Question

The enthalpy of formation of $C{{H}_{4}},\text{ }{{\text{C}}_{2}}{{H}_{6}}$ and ${{C}_{4}}{{H}_{10}}$ are -74.8, -84.7 and -126.1 KJ/mol. Arrange them in the order of their efficiency as fuel/g. (Enthalpy of formation of $C{{O}_{2}}\text{(g) and }{{\text{H}}_{2}}O(g)$ are -393.5 and -285.8 KJ/mol respectively)

Answer 1

The heat of formation is the energy required for the formation of the bonds and the heat evolved by burning of that substance is the heat of dissociation. To solve this, you have to find the change in energy in form of heat i.e. $\Delta H$ . Use the relation $\Delta H=\Delta {{H}_{f(product)}}-\Delta {{H}_{f(react\text{ant})}}$ to find the required answer.

Complete answer:
Before answering this, let us discuss what enthalpy of formation is.
We know that the standard condition is 1atm pressure and 25 degree Celsius temperature.
At these standard conditions we can define the standard enthalpy of formation as the change of enthalpy of during the formation of 1 mole of the substance where all the substances are in their standard states.
Now, here let us calculate the efficiency as fuel/g one by one.
At first, we have $C{{H}_{4}}$ . We can write its reaction of combustion as-
$C{{H}_{4}}(g)+2{{O}_{2}}(g)\to C{{O}_{2}}(g)+2{{H}_{2}}O(l)$
Now, we know that enthalpy of formation is equal to the difference of enthalpy of formation of reactant and product. So, for $C{{H}_{4}}$ we can write that-
$\Delta H=\Delta {{H}_{C{{O}_{2}}(g)}}+2\Delta {{H}_{{{H}_{2}}O(g)}}-\Delta {{H}_{C{{H}_{4}}(g)}}-2\Delta {{H}_{{{O}_{2}}(g)}}$
We can see that enthalpy of formation of carbon dioxide, water and methane gas is given to us. So, putting the values in the above equation we will get -
$\begin{aligned} & \Rightarrow \Delta H=-393.5+2\times -285.8-(-74.8)-2\times 0 \\\ & \Rightarrow \Delta H=-890.3KJ/mol \\\ \end{aligned}$
Now, we know that the molecular weight of $C{{H}_{4}}$ is 16 g/mol.
Therefore, efficiency = $\Delta {{H}_{comb/g}}=\dfrac{890.3KJ/mol}{16g/mol}=55.64\text{ }KJ/g$
Now, similarly for ${{C}_{2}}{{H}_{6}}$ we can write the combustion reaction as-
${{C}_{2}}{{H}_{6}}(g)+\dfrac{7}{2}{{O}_{2}}(g)\to 2C{{O}_{2}}(g)+3{{H}_{2}}O(l)$
Therefore, $\Delta H=2\Delta {{H}_{C{{O}_{2}}(g)}}+3\Delta {{H}_{{{H}_{2}}O(g)}}-\Delta {{H}_{{{C}_{2}}{{H}_{6}}(g)}}-\dfrac{7}{2}\Delta {{H}_{{{O}_{2}}(g)}}$
Now, putting the values, we will get-
$\begin{aligned} & \Rightarrow \Delta H=2\times -393.5+3\times -285.8-\left( -84.7 \right)-\dfrac{7}{2}\times 0 \\\ & \Rightarrow \Delta H=-1559.1KJ/mol \\\ \end{aligned}$
We know the molecular weight of ${{C}_{2}}{{H}_{6}}$ is 30 g/mol.
Therefore, efficiency = $\Delta {{H}_{comb/g}}=\dfrac{1559.1KJ/mol}{30g/mol}=51.97\text{ }KJ/g$
And lastly, we have ${{C}_{4}}{{H}_{10}}$. We can write the combustion reaction as -
${{C}_{4}}{{H}_{10}}(g)+\dfrac{13}{2}{{O}_{2}}(g)\to 4C{{O}_{2}}(g)+5{{H}_{2}}O(l)$
Therefore, $\Delta H=4\Delta {{H}_{C{{O}_{2}}(g)}}+5\Delta {{H}_{{{H}_{2}}O(g)}}-\Delta {{H}_{{{C}_{4}}{{H}_{10}}(g)}}-\dfrac{13}{2}\Delta {{H}_{{{O}_{2}}(g)}}$
Now, putting the values, we will get-
$\begin{aligned} & \Rightarrow \Delta H=4\times -393.5+5\times -285.8-\left( -126.1 \right)-\dfrac{13}{2}\times 0 \\\ & \Rightarrow \Delta H=-2876.9KJ/mol \\\ \end{aligned}$
We know the molecular weight of ${{C}_{4}}{{H}_{10}}$ is 58 g/mol.
Therefore, efficiency = $\Delta {{H}_{comb/g}}=\dfrac{2876.9KJ/mol}{58g/mol}=49.6\text{ }KJ/g$
Therefore, if we arrange them in the order of increasing efficiency as fuel in KJ/g, the order will be- $C{{H}_{4}}$ > ${{C}_{2}}{{H}_{6}}$ > ${{C}_{4}}{{H}_{10}}$ . And, this is the required answer.

Note:
We know that in a crystalline solid, when the ions are combined, it causes energy release. We know this energy by the name lattice energy, the change in energy in this process if basically the lattice enthalpy. However, to be more appropriate about the energy required for the formation of the crystal lattice, we use the term formation enthalpy. There are two factors that affect the lattice energy of any substance. Firstly it is the charge on the ions and second is the radius of the ions.