Chapter 4, Problem 4.126P

(a)

Interpretation Introduction

Interpretation:

The percent yield of the formation of ${\text{NH}}_3$ is to be calculated.

Concept introduction:

Haber process is one of the industrial methods employed for the preparation of ammonia $\left({\text{NH}}_3\right)$. In this method, nitrogen and hydrogen molecule react in the presence of an iron catalyst to form ammonia gas. High temperature and pressure condition is required for the reaction. The balanced chemical equation for the preparation of ${\text{NH}}_3$ is:

${\text{N}}_2\left(g\right)+3{\text{H}}_2\left(g\right)\to 2{\text{NH}}_3\left(g\right)$

Combination redox reactions are the reactions in which two or more reactants combine to form a single product. The formation of ammonia from nitrogen and hydrogen molecule is an example of a combination redox reaction.

Answer to Problem 4.126P

The percent yield of the formation of ${\text{NH}}_3$ is $61.7\%$.

Explanation of Solution

The mass of ${\text{N}}_2$ is $20\text{g}$.

The mass of ${\text{H}}_2$ is $10\text{g}$.

The mass of ${\text{NH}}_3$ at equilibrium is $15\text{g}$.

Nitrogen molecule $\left({\text{N}}_2\right)$ reacts with hydrogen molecule $\left({\text{H}}_2\right)$ to form ammonia molecule $\left({\text{NH}}_3\right)$. The balanced chemical equation of the combination redox reaction is:

${\text{N}}_2\left(g\right)+3{\text{H}}_2\left(g\right)\to 2{\text{NH}}_3\left(g\right)$

One mole of ${\text{N}}_2$ reacts with three moles of ${\text{H}}_2$ to give two moles of ${\text{NH}}_3$.

The molecular mass of ${\text{N}}_2$ is $28.02\text{g/mol}$ and the molecular mass of ${\text{H}}_2$ is $2.016\text{g/mol}$.

The formula to calculate moles of ${\text{NH}}_3$ when ${\text{N}}_2$ is limiting reagent is:

$\text{Moles}\text{of}{\text{NH}}_3=\left[\left(\frac{\text{mass of}{\text{N}}_2\left(\text{g}\right)}{\text{molecular mass}\text{of}{\text{N}}_2\left(\text{g/mol}\right)}\right)\left(\frac{2\text{mol}{\text{NH}}_3}{1\text{mol}{\text{N}}_2}\right)\right]$                       (1)

Substitute $20\text{g}$ for the mass of ${\text{N}}_2$ and $28.02\text{g/mol}$ for molecular mass of ${\text{N}}_2$ in the equation (1).

$\begin{array}{c}\text{Moles}\text{of}{\text{NH}}_3=\left[\left(\frac{20\text{g}}{28.02\text{g/mol}}\right)\left(\frac{2\text{mol}{\text{NH}}_3}{1\text{mol}{\text{N}}_2}\right)\right]\\ =\left[\left(0.713775\text{mol}\right)\left(\frac{2\text{mol}{\text{NH}}_3}{1\text{mol}{\text{N}}_2}\right)\right]\\ =1.42755\text{mol}\end{array}$

The formula to calculate moles of ${\text{NH}}_3$ when ${\text{H}}_2$ is limiting reagent is:

$\text{Moles}\text{of}{\text{NH}}_3=\left[\left(\frac{\text{mass of}{\text{H}}_2\left(\text{g}\right)}{\text{molecular mass}\text{of}{\text{H}}_2\left(\text{g/mol}\right)}\right)\left(\frac{2\text{mol}{\text{NH}}_3}{3\text{mol}{\text{H}}_2}\right)\right]$                        (2)

Substitute $10\text{g}$ for the mass of ${\text{H}}_2$ and $2.016\text{g/mol}$ for molecular mass of ${\text{H}}_2$ in the equation (2).

$\begin{array}{c}\text{Moles}\text{of}{\text{NH}}_3=\left[\left(\frac{10\text{g}}{2.016\text{g/mol}}\right)\left(\frac{2\text{mol}{\text{NH}}_3}{3\text{mol}{\text{H}}_2}\right)\right]\\ =\left[\left(4.9603\text{mol}\right)\left(\frac{2\text{mol}{\text{NH}}_3}{3\text{mol}{\text{H}}_2}\right)\right]\\ =3.306878\text{mol}\end{array}$

${\text{N}}_2$ is limiting reagent in the reaction as the moles of ${\text{NH}}_3$ produced is less in this case as compared to when ${\text{H}}_2$ is the limiting agent.

The molecular mass of ${\text{NH}}_3$ is $17.03\text{g/mol}$.

The formula to calculate the mass of ${\text{NH}}_3$ is:

$\text{Mass}\text{of}{\text{NH}}_3=\left(\text{moles of}{\text{NH}}_3\left(\text{mol}\right)\right)\left({\text{molecular mass of NH}}_3\left(\text{g/mol}\right)\right)$ (3)

Substitute $1.42755\text{mol}$ for moles of ${\text{NH}}_3$ and $17.03\text{g/mol}$ for molecular mass of ${\text{NH}}_3$ in equation (3).

$\begin{array}{c}\text{Mass}\text{of}\text{Fe}=\left(1.42755\text{mol}\right)\left(17.03\text{g/mol}\right)\\ =24.311\text{g}\end{array}$

The expression to calculate the percent yield of the reaction is:

$\%\text{yield}=\left(\frac{\text{actual yield}\left(\text{g}\right)}{\text{theoretical yield}\left(\text{g}\right)}\right)\left(100\right)$ (4)

Substitute $15\text{g}$ for actual yield and $24.311\text{g}$ for theoretical yield in the equation (4).

$\begin{array}{c}\%\text{yield}=\left(\frac{15\text{g}}{24.311\text{g}}\right)\left(100\right)\\ =61.7\%\end{array}$

Conclusion

The percent yield of the formation of ${\text{NH}}_3$ is $61.7\%$.

(b)

Interpretation Introduction

Interpretation:

The moles of ${\text{N}}_2$ and ${\text{H}}_2$ that are present at equilibrium is to be calculated.

Concept introduction:

The redox reaction can be classified into three types depending upon the number of reactants and products as follows:

1. Combination redox reaction

2. Decomposition redox reaction

3. Displacement redox reactions

Combination redox reactions are the reactions in which two or more reactants combine to form a single product. In displacement redox reactions, substances on both sides of the equation remain the same but the atoms exchange places in order to form the product while in decomposition reaction, one compound decomposes to form one or more product.

Answer to Problem 4.126P

The moles of ${\text{N}}_2$ and ${\text{H}}_2$ that are present at equilibrium is $0.273\text{mol}$ and $3.64\text{mol}$ respectively.

Explanation of Solution

The mass of ${\text{N}}_2$ is $20\text{g}$.

The mass of ${\text{H}}_2$ is $10\text{g}$.

The mass of ${\text{NH}}_3$ at equilibrium is $15\text{g}$.

The molecular mass of ${\text{N}}_2$ is $28.02\text{g/mol}$ and the molecular mass of ${\text{H}}_2$ is $2.0916\text{g/mol}$.

The formula to calculate moles of ${\text{N}}_2$ initially present in the reaction is:

$\text{Moles}\text{of}{\text{N}}_2=\left(\frac{\text{mass of}{\text{N}}_2\left(\text{g}\right)}{\text{molecular mass}\text{of}{\text{N}}_2\left(\text{g/mol}\right)}\right)$                                                    (5)

Substitute $20\text{g}$ for the mass of ${\text{N}}_2$ and $28.02\text{g/mol}$ for molecular mass of ${\text{N}}_2$ in the equation (5).

$\begin{array}{c}\text{Moles}\text{of}{\text{N}}_2=\left(\frac{20\text{g}}{28.02\text{g/mol}}\right)\\ =0.71378\text{mol}\end{array}$

The formula to calculate moles of ${\text{H}}_2$ initially present in the reaction is:

$\text{Moles}\text{of}{\text{H}}_2=\left(\frac{\text{mass of}{\text{H}}_2\left(\text{g}\right)}{\text{molecular mass}\text{of}{\text{H}}_2\left(\text{g/mol}\right)}\right)$                                                     (6)

Substitute $10\text{g}$ for the mass of ${\text{H}}_2$ and $2.016\text{g/mol}$ for molecular mass of ${\text{H}}_2$ in the equation (6).

$\begin{array}{c}\text{Moles}\text{of}{\text{H}}_2=\left(\frac{10\text{g}}{2.016\text{g/mol}}\right)\\ =4.9603\text{mol}\end{array}$

The molecular mass of ${\text{NH}}_3$ is $17.03\text{g/mol}$.

The formula to calculate moles of ${\text{N}}_2$ to produce $15\text{g}{\text{NH}}_3$ is:

$\text{Moles}\text{of}{\text{N}}_2=\left[\left(\frac{\text{mass of}{\text{NH}}_3\left(\text{g}\right)}{\text{molecular mass}\text{of}{\text{NH}}_3\left(\text{g/mol}\right)}\right)\left(\frac{1\text{mol}{\text{N}}_2}{2\text{mol}{\text{NH}}_3}\right)\right]$                       (7)

Substitute $15\text{g}$ for the mass of ${\text{NH}}_3$ and $17.03\text{g/mol}$ for molecular mass of ${\text{NH}}_3$ in the equation (7).

$\begin{array}{c}\text{Moles}\text{of}{\text{N}}_2=\left[\left(\frac{15\text{g}}{17.03\text{g/mol}}\right)\left(\frac{1\text{mol}{\text{N}}_2}{2\text{mol}{\text{NH}}_3}\right)\right]\\ =0.4404\text{mol}\end{array}$

The formula to calculate moles of ${\text{H}}_2$ to produce $15\text{g}{\text{NH}}_3$ is:

$\text{Moles}\text{of}{\text{H}}_2=\left[\left(\frac{\text{mass of}{\text{NH}}_3\left(\text{g}\right)}{\text{molecular mass}\text{of}{\text{NH}}_3\left(\text{g/mol}\right)}\right)\left(\frac{3\text{mol}{\text{H}}_2}{2\text{mol}{\text{NH}}_3}\right)\right]$                       (8)

Substitute $15\text{g}$ for the mass of ${\text{NH}}_3$ and $17.03\text{g/mol}$ for molecular mass of ${\text{NH}}_3$ in the equation (8).

$\begin{array}{c}\text{Moles}\text{of}{\text{H}}_2=\left[\left(\frac{15\text{g}}{17.03\text{g/mol}}\right)\left(\frac{3\text{mol}{\text{H}}_2}{2\text{mol}{\text{NH}}_3}\right)\right]\\ =1.3212\text{mol}\end{array}$

The formula to calculate moles of ${\text{N}}_2$ at equilibrium is:

${\text{Moles of N}}_2\text{ at equilibrium}=\left(\text{Initial moles of}{\text{N}}_2\right)-\left(\text{Reacted mole of}{\text{N}}_2\right)$ (9)

Substitute $0.71378\text{mol}$ for an initial mole of ${\text{N}}_2$ and $0.4404\text{mol}$ for a reacted mole of ${\text{N}}_2$ in the equation (9).

$\begin{array}{c}{\text{Moles of N}}_2\text{ at equilibrium}=\left(0.71378\text{mol}\right)-\left(0.4404\text{mol}\right)\\ =0.27338\text{mol}\\ \approx 0.273\text{mol}\end{array}$

The formula to calculate moles of ${\text{H}}_2$ at equilibrium is:

${\text{Moles of H}}_2\text{ at equilibrium}=\left(\text{Initial moles of}{\text{H}}_2\right)-\left(\text{Reacted mole of}{\text{H}}_2\right)$ (10)

Substitute $4.9603\text{mol}$ for an initial mole of ${\text{H}}_2$ and $1.3212\text{mol}$ for a reacted mole of ${\text{H}}_2$ in the equation (10).

$\begin{array}{c}{\text{Moles of N}}_2\text{ at equilibrium}=\left(4.9603\text{mol}\right)-\left(1.3212\text{mol}\right)\\ =3.6391\text{mol}\\ \approx 3.64\text{mol}\end{array}$

Conclusion

The moles of ${\text{N}}_2$ and ${\text{H}}_2$ that are present at equilibrium is $0.273\text{mol}$ and $3.64\text{mol}$ respectively.