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If \begin{array}{l}
\mathrm{C}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+94.2 \mathrm{kcal} \\
\mathrm{H}_{2}+\dfrac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{H}_{2} \mathrm{O}+68.3 \mathrm{kcal} \\
\mathrm{CH}_{4}+2 \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}+210.8 \mathrm{kcal}
\end{array}
Thus, the heat of formation of methane:
(A) 47.3kcal
(B) 20kcal
(C) 45.9kcal
(D) 47.3kcal

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Last updated date: 29th Feb 2024
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IVSAT 2024
Answer
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Hint: 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. We have to add all the reactions with their respective enthalpies and modify them so that we get methane in the product side.

Complete step by step solution:
-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 us approach to the numerical:
\begin{array}{l}
\mathrm{C}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+94.2 \mathrm{kcal} \\
\mathrm{H}_{2}+\dfrac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{H}_{2} \mathrm{O}+68.3 \mathrm{kcal} \\
\mathrm{CH}_{4}+2 \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}+210.8 \mathrm{kcal}
\end{array}
-Here, as it clearly seen that all the reactions are exothermic, thus for the convention all these energies are negative in nature.
-Now, multiplying the equation by 2, we get:
\begin{array}{l}
\mathrm{H}_{2}+\dfrac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{H}_{2} \mathrm{O} \\
2 \mathrm{H}_{2}+\mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}
\end{array}
Thus, we will multiply the enthalpy of formation of water given to us by 2, we get -136.6kcal.
-Now, let us reverse the last equation then, the energy is changed from negative sign to positive thus, +210.8kcal.
\mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{CH}_{4}+2 \mathrm{O}_{2}
Now, let us write all the final reactions we have get:
\begin{array}{l}
\mathrm{C}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} \\
2 \mathrm{H}_{2}+\mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} \\
\mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{CH}_{4}+2 \mathrm{O}_{2}
\end{array}
Adding them all up we get as:
$C$ $+$ $2{H_2}$ → ${{C}{H_4}}$
Now calculating the energy of formation = $-94.2kcal$ $-$ $136.6kcal$ $+$ $210.8kcal$
$=$ $-20.0kcal$

Clearly, the answer is (B).

Note: Negative heat of formation indicates the energy of products is lower than the energy of reactants. Hence, during the reaction, energy is released and the reaction is exothermic.