
What will be the heat of formation of methane, if the heat of combustion of carbon is \[{{' - 'xkJ}}\], heat of formation of water is \[{{' - 'ykJ}}\] and heat of combustion of methane is \[{{' - 'zkJ}}\]?
A. \[\left( {{{ - x - y + z}}} \right){{kJ}}\]
B. \[\left( {{{ z - x + 2y}}} \right){{kJ}}\]
C. \[\left( {{{ - x - 2y - z}}} \right){{kJ}}\]
D. \[\left( {{{ x - 2y + z}}} \right){{kJ}}\]
Answer
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Hint: This can be solved using Hess’s law of constant heat summation. Here given reactions to find the heat of formation of methane are, heat of combustion of carbon, heat of formation of water and heat of combustion of methane. The complete balanced reaction is to be written along with the enthalpy values and these are added to find the solution.
Complete step by step answer:
This is solved as per Hess’s law of constant heat summation. It states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes.
Enthalpy is a measure of energy in a thermodynamic system.
Hess’s law is one of the applications of the first law of thermodynamics.
The enthalpy change in a chemical or physical process is the same whether the process is carried out in one step or in several steps.
Methane is \[{{C}}{{{H}}_{{4}}}\]. It is formed from hydrogen and carbon so, the heat of formation of methane can be written as,
\[{{C + 2}}{{{H}}_{{2}}}\, \to {{C}}{{{H}}_{{4}}}\]
From given data, we have
Heat of combustion of carbon, (combustion is the reaction with oxygen)
\[{{C + }}{{{O}}_{{2}}}\, \to {{C}}{{{O}}_2}\,\,{{ - x}}\,{{kJ}}\] \[...\left( {{i}} \right)\]
Heat of formation of hydrogen,
\[{{{H}}_{{2}}}{{ + 2}}{{{O}}_{{2}}} \to {{{H}}_{{2}}}{{O}}\,\,{{ - ykJ\; \; }}...\left( {{{ii}}} \right)\]
Heat of combustion of methane,
\[{{C}}{{{H}}_{{4}}}{{ + 2}}{{{O}}_{{2}}} \to {{C}}{{{O}}_{{2}}}{{ + 2}}{{{H}}_{{2}}}{{O}}\,\,{{ + zkJ\; \; \;}}...\left( {{{iii}}} \right)\]
The required for the formation of methane
\[{{C + 2}}{{{H}}_{{2}}}\, \to {{C}}{{{H}}_{{4}}}\,\,\,{{ + }}\,{{H}}\]
Where H is the heat of formation of the \[{{C}}{{{H}}_{{4}}}\]
To get the required equation,
\[\left( {{i}} \right){{ + 2 \times }}\left( {{{ii}}} \right){{ - }}\left( {{{iii}}} \right)\]
Thus, we obtain
\[{{H}}\,{{ = }}\left[ {\left( {{{ - x}}} \right){{ + }}\left( {{{ - 2y}}} \right){{ - }}\left( {{{ - z}}} \right)} \right]\]
Thus, the heat of formation of methane is \[{{\;}}\left( {{{ x - 2y + z}}} \right){{kJ}}\].
So, the correct answer is D.
Additional information:
The enthalpy change for an exothermic reaction is negative here energy is released from the system. The enthalpy change for an endothermic reaction is negative here energy is absorbed by the system. Hess’s law can be used to find the enthalpy change, physical change, enthalpy change of chemical reaction Bond energy, and Lattice energy Enthalpy of formation.
Note: The part of the universe under study is called a system and the part excluding the system is called surroundings. Based on the exchange of matter and heat thermodynamic systems are of three types they are open system, closed system, isolated system.
Complete step by step answer:
This is solved as per Hess’s law of constant heat summation. It states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes.
Enthalpy is a measure of energy in a thermodynamic system.
Hess’s law is one of the applications of the first law of thermodynamics.
The enthalpy change in a chemical or physical process is the same whether the process is carried out in one step or in several steps.
Methane is \[{{C}}{{{H}}_{{4}}}\]. It is formed from hydrogen and carbon so, the heat of formation of methane can be written as,
\[{{C + 2}}{{{H}}_{{2}}}\, \to {{C}}{{{H}}_{{4}}}\]
From given data, we have
Heat of combustion of carbon, (combustion is the reaction with oxygen)
\[{{C + }}{{{O}}_{{2}}}\, \to {{C}}{{{O}}_2}\,\,{{ - x}}\,{{kJ}}\] \[...\left( {{i}} \right)\]
Heat of formation of hydrogen,
\[{{{H}}_{{2}}}{{ + 2}}{{{O}}_{{2}}} \to {{{H}}_{{2}}}{{O}}\,\,{{ - ykJ\; \; }}...\left( {{{ii}}} \right)\]
Heat of combustion of methane,
\[{{C}}{{{H}}_{{4}}}{{ + 2}}{{{O}}_{{2}}} \to {{C}}{{{O}}_{{2}}}{{ + 2}}{{{H}}_{{2}}}{{O}}\,\,{{ + zkJ\; \; \;}}...\left( {{{iii}}} \right)\]
The required for the formation of methane
\[{{C + 2}}{{{H}}_{{2}}}\, \to {{C}}{{{H}}_{{4}}}\,\,\,{{ + }}\,{{H}}\]
Where H is the heat of formation of the \[{{C}}{{{H}}_{{4}}}\]
To get the required equation,
\[\left( {{i}} \right){{ + 2 \times }}\left( {{{ii}}} \right){{ - }}\left( {{{iii}}} \right)\]
Thus, we obtain
\[{{H}}\,{{ = }}\left[ {\left( {{{ - x}}} \right){{ + }}\left( {{{ - 2y}}} \right){{ - }}\left( {{{ - z}}} \right)} \right]\]
Thus, the heat of formation of methane is \[{{\;}}\left( {{{ x - 2y + z}}} \right){{kJ}}\].
So, the correct answer is D.
Additional information:
The enthalpy change for an exothermic reaction is negative here energy is released from the system. The enthalpy change for an endothermic reaction is negative here energy is absorbed by the system. Hess’s law can be used to find the enthalpy change, physical change, enthalpy change of chemical reaction Bond energy, and Lattice energy Enthalpy of formation.
Note: The part of the universe under study is called a system and the part excluding the system is called surroundings. Based on the exchange of matter and heat thermodynamic systems are of three types they are open system, closed system, isolated system.
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