Q9.74CP - Chemistry: Molecular Nature Of Matter And Change

Question

Acetylene gas (ethyne; HC≡CH) burns in an oxyacetylene torch to produce carbon dioxide and water vapour. The heat of reaction for the combustion of acetylene is 1259 kJ/mol. 

(a) Calculate the C≡C bond energy, and compare your value with that in Table 9.2, p. 353. 

(b) When 500.0 g of acetylene burns, how many kilojoules of heat are given off? 

(c) How many grams of ${\mathbf{CO}}_2$form? 

(d) How many litres of  ${\mathbf{O}}_{\mathbf{2}}$ at 298 K and 18.0 atm are consumed?

Step-by-Step Solution

Answer

The intermolecular force between the molecules are:

1. The bond energy of the ethyne is 800 kJ/mole.
2. The heat produced after the combustion is 41.82 kJ.
3. Mass of ${\mathbf{CO}}_2$ produces after the combustion of ethyne= 88 g.
4. The litres of at 298 k when pressure is 18 atm are 344 litres.

Step 1: Intermolecular Forces

Intermolecular forces are the interaction which are formed by the attraction of the two having opposite charges (partial positive and partial negative charge). The opposite charge are formed by the presence of the electron-negative atom in the molecule. Due to the presence of an electron-negative atom, there is an induced partial positive charge is generated on the electron-positive charge (or less electron-negative charge) atom. There will be an attraction between the both oppositely charges to form a bond.

 

Bond energy can be defined as the energy required for the formation of the bond or the energy required to break the bond. The energy is then stored in the bond. 

Step 2: Bond Energy Explanation

The combustion of the ethyne in the presence of oxygen gives carbon dioxide.

The bond energy of reaction can be calculated as,

                                                              

 $\begin{array}{c}{{\mathrm{DH}}^{\mathrm{o}}}_{\mathrm{rxn}}= {\mathrm{DH}}^{\mathrm{o}}{{}_{\mathrm{bond}}}_{\mathrm{break}} + {\mathrm{DH}}^{\mathrm{o}}{{}_{\mathrm{bond}}}_{\mathrm{formed}}\\ {{\mathrm{DH}}^{\mathrm{o}}}_{\mathrm{rxn}}=[2{\mathrm{BE}}_{\mathrm{C}-\mathrm{H}} +{\mathrm{BE}}_{\mathrm{C}º\mathrm{C}}+\frac{5}{2}{\mathrm{BE}}_{\mathrm{O}=\mathrm{O}}\left]+\right[4\left(-{\mathrm{BE}}_{\mathrm{C}=\mathrm{O}}\right)+2\left(-{\mathrm{BE}}_{\mathrm{O}-\mathrm{H}}\right)]\\ -1259\mathrm{kJ}=[2\left(413\right)+{\mathrm{BE}}_{{}_{\mathrm{C}º\mathrm{C}}}+\frac{5}{2}\left(498\right)\left]+\right[4\left(-799\right)+2\left(-467\right)]\\ -1259\mathrm{kJ}=[826++{\mathrm{BE}}_{\mathrm{C}º\mathrm{C}}+1245\left]+\right[-4130]\mathrm{kJ}\\ -1259\mathrm{kJ}=-2059+{\mathrm{BE}}_{\mathrm{C}º\mathrm{C}} \mathrm{kJ}\\ {\mathrm{BE}}_{\mathrm{C}º\mathrm{C}}=800.0\mathrm{kJ}/\mathrm{mol}\end{array}$

Step 3: Heat Energy Explanation

You know the 500 g of ethyne releases 800 kJ of heat.

Here, we have to calculate heat liberated by 1 mole of ethyne

 $1\mathrm{mole} \mathrm{of} \mathrm{ethyne}\times \frac{26\mathrm{g}}{1\mathrm{mole} \mathrm{ethyne}}=26\mathrm{g}/\mathrm{mole}$

$\mathbf{\Delta H}\mathbf{=}\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{1}\mathbf{ }\mathbf{mole}\mathbf{ }\mathbf{ethyne}\mathbf{\times }\frac{\mathrm{Energy} \mathrm{of} \mathrm{ethyne}}{\mathrm{Mass} \mathrm{of} \mathrm{ethyne}}$

After, filling the values 

 

 $\mathrm{\Delta H}=26.14\mathrm{g} \mathrm{ethyne}\times \frac{800\mathrm{kJ}}{500\mathrm{g} \mathrm{ethyne}}=41.82\mathrm{kJ}$

 

The heat produced after the combustion is 41.82 kJ.

Step 4: Carbon Dioxide Formation

The combustion of 1 mole of ethyne produces 2 moles of carbon dioxide and a water molecule.

The molar mass of ${\mathrm{CO}}_2$ = 44 g/mole

2 mole of ${\mathrm{CO}}_2$ = 2mole × 44 g/mole

Mass of  ${\mathrm{CO}}_2$= 88 g

Step 5: Carbon Dioxide Formation

In the combustion of ethyne required 2.5 moles of ${\mathbf{O}}_{\mathbf{2}}$ molecule.

Temperature = 298k

Pressure = 18atm

 

According to the Ideal gas equation, 

$\mathbf{PV}\mathbf{=}\mathbf{nRT}$ 

Where,

$\mathrm{P}=\mathrm{Pressure},\mathrm{V}=\mathrm{Volume},\mathrm{n}=\mathrm{Number} \mathrm{of} \mathrm{moles}$ 

 $\mathrm{T}=\mathrm{Temperature} \mathrm{and} \mathrm{R}=\mathrm{Gas} \mathrm{constant}=8.314\mathrm{J}/\mathrm{mole}/\mathrm{k}$

 

 $\begin{array}{c}\mathrm{Volume} \mathrm{of} \mathrm{consumed}=\frac{\mathrm{nRT}}{\mathrm{P}}\\ \mathrm{V}=\frac{5}{2}\times 8.314\times \frac{298}{18}\\ \mathrm{V}=344 \mathrm{litre}\end{array}$

 

Hence, the litres of at 298 k when pressure is 18 atm are 344 litres.