Chapter 14, Problem 14.33QE

(a)

Interpretation Introduction

Interpretation:

If $0.005\text{ mmol}$ of $\text{HI}$, $0.020\text{ mmol}$ of ${\text{H}}_2$, and $0.010\text{ mmol}$ of ${\text{I}}_2$ is place in $1.0\text{ L}$ container then thedirection of the following reaction has to be determined.

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

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:

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

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

  $\begin{array}{c}\left[\text{HI}\right]=\left(\frac{\text{0}\text{.005 mmol}}{1.0\text{ L}}\right)\left(\frac{1.0\times {10}^{-3}\text{ mol}}{1\text{ mmol}}\right)\\ =5.0\times {10}^{-6}M\end{array}$

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

  $\begin{array}{c}\left[{\text{H}}_2\right]=\left(\frac{\text{0}\text{.020 mmol}}{1.0\text{ L}}\right)\left(\frac{1.0\times {10}^{-3}\text{ mol}}{1\text{ mmol}}\right)\\ =2.0\times {10}^{-5}M\end{array}$

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

  $\begin{array}{c}\left[{\text{I}}_2\right]=\left(\frac{\text{0}\text{.010 mmol}}{1.0\text{ L}}\right)\left(\frac{1.0\times {10}^{-3}\text{ mol}}{1\text{ mmol}}\right)\\ =1.0\times {10}^{-5}M\end{array}$

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

  $Q=\frac{\left[{\text{H}}_2\right]\left[{\text{I}}_2\right]}{{\left[\text{HI}\right]}^2}$        (2)

Substitute $2.0\times {10}^{-5}M$ for $\left[{\text{H}}_2\right]$, $1.0\times {10}^{-5}M$ for $\left[{\text{I}}_2\right]$, and $5.0\times {10}^{-6}M$ for $\left[\text{HI}\right]$ in equation (2).

  $\begin{array}{c}Q=\frac{\left(2.0\times {10}^{-5}M\right)\left(1.0\times {10}^{-5}M\right)}{{\left(5.0\times {10}^{-6}M\right)}^2}\\ =8.0\end{array}$

The value of $K_{\text{c}}$ for a given reaction is $0.010$. 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.

(b)

Interpretation Introduction

Interpretation:

If $0.020\text{ mmol}$ of $\text{HI}$, $0.20\text{ mmol}$ of ${\text{H}}_2$, and $0.20\text{ mmol}$ of ${\text{I}}_2$ is place in $1.0\text{ L}$ container then the direction of the following reaction has to be determined.

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

Concept Introduction:

Refer to part (a).

Explanation of Solution

The given reaction occurs is as follows:

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

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

  $\begin{array}{c}\left[\text{HI}\right]=\left(\frac{0.020\text{ mmol}}{1.0\text{ L}}\right)\left(\frac{1.0\times {10}^{-3}\text{ mol}}{1\text{ mmol}}\right)\\ =2.0\times {10}^{-5}M\end{array}$

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

  $\begin{array}{c}\left[{\text{H}}_2\right]=\left(\frac{\text{0}\text{.20 mmol}}{1.0\text{ L}}\right)\left(\frac{1.0\times {10}^{-3}\text{ mol}}{1\text{ mmol}}\right)\\ =2.0\times {10}^{-4}M\end{array}$

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

  $\begin{array}{c}\left[{\text{I}}_2\right]=\left(\frac{\text{0}\text{.20 mmol}}{1.0\text{ L}}\right)\left(\frac{1.0\times {10}^{-3}\text{ mol}}{1\text{ mmol}}\right)\\ =2.0\times {10}^{-4}M\end{array}$

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

  $Q=\frac{\left[{\text{H}}_2\right]\left[{\text{I}}_2\right]}{{\left[\text{HI}\right]}^2}$        (2)

Substitute $2.0\times {10}^{-4}M$ for $\left[{\text{H}}_2\right]$, $2.0\times {10}^{-4}M$ for $\left[{\text{I}}_2\right]$, and $2.0\times {10}^{-5}M$ for $\left[\text{HI}\right]$ in equation (2).

  $\begin{array}{c}Q=\frac{\left(2.0\times {10}^{-4}M\right)\left(2.0\times {10}^{-4}M\right)}{{\left(2.0\times {10}^{-5}M\right)}^2}\\ =1\times {10}^2\end{array}$

The value of $K_{\text{c}}$ for a given reaction is $0.010$. 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 $1.05\text{ mmol}$ of $\text{HI}$, $0.10\text{ mmol}$ of ${\text{H}}_2$, and $0.090\text{ mmol}$ of ${\text{I}}_2$ is place in $1.0\text{ L}$ container then the direction of the following reaction has to be determined.

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

Concept Introduction:

Refer to part (a).

Explanation of Solution

The given reaction occurs is as follows:

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

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

  $\begin{array}{c}\left[\text{HI}\right]=\left(\frac{1.05\text{ mmol}}{1.0\text{ L}}\right)\left(\frac{1.0\times {10}^{-3}\text{ mol}}{1\text{ mmol}}\right)\\ =1.05\times {10}^{-3}M\end{array}$

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

  $\begin{array}{c}\left[{\text{H}}_2\right]=\left(\frac{\text{0}\text{.10 mmol}}{1.0\text{ L}}\right)\left(\frac{1.0\times {10}^{-3}\text{ mol}}{1\text{ mmol}}\right)\\ =1.0\times {10}^{-4}M\end{array}$

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

  $\begin{array}{c}\left[{\text{I}}_2\right]=\left(\frac{\text{0}\text{.090 mmol}}{1.0\text{ L}}\right)\left(\frac{1.0\times {10}^{-3}\text{ mol}}{1\text{ mmol}}\right)\\ =9.0\times {10}^{-5}M\end{array}$

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

  $Q=\frac{\left[{\text{H}}_2\right]\left[{\text{I}}_2\right]}{{\left[\text{HI}\right]}^2}$        (2)

Substitute $1.0\times {10}^{-4}M$ for $\left[{\text{H}}_2\right]$, $9.0\times {10}^{-5}M$ for $\left[{\text{I}}_2\right]$, and $1.05\times {10}^{-3}M$ for $\left[\text{HI}\right]$ in equation (2).

  $\begin{array}{c}Q=\frac{\left(1.0\times {10}^{-4}M\right)\left(9.0\times {10}^{-5}M\right)}{{\left(1.05\times {10}^{-3}M\right)}^2}\\ =8\times {10}^{-3}\end{array}$

The value of $K_{\text{c}}$ for a given reaction is $0.010$. 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.

(d)

Interpretation Introduction

Interpretation:

If $2.00\text{ mmol}$ of $\text{HI}$, $0.050\text{ mmol}$ of ${\text{H}}_2$, and $0.080\text{ mmol}$ of ${\text{I}}_2$ is place in $1.0\text{ L}$ container then the direction of the following reaction has to be determined.

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

Concept Introduction:

Refer to part (a).

Explanation of Solution

The given reaction occurs is as follows:

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

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

  $\begin{array}{c}\left[\text{HI}\right]=\left(\frac{2.00\text{ mmol}}{1.0\text{ L}}\right)\left(\frac{1.0\times {10}^{-3}\text{ mol}}{1\text{ mmol}}\right)\\ =2.00\times {10}^{-3}M\end{array}$

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

  $\begin{array}{c}\left[{\text{H}}_2\right]=\left(\frac{\text{0}\text{.050 mmol}}{1.0\text{ L}}\right)\left(\frac{1.0\times {10}^{-3}\text{ mol}}{1\text{ mmol}}\right)\\ =5.0\times {10}^{-5}M\end{array}$

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

  $\begin{array}{c}\left[{\text{I}}_2\right]=\left(\frac{0.080\text{ mmol}}{1.0\text{ L}}\right)\left(\frac{1.0\times {10}^{-3}\text{ mol}}{1\text{ mmol}}\right)\\ =8.0\times {10}^{-5}M\end{array}$

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

  $Q=\frac{\left[{\text{H}}_2\right]\left[{\text{I}}_2\right]}{{\left[\text{HI}\right]}^2}$        (2)

Substitute $5.0\times {10}^{-5}M$ for $\left[{\text{H}}_2\right]$, $8.0\times {10}^{-5}M$ for $\left[{\text{I}}_2\right]$, and $2.00\times {10}^{-3}M$ for $\left[\text{HI}\right]$ in equation (2).

  $\begin{array}{c}Q=\frac{\left(5.0\times {10}^{-5}M\right)\left(8.0\times {10}^{-5}M\right)}{{\left(2.00\times {10}^{-3}M\right)}^2}\\ =1\times {10}^{-3}\end{array}$

The value of $K_{\text{c}}$ for a given reaction is $0.010$. 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.