

  $\text{CO}\left(\text{g}\right)+{\text{H}}_2\text{O}\left(\text{g}\right)\rightleftharpoons {\text{CO}}_{\text{2}}\left(\text{g}\right){\text{+H}}_2\left(\text{g}\right)$        (1)

The concentration of $\text{CO}$ can be calculated as follows:

  $\begin{array}{c}\left[\text{CO}\right]=\left(\frac{\text{0}\text{.50 mol}}{1.0\text{ L}}\right)\\ =5.0\times {10}^{-1}M\end{array}$

The concentration of ${\text{H}}_2\text{O}$ can be calculated as follows:

  $\begin{array}{c}\left[{\text{H}}_2\text{O}\right]=\left(\frac{\text{0}\text{.40 mol}}{1.0\text{ L}}\right)\\ =4.0\times {10}^{-1}M\end{array}$

The concentration of ${\text{CO}}_2$ can be calculated as follows:

  $\begin{array}{c}\left[{\text{CO}}_2\right]=\left(\frac{\text{0}\text{.80 mol}}{1.0\text{ L}}\right)\\ =8.0\times {10}^{-1}M\end{array}$

The concentration of ${\text{H}}_2$ can be calculated as follows:

  $\begin{array}{c}\left[{\text{H}}_2\right]=\left(\frac{\text{0}\text{.90 mol}}{1.0\text{ L}}\right)\\ =9.0\times {10}^{-1}M\end{array}$

The expression to calculate $Q$ for the chemical equation (1) is as follows:

  $Q=\frac{\left[{\text{CO}}_2\right]\left[{\text{H}}_2\right]}{\left[\text{CO}\right]\left[{\text{H}}_2\text{O}\right]}$        (2)

Substitute $5.0\times {10}^{-1}M$ for $\left[\text{CO}\right]$, $4.0\times {10}^{-1}M$ for $\left[{\text{H}}_2\text{O}\right]$, $8.0\times {10}^{-1}M$ for $\left[{\text{CO}}_2\right]$,and $9.0\times {10}^{-1}M$ for $\left[{\text{H}}_2\right]$ in equation (2).

  $\begin{array}{c}Q=\frac{\left(8.0\times {10}^{-1}M\right)\left(9.0\times {10}^{-1}M\right)}{\left(5.0\times {10}^{-1}M\right)\left(4.0\times {10}^{-1}M\right)}\\ =3.6\end{array}$

The value of $K_{\text{c}}$ for a given reaction is $5.0$. Since the value of $Q$ is less than the value of $K_{\text{c}}$, therefore, the given reaction proceeds in a forward or right direction.

(b)

Interpretation Introduction

Interpretation:

If $0.01\text{ mol}$ of $\text{CO}$, $0.02\text{mol}$ of ${\text{H}}_2\text{O}$, $0.03\text{ mol}$ of ${\text{CO}}_2$, and $0.04\text{ mol}$ of ${\text{H}}_2$ is place in $1.0\text{ L}$ container then the direction of the following reaction has to be determined.

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

Concept Introduction:

Refer to part (a).

Explanation of Solution

The given reaction occurs is as follows:

  $\text{CO}\left(\text{g}\right)+{\text{H}}_2\text{O}\left(\text{g}\right)\rightleftharpoons {\text{CO}}_{\text{2}}\left(\text{g}\right){\text{+H}}_2\left(\text{g}\right)$        (1)

The concentration of $\text{CO}$ can be calculated as follows:

  $\begin{array}{c}\left[\text{CO}\right]=\left(\frac{\text{0}\text{.01 mol}}{1.0\text{ L}}\right)\\ =1\times {10}^{-2}M\end{array}$

The concentration of ${\text{H}}_2\text{O}$ can be calculated as follows:

  $\begin{array}{c}\left[{\text{H}}_2\text{O}\right]=\left(\frac{\text{0}\text{.02 mol}}{1.0\text{ L}}\right)\\ =2\times {10}^{-2}M\end{array}$

The concentration of ${\text{CO}}_2$ can be calculated as follows:

  $\begin{array}{c}\left[{\text{CO}}_2\right]=\left(\frac{\text{0}\text{.03 mol}}{1.0\text{ L}}\right)\\ =3\times {10}^{-2}M\end{array}$

The concentration of ${\text{H}}_2$ can be calculated as follows:

  $\begin{array}{c}\left[{\text{H}}_2\right]=\left(\frac{\text{0}\text{.04 mol}}{1.0\text{ L}}\right)\\ =4\times {10}^{-2}M\end{array}$

The expression to calculate $Q$ for the chemical equation (1) is as follows:

  $Q=\frac{\left[{\text{CO}}_2\right]\left[{\text{H}}_2\right]}{\left[\text{CO}\right]\left[{\text{H}}_2\text{O}\right]}$        (2)

Substitute $1\times {10}^{-2}M$ for $\left[\text{CO}\right]$, $2\times {10}^{-2}M$ for $\left[{\text{H}}_2\text{O}\right]$, $3\times {10}^{-2}M$ for $\left[{\text{CO}}_2\right]$, and $4\times {10}^{-2}M$ for $\left[{\text{H}}_2\right]$ in equation (2).

  $\begin{array}{c}Q=\frac{\left(3\times {10}^{-2}M\right)\left(4\times {10}^{-2}M\right)}{\left(1\times {10}^{-2}M\right)\left(2\times {10}^{-2}M\right)}\\ =6\end{array}$

The value of $K_{\text{c}}$ for a given reaction is $5.0$. Since the value of $Q$ is greater than the value of $K_{\text{c}}$, therefore, the given reaction proceeds in a backward or left direction.

(c)

Interpretation Introduction

Interpretation:

If $1.22\text{ mol}$ of $\text{CO}$, $1.22\text{ mol}$ of ${\text{H}}_2\text{O}$, $2.78\text{ mol}$ of ${\text{CO}}_2$, and $2.78\text{ mol}$ of ${\text{H}}_2$ is place in $1.0\text{ L}$ container then the direction of the following reaction has to be determined.

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

Concept Introduction:

Refer to part (a).

Explanation of Solution

The given reaction occurs is as follows:

  $\text{CO}\left(\text{g}\right)+{\text{H}}_2\text{O}\left(\text{g}\right)\rightleftharpoons {\text{CO}}_{\text{2}}\left(\text{g}\right){\text{+H}}_2\left(\text{g}\right)$        (1)

The concentration of $\text{CO}$ can be calculated as follows:

  $\begin{array}{c}\left[\text{CO}\right]=\left(\frac{\text{1}\text{.22 mol}}{1.0\text{ L}}\right)\\ =1.22M\end{array}$

The concentration of ${\text{H}}_2\text{O}$ can be calculated as follows:

  $\begin{array}{c}\left[{\text{H}}_2\text{O}\right]=\left(\frac{\text{1}\text{.22 mol}}{1.0\text{ L}}\right)\\ =1.22M\end{array}$

The concentration of ${\text{CO}}_2$ can be calculated as follows:

  $\begin{array}{c}\left[{\text{CO}}_2\right]=\left(\frac{\text{2}\text{.78 mol}}{1.0\text{ L}}\right)\\ =2.78M\end{array}$

The concentration of ${\text{H}}_2$ can be calculated as follows:

  $\begin{array}{c}\left[{\text{H}}_2\right]=\left(\frac{\text{2}\text{.78 mol}}{1.0\text{ L}}\right)\\ =2.78M\end{array}$

The expression to calculate $Q$ for the chemical equation (1) is as follows:

  $Q=\frac{\left[{\text{CO}}_2\right]\left[{\text{H}}_2\right]}{\left[\text{CO}\right]\left[{\text{H}}_2\text{O}\right]}$        (2)

Substitute $\text{1}\text{.22 }M$ for $\left[\text{CO}\right]$, $\text{1}\text{.22 }M$ for $\left[{\text{H}}_2\text{O}\right]$, $\text{2}\text{.78 }M$ for $\left[{\text{CO}}_2\right]$, and $\text{2}\text{.78 }M$ for $\left[{\text{H}}_2\right]$ in equation (2).

  $\begin{array}{c}Q=\frac{\left(\text{2}\text{.78 }M\right)\left(\text{2}\text{.78 }M\right)}{\left(\text{1}\text{.22 }M\right)\left(\text{1}\text{.22 }M\right)}\\ =5.19\end{array}$

The value of $K_{\text{c}}$ for a given reaction is $5.0$. Since the value of $Q$ is greater than the value of $K_{\text{c}}$, therefore, the given reaction proceeds in a backward or left direction.

(d)

Interpretation Introduction

Interpretation:

If $0.61\text{ mol}$ of $\text{CO}$, $1.22\text{ mol}$ of ${\text{H}}_2\text{O}$, $1.39\text{ mol}$ of ${\text{CO}}_2$, and $2.39\text{ mol}$ of ${\text{H}}_2$ is place in $1.0\text{ L}$ container then the direction of the following reaction has to be determined.

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

Concept Introduction:

Refer to part (a).

Explanation of Solution

The given reaction occurs is as follows:

  $\text{CO}\left(\text{g}\right)+{\text{H}}_2\text{O}\left(\text{g}\right)\rightleftharpoons {\text{CO}}_{\text{2}}\left(\text{g}\right){\text{+H}}_2\left(\text{g}\right)$        (1)

The concentration of $\text{CO}$ can be calculated as follows:

  $\begin{array}{c}\left[\text{CO}\right]=\left(\frac{\text{0}\text{.61 mol}}{1.0\text{ L}}\right)\\ =0.61M\end{array}$

The concentration of ${\text{H}}_2\text{O}$ can be calculated as follows:

  $\begin{array}{c}\left[{\text{H}}_2\text{O}\right]=\left(\frac{\text{1}\text{.22 mol}}{1.0\text{ L}}\right)\\ =1.22M\end{array}$

The concentration of ${\text{CO}}_2$ can be calculated as follows:

  $\begin{array}{c}\left[{\text{CO}}_2\right]=\left(\frac{\text{1}\text{.39 mol}}{1.0\text{ L}}\right)\\ =1.39M\end{array}$

The concentration of ${\text{H}}_2$ can be calculated as follows:

  $\begin{array}{c}\left[{\text{H}}_2\right]=\left(\frac{\text{2}\text{.39 mol}}{1.0\text{ L}}\right)\\ =2.39M\end{array}$

The expression to calculate $Q$ for the chemical equation (1) is as follows:

  $Q=\frac{\left[{\text{CO}}_2\right]\left[{\text{H}}_2\right]}{\left[\text{CO}\right]\left[{\text{H}}_2\text{O}\right]}$        (2)

Substitute $\text{0}\text{.61 }M$ for $\left[\text{CO}\right]$, $\text{1}\text{.22 }M$ for $\left[{\text{H}}_2\text{O}\right]$, $\text{1}\text{.39 }M$ for $\left[{\text{CO}}_2\right]$, and $\text{2}\text{.39 }M$ for $\left[{\text{H}}_2\right]$ in equation (2).

  $\begin{array}{c}Q=\frac{\left(\text{1}\text{.39 }M\right)\left(\text{2}\text{.39 }M\right)}{\left(\text{0}\text{.61 }M\right)\left(\text{1}\text{.22 }M\right)}\\ =8.9\end{array}$

The value of $K_{\text{c}}$ for a given reaction is $5.0$. Since the value of $Q$ is greater than the value of $K_{\text{c}}$, therefore, the given reaction proceeds in a backward or left direction.