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Question: The standard heats of formation of \(C{{H}_{4}}(g),C{{O}_{2}}(g)\text{ and }{{\text{H}}_{2}}O(g)\) a...

The standard heats of formation of CH4(g),CO2(g) and H2O(g)C{{H}_{4}}(g),C{{O}_{2}}(g)\text{ and }{{\text{H}}_{2}}O(g) are: –76.2, –398.8, –241.6 KJ/mol respectively. Calculate the amount of heat evolved by burning 1m3{{m}^{3}} of methane measured under normal conditions.

Explanation

Solution

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. ΔH\Delta H . Use the relation ΔH=ΔHf(product)−ΔHf(reactant)\Delta H=\Delta {{H}_{f(product)}}-\Delta {{H}_{f(react\text{ant})}} to find the required answer.

Complete step by step answer:
To answer this, firstly let us discuss the meaning of the term heat of formation.
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.
The term lattice enthalpy if used for both i.e. the change in energy when 1 mole of any substance is formed from their corresponding ionic forms or we can also say that it is the energy required for separating 1 mole of a solid crystal into its corresponding ions.

However, to be more appropriate about the energy required for breaking of the crystal lattice and the formation of the crystal lattice, we use the terms lattice dissociation enthalpy and lattice formation enthalpy i.e. basically the heat of dissociation and formation respectively.
Now, let’s see the question given to us where we have to calculate the amount of heat evolved i.e. ΔH\Delta H when methane is burnt under normal conditions.
So, firstly let’s write down the reaction-
CH4+2O2→CO2+2H2OC{{H}_{4}}+2{{O}_{2}}\to C{{O}_{2}}+2{{H}_{2}}O

We can calculate ΔH\Delta H by subtracting the heat of formation of reactants from that of the product. We can write it as-
ΔH=ΔHf(product)−ΔHf(reactant)\Delta H=\Delta {{H}_{f(product)}}-\Delta {{H}_{f(react\text{ant})}}
The value of heat of formations of water, methane and carbon dioxide is given to us and that of oxygen will be zero as it exists as O2{{O}_{2}} .

So, we can write that-
ΔH=[ΔHf CO2+2×ΔHf H2O]−ΔHf CH4\Delta H=\left[ \Delta {{H}_{f\text{ C}{{\text{O}}_{2}}}}+2\times \Delta {{H}_{f\text{ }{{\text{H}}_{2}}O}} \right]-\Delta {{H}_{f\text{ }C{{H}_{4}}}}
Now, putting the values we will get-
ΔH=[−398.8−2×241.6]−(−76.2)KJ/mol=−805.8 KJ/mol\Delta H=\left[ -398.8-2\times 241.6 \right]-\left( -76.2 \right)KJ/mol=-805.8\text{ }KJ/mol
Now, this value is for 1 mole of methane i.e. 22.4 L.

Therefore, for 1 m3{{m}^{3}} i.e. 1000L, heat of formation = 805.822.4×1000 KJ/mol = -35.97×103 KJ/mol\dfrac{805.8}{22.4}\times 1000\text{ KJ/mol = -35}\text{.97}\times \text{1}{{\text{0}}^{3}}\text{ KJ/mol}
Therefore, the amount of heat evolved by burning 1m3{{m}^{3}} of methane measured under normal conditions is -35.97×103 KJ/mol\text{-35}\text{.97}\times \text{1}{{\text{0}}^{3}}\text{ KJ/mol} and this is the required answer.

Note: 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. Higher is the charge on each ion, higher will be the lattice energy. Lower the internuclear distances among the ions in an ionic crystal, higher will be their lattice energy i.e. closer the two ions are, higher is the force of attraction between them.