Chapter 13, Problem 13.127AP

(a)

Interpretation Introduction

Interpretation: The reaction between Ammonia and Ammonium nitrate is given. Various questions based on the given reaction have to be answered.

Concept introduction: According to the rate law the rate of reaction depends on the concentration of reactants that are involved in the reaction. Therefore, for a reaction like $\text{A}\to \text{B}$, the rate law equation is given as,

$\text{Rate}=k{\left[\text{A}\right]}^m$

Where,

• $k$ is the rate constant.
• $m$ is the order of the reaction.

To determine: The rate law and the unit of rate constant for the given reaction.

Answer to Problem 13.127AP

Solution

The rate law expression for the given reaction is,

$\text{Rate}=k{\left[{\text{NH}}_{\text{3}}\right]}^1{\left[{\text{HNO}}_{\text{2}}\right]}^2$

The unit of rate constant is $\underset{\_}{M^{-2}{s}^{-1}}$.

Explanation of Solution

Explanation

The given reaction is,

${\text{NH}}_3\left(g\right)+{\text{HNO}}_2\left(g\right)\to {\text{NH}}_4{\text{NO}}_2\left(g\right)\to {\text{N}}_2\left(g\right)+2{\text{H}}_2\text{O}\left(l\right)$

As the reaction is first order with respect to Ammonia and second order in terms of Nitrous acid, therefore rate law is,

$\text{Rate}=k{\left[{\text{NH}}_{\text{3}}\right]}^1{\left[{\text{HNO}}_{\text{2}}\right]}^2$

The unit of rate is $\text{M}/\text{s}$ and the unit of concentration is $\text{M}$. Substitute these units in above rate law equation as,

$\begin{array}{c}\text{Rate}=k{\left[{\text{NH}}_{\text{3}}\right]}^1{\left[{\text{HNO}}_{\text{2}}\right]}^2\\ \frac{\text{M}}{\text{s}}=k{\left[\text{M}\right]}^1{\left[\text{M}\right]}^2\\ \frac{\text{M}}{\text{s}}=k{\left[\text{M}\right]}^3\\ k=\underset{\_}{M^{-2}{s}^{-1}}\end{array}$

(b)

Interpretation Introduction

To determine: The similarity between given rate law expression and the rate law equation obtained in part (a).

Answer to Problem 13.127AP

Solution

The given rate law expression is similar to the rate law equation obtained in part (a) as,

$\text{Rate}=k{\left[{\text{NH}}_{\text{3}}\right]}^1{\left[{\text{HNO}}_{\text{2}}\right]}^2$

Explanation of Solution

Explanation

The given rate law equation is,

$\text{Rate}=k\left[{\text{NH}}_{\text{4}}{}^{+}\right]\left[{\text{NO}}_{\text{2}}{}^{-}\right]\left[{\text{HNO}}_{\text{2}}\right]$

The acid base reaction between ${\text{NH}}_{\text{4}}{}^{+}$ and ${\text{NO}}_{\text{2}}{}^{-}$ is given as,

${\text{NH}}_4{}^{+}+{\text{NO}}_2{}^{-}\rightleftharpoons {\text{NH}}_3+{\text{HNO}}_2$

Here, Ammonium ion acts as an acid because it can easily donate electron to ${\text{NO}}_{\text{2}}{}^{-}$ that acts as a base and gives the respective conjugate base and conjugate acid. So, substitute concentration of Ammonia and ${\text{HNO}}_2$ in above equation and this will give the final rate law as,

$\text{Rate}=k{\left[{\text{NH}}_{\text{3}}\right]}^1{\left[{\text{HNO}}_{\text{2}}\right]}^2$

The equation for this rate law is similar to the given expression for rate law.

(c)

Interpretation Introduction

To determine: The value of $\Delta H{°}_{\text{rxn}}$.

Answer to Problem 13.127AP

Solution

The value of $\Delta H{°}_{\text{rxn}}$ is $\underset{\_}{-482.4kJ/mol}$.

Explanation of Solution

Explanation

The reactions are given as,

$\begin{array}{c}{\text{NH}}_3\left(g\right)+{\text{HNO}}_2\left(aq\right)\to {\text{NH}}_4{\text{NO}}_2\left(aq\right)\\ {\text{NH}}_4{\text{NO}}_2\left(aq\right)\to {\text{N}}_2\left(g\right)+{\text{H}}_2\text{O}\left(l\right)\end{array}$

Add both the reactions to obtain the overall reaction as,

${\text{NH}}_3\left(g\right)+{\text{HNO}}_2\left(aq\right)\to {\text{N}}_2\left(g\right)+{\text{H}}_2\text{O}\left(l\right)$

The heat of reaction is given as,

$\Delta H{°}_{\text{rxn}}=\sum m\Delta H{°}_{\text{Products}}-\sum n\Delta H{°}_{\text{Reactants}}$

Where,

• $m$ is the coefficient of reactants.
• $n$ is the coefficients of products.
• $\Delta H{°}_{\text{Reactants}}$ is the enthalpy of reactants.
• $\Delta H{°}_{\text{Products}}$ is the enthalpy of products.

The heat of reaction for the above reaction is,

$\begin{array}{c}\Delta H°=\sum m\Delta H{°}_{\text{Products}}-\sum n\Delta H{°}_{\text{Reactants}}\\ =\left({\left(\Delta H_{\text{f}}°\right)}_{{\text{N}}_2}+2\times {\left(\Delta H_{\text{f}}°\right)}_{{\text{H}}_2\text{O}}\right)-\left(\left(\Delta {\left(H_{\text{f}}°\right)}_{{\text{NH}}_3}\right)+\left({\left(\Delta H_{\text{f}}°\right)}_{{\text{HNO}}_2}\right)\right)\end{array}$

As referred from Appendix $4$, the value of ${\left(\Delta H_{\text{f}}°\right)}_{{\text{N}}_2}$ is $0.0\text{kJ}/\text{mol}$, ${\left(\Delta H_{\text{f}}°\right)}_{{\text{H}}_2\text{O}}$ is $-285.8\text{kJ}/\text{mol}$, $\Delta {\left(H_{\text{f}}°\right)}_{{\text{NH}}_3}$ is $-46.1\text{kJ}/\text{mol}$ and ${\left(\Delta H_{\text{f}}°\right)}_{{\text{HNO}}_2}$ is $-43.1\text{kJ}/\text{mol}$.

Substitute these values in above equation as,

$\begin{array}{c}\Delta H°=\left({\left(\Delta H_{\text{f}}°\right)}_{{\text{N}}_2}+2\times {\left(\Delta H_{\text{f}}°\right)}_{{\text{H}}_2\text{O}}\right)-\left(\left(\Delta {\left(H_{\text{f}}°\right)}_{{\text{NH}}_3}\right)+\left({\left(\Delta H_{\text{f}}°\right)}_{{\text{HNO}}_2}\right)\right)\\ =\left(0.0\text{kJ}/\text{mol}+2\times \left(-285.8\text{kJ}/\text{mol}\right)\right)-\left(\left(-46.1\text{kJ}/\text{mol}\right)+\left(-43.1\text{kJ}/\text{mol}\right)\right)\\ =-571.6\text{kJ}/\text{mol}-\left(-89.2\text{kJ}/\text{mol}\right)\\ =\underset{\_}{-482.4kJ/mol}\end{array}$

(d)

Interpretation Introduction

To determine: The reaction profile graph for the given reaction having lower activation energy for step (1) and higher activation energy for step (2).

Answer to Problem 13.127AP

Solution

The reaction profile graph for the given reaction having lower activation energy for step (1) and higher activation energy for step (2) is shown as,

Figure 1

Explanation of Solution

Explanation

The reactions are given as,

${\text{NH}}_3\left(g\right)+{\text{HNO}}_2\left(aq\right)\to {\text{NH}}_4{\text{NO}}_2\left(aq\right)$ (1)

${\text{NH}}_4{\text{NO}}_2\left(aq\right)\to {\text{N}}_2\left(g\right)+{\text{H}}_2\text{O}\left(l\right)$ (2)

The graph is plotted between energy taken at the y-axis and the progress of the reaction at the x-axis. The activation energy is lower first step and higher for second step in the below graph and as ${\text{NH}}_4{\text{NO}}_2$ acts as an intermediate, therefore it has been shown at the valley in the graph. Due to the presence of intermediate two peaks are obtained in the graph.

Figure 1

Conclusion

1. a. The rate law expression for the given reaction is,

$\text{Rate}=k{\left[{\text{NH}}_{\text{3}}\right]}^1{\left[{\text{HNO}}_{\text{2}}\right]}^2$

The unit of rate constant is $\underset{\_}{M^{-2}{s}^{-1}}$.

1. b. The given rate law expression is similar to the rate law equation obtained in part (a) as,

$\text{Rate}=k{\left[{\text{NH}}_{\text{3}}\right]}^1{\left[{\text{HNO}}_{\text{2}}\right]}^2$

1. c. The value of $\Delta H{°}_{\text{rxn}}$ is $\underset{\_}{-482.4kJ/mol}$.
2. d. The reaction profile graph for the given reaction having lower activation energy for step (1) and higher activation energy for step (2) is shown as,

Figure 1