Chapter 9, Problem 9.73P

(a)

Interpretation Introduction

Interpretation:

The carbon-carbon triple bond energy is to be calculated. Also, the calculated value and the theoretical value are to be compared.

Concept introduction:

The heat of the reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is defined as the heat released or absorbed during a chemical reaction as a result of the difference in the bond energies (BE) of reactant and product in the reaction. $\Delta H_{\text{rxn}}^{°}$ is negative for exothermic reaction and $\Delta H_{\text{rxn}}^{°}$ is positive for an endothermic reaction.

The formula to calculate $\Delta H_{\text{rxn}}^{°}$ of reaction is as follows:

$\Delta H_{\text{rxn}}^{°}=\sum \Delta H_{\text{reactant bond broken}}^{°}+\sum \Delta H_{\text{product bond formed}}^{°}$

Or,

$\Delta H_{\text{rxn}}^{°}=\sum {\text{BE}}_{\text{reactant bond broken}}-\sum {\text{BE}}_{\text{product bond formed}}$

The bond energy of reactants is positive and the bond energy of products is negative.

Answer to Problem 9.73P

The carbon-carbon triple bond energy is $800\text{kJ/mol}$ and the value is close to the theoretical value.

Explanation of Solution

The given chemical equation for the combustion of ethyne is as follows:

The formula to the enthalpy of the given reaction is as follows:

$\Delta H_{\text{rxn}}^{°}=\left({\text{2BE}}_{\text{C}-\text{H}}+5{\text{/2BE}}_{\text{O=O}}+{\text{BE}}_{\text{C-C triple bond }}\right)-\left({\text{4BE}}_{\text{C=O}}+{\text{2BE}}_{\text{O}-\text{H}}\right)$        (1)

Rearrange the equation (1) to calculate ${\text{BE}}_{\text{C-C triple bond }}$.

${\text{BE}}_{\text{C-C triple bond }}=\left(\Delta H_{\text{rxn}}^{°}-{\text{2BE}}_{\text{C}-\text{H}}-5{\text{/2BE}}_{\text{O=O}}\right)+\left({\text{4BE}}_{\text{C=O}}+{\text{2BE}}_{\text{O}-\text{H}}\right)$        (2)

Substitute $413\text{kJ/mol}$ for ${\text{BE}}_{\text{C}-\text{H}}$, $498\text{kJ/mol}$ for ${\text{BE}}_{\text{O=O}}$, $799\text{kJ/mol}$ for ${\text{BE}}_{\text{C=O}}$, $–1259\text{ kJ}$ for $\Delta H_{\text{rxn}}^{°}$ and $467\text{kJ/mol}$ for ${\text{BE}}_{\text{O}-\text{H}}$ in the equation (2).

$\begin{array}{c}{\text{BE}}_{\text{C-C triple bond }}=\left[\begin{array}{l}\left(\left(–1259\text{ kJ}\right)-\left(\text{2 mol}\right)\left(413\text{kJ/mol}\right)-\left(\text{5/2 mol}\right)\left(498\text{kJ/mol}\right)\right)-\\ \left(\text{4 mol}\right)\left(799\text{kJ/mol}\right)+\left(\text{2 mol}\right)\left(467\text{kJ/mol}\right)\end{array}\right]\\ =800\text{kJ/mol}\end{array}$

The value is close to the theoretical value $839\text{kJ/mol}$.

Conclusion

The carbon-carbon triple bond energy is $800\text{kJ/mol}$ and the value is close to the theoretical value.

(b)

Interpretation Introduction

Interpretation:

The heat released when acetylene burn is to be determined.

Concept introduction:

The expression to calculate the moles of solute when given mass and molecular mass of compound are given is as follows:

$\text{Moles of compound}\left(\text{mol}\right)=\left[\begin{array}{c}\text{given mass}\text{of compound}\left(\text{g}\right)\\ \left(\frac{\text{1mole}\text{of compound}\left(\text{mol}\right)}{\text{molecular mass of compound}\left(\text{g}\right)}\right)\end{array}\right]$

Stoichiometry of a reaction is utilized to determine the amount of any species in the reaction by the relationship between the reactants and products.

Consider the general reaction,

$\text{A}+2\text{B}\to 3\text{C}$

One mole of $\text{A}$ reacts with two moles of $\text{B}$ to produce three moles of $\text{C}$. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.

Answer to Problem 9.73P

The heat released when acetylene burn is $–2.417\times {10}^4\text{kJ}$.

Explanation of Solution

The formula to calculate the moles of acetylene is as follows:

$\text{Moles of acetylene}=\left[\frac{\text{given mass}\text{of acetylene}}{\text{molecular mass of acetylene}}\right]$        (3)

Substitute $500\text{g}$ for the mass of acetylene and $26.09\text{ g/mol}$ for the molar mass of acetylene in the equation (3).

$\begin{array}{c}\text{Moles of acetylene}=\left[\frac{500\text{g}}{26.09\text{ g/mol}}\right]\\ =19.2012\text{mol}\end{array}$

The heat released by one mole is $-1259\text{ kJ/mol}$. Therefore the heat released by $19.2012\text{mol}$ is calculated as follows:

$\begin{array}{c}\text{Heat released}=\left(-1259\text{ kJ/mol}\right)\left(19.2012\text{mol}\right)\\ =–2.417\times {10}^4\text{kJ}\end{array}$

Conclusion

The heat released when acetylene burn is $–2.417\times {10}^4\text{kJ}$.

(c)

Interpretation Introduction

Interpretation:

The mass of ${\text{CO}}_2$ formed is to be determined.

Concept introduction:

The expression to calculate the moles of solute when given mass and molecular mass of compound are given is as follows:

$\text{Moles of compound}\left(\text{mol}\right)=\left[\begin{array}{c}\text{given mass}\text{of compound}\left(\text{g}\right)\\ \left(\frac{\text{1mole}\text{of compound}\left(\text{mol}\right)}{\text{molecular mass of compound}\left(\text{g}\right)}\right)\end{array}\right]$

Stoichiometry of a reaction is utilized to determine the amount of any species in the reaction by the relationship between the reactants and products.

Consider the general reaction,

$\text{A}+2\text{B}\to 3\text{C}$

One mole of $\text{A}$ reacts with two moles of $\text{B}$ to produce three moles of $\text{C}$. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.

Answer to Problem 9.73P

The mass of ${\text{CO}}_2$ formed is $1690\text{g}$.

Explanation of Solution

The formula to calculate the mass of ${\text{CO}}_2$ is as follows:

$\text{Mass of}{\text{CO}}_2=\text{moles of}\text{acetylene}\left(\frac{1\text{mol}{\text{CO}}_2}{1\text{mol}\text{acetylene}}\right)\left({\text{molar mass of CO}}_2\right)$        (4)

Substitute $19.2012\text{mol}$ for moles of acetylene and $44.01\text{ g/mol}$ for molar mass of ${\text{CO}}_2$ in the equation (4).

$\begin{array}{c}\text{Mass of}{\text{CO}}_2=19.2012\text{mol}\left(\frac{2\text{mol}{\text{CO}}_2}{1\text{mol}\text{acetylene}}\right)\left(44.01\text{ g/mol}\right)\\ =1690.092\text{g}\\ \approx 1690\text{g}\end{array}$

Conclusion

The mass of ${\text{CO}}_2$ formed is $1690\text{g}$.

(d)

Interpretation Introduction

Interpretation:

The volume of ${\text{O}}_{\text{2}}$ at $298\text{ K}$ and $18.0\text{ atm}$ are consumed is to be determined.

Concept introduction:

The ideal gas equation can be expressed as follows,

$PV=nRT$        (5)

Here,

$P$ is the pressure.

$V$ is the volume.

$T$ is the temperature.

$n$ is the mole of the gas.

$R$ is the gas constant.

Answer to Problem 9.73P

The volume of ${\text{O}}_{\text{2}}$ at $298\text{ K}$ and $18.0\text{ atm}$ are consumed is $65.2\text{L}$.

Explanation of Solution

The formula to calculate the moles of ${\text{O}}_2$ is as follows:

$\text{Moles of}{\text{O}}_2=\text{moles of}\text{acetylene}\left(\frac{\frac{5}{2}\text{mol}{\text{O}}_2}{1\text{mol}\text{acetylene}}\right)$        (6)

Substitute $19.2012\text{mol}$ for moles of acetylene in the equation (6).

$\begin{array}{c}\text{Moles of}{\text{O}}_2=19.2012\text{mol}\left(\frac{\frac{5}{2}\text{mol}{\text{O}}_2}{1\text{mol}\text{acetylene}}\right)\\ =48.0030722\text{ mol }\end{array}$

Rearrange the equation (5) to calculate the volume of ${\text{O}}_2$.

$V_{{\text{O}}_2}=\frac{nRT}{P}$        (7)

Substitute the value $18\text{atm}$ for $P$, $298\text{K}$ for $T$, $48.0030722\text{ mol }$ for $n$ and $0.0821\text{ L}\cdot \text{atm/K}\cdot \text{mol}$ for $R$ in the equation (7).

$\begin{array}{c}{V}_{{\text{O}}_2}=\frac{\left(48.0030722\text{ mol }\right)\left(0.0821\text{ L}\cdot \text{atm/K}\cdot \text{mol}\right)\left(298\text{K}\right)}{18\text{atm}}\\ =65.2145\text{ L}\\ \approx 65.2\text{L}\end{array}$

Conclusion

The volume of ${\text{O}}_{\text{2}}$ at $298\text{ K}$ and $18.0\text{ atm}$ are consumed is $65.2\text{L}$.