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Question: The enthalpy of combustion of methane, graphite and dihydrogen at 298K are −890.3 \(kJmo{l^{ - 1}}\)...

The enthalpy of combustion of methane, graphite and dihydrogen at 298K are −890.3 kJmol−1kJmo{l^{ - 1}} ,−393.5 kJmol−1kJmo{l^{ - 1}} ,−285.8 kJmol−1kJmo{l^{ - 1}} respectively. Enthalpy of formation of CH4(g)C{H_4}(g) ​ will be:

Explanation

Solution

To solve this question you must be aware of the concepts of thermodynamics briefly at least. Standard enthalpy of combustion is defined as the enthalpy change when one mole of a compound is completely burnt in oxygen with all the reactants and products in their standard state under standard conditions i.e. 298K and 1 bar pressure. So firstly, write all the combustion equations used in the question i.e. of methane, graphite and dihydrogen at 298K. And then form the equation of formation of CH4(g)C{H_4}(g) . And then calculate its enthalpy of formation.

Complete step by step answer:
Step 1: In this step we are going to write the equations of combustion of methane, graphite and dihydrogen at 298K:
Combustion of Methane: CH4(g)C{H_4}\left( g \right) +   + \,\, 2O2(g)2{O_2}\left( g \right) →  \to \, CO2(g)C{O_2}(g) ++ 2H2O(g)2{H_2}O\left( g \right)
Here, △H1=\vartriangle {H_1} = −890.3 kJmol−1kJmo{l^{ - 1}}
Combustion of Methane: C(s)C\left( s \right) +   + \,\, O2(g){O_2}\left( g \right) →  \to \, CO2(g)C{O_2}(g)
Here, △H2=\vartriangle {H_2} = −393.5 kJmol−1kJmo{l^{ - 1}}
Combustion of Methane: 2H2(g)2{H_2}\left( g \right) +   + \,\, O2(g){O_2}\left( g \right) →  \to \, H2O(g){H_2}O(g)
Here, △H3=\vartriangle {H_3} = −285.8 kJmol−1kJmo{l^{ - 1}}
Step 2: In this step we will write the desired equation of formation of CH4(g)C{H_4}(g) by using the above equations:
C(s)C\left( s \right) ++ 2H2(g)2{H_2}\left( g \right) →\to CH4(g)C{H_4}(g)
Step 3: In this step we will write the desired equation of enthalpy formation of CH4(g)C{H_4}(g) by using the above equation:
\vartriangle {H_{f\,}} = \,$$$\vartriangle {H_2}\, + \,$$ 2 \times \vartriangle {H_3}, - \vartriangle {H_1} \vartriangle {H_{f,}} = ,[−393.5+2(−285.8)−(−890.3)][-393.5 + 2(-285.8) - (-890.3)]kJmo{l^{ - 1}} \vartriangle {H_{f,}} = ,−74.8-74.8kJmo{l^{ - 1}}Hence,requiredenthalpyofformationof Hence, required enthalpy of formation ofC{H_4}(g)==\vartriangle {H_{f,}} = ,−74.8-74.8kJmo{l^{ - 1}}$ and this is our answer.

Note: The enthalpy of combustion of a substance is defined as the heat exchange when 1 mole of substance is completely burnt in oxygen. Standard enthalpy of combustion is the amount evolved when one mole of the substance under standard conditions is completely burnt to form the products also under standard conditions. Enthalpies of combustion can be used to compare which fuels or substances release the most energy when they are burned. They can be calculated using a bomb calorimeter.