
The value of $\Delta {\text{H}}$ for the combustion of ${{\text{C}}_{\left( {\text{s}} \right)}}$ is $ - 94.4{\text{Kcal}}$. The heat of formation of ${\text{C}}{{\text{O}}_{2\left( {\text{g}} \right)}}$ is:
A. $ - 49.5{\text{Kcal}}$
B. $ - 94.4{\text{Kcal}}$
C. $ - 188.8{\text{Kcal}}$
D. More data required
Answer
580.5k+ views
Hint: Standard enthalpy of formation is the amount of heat absorbed or released when one mole of the substance is formed at ${\text{2}}{{\text{5}}^ \circ }{\text{C}}$ and $100{\text{kPa}}$ from its elements in their standard states. Combustion is a process in which a substance reacts with the oxygen of air to give heat and light is called combustion.
Complete step by step answer:
Enthalpy change is the change of heat content of the system at constant pressure. Standard heat of formation is called standard enthalpy of formation. Its unit is ${\text{kJ}}.{\text{mo}}{{\text{l}}^{ - 1}}$. Formation reactions and their enthalpy change values may be manipulated to determine the balanced equation and the enthalpy change value of chemical reactions.
It can be calculated by subtracting the enthalpy change of reactants, $\Delta {{\text{H}}_{\text{r}}}$ from the enthalpy change of products, $\Delta {{\text{H}}_{\text{p}}}$.
So when hydrocarbons are involved in the reaction which gives carbon dioxide and water molecules, it is called combustion. Consider the reaction of formation of ${\text{C}}{{\text{O}}_{2\left( {\text{g}} \right)}}$.
${{\text{C}}_{\left( {\text{s}} \right)}}{\text{ + }}{{\text{O}}_{2\left( {\text{g}} \right)}}{\text{C}}{{\text{O}}_{2\left( {\text{g}} \right)}}$
This is also a combustion reaction since carbon dioxide is formed and the carbon is burnt in the presence of oxygen. Thus the enthalpy of both will be the same.
i.e. the enthalpy of formation of ${\text{C}}{{\text{O}}_{2\left( {\text{g}} \right)}}$ is same as that of the enthalpy of combustion of ${{\text{C}}_{\left( {\text{s}} \right)}}$ which is equal to $ - 94.4{\text{Kcal}}$.
Hence the correct option is B.
Note:
For complex organic compounds, it is difficult to directly measure its formation. Instead, the compound is combusted and the products analyzed to determine the enthalpy change of formation, for the original compound. Also the standard heat of combustion is the enthalpy change for the combustion of one mole of compound.
Complete step by step answer:
Enthalpy change is the change of heat content of the system at constant pressure. Standard heat of formation is called standard enthalpy of formation. Its unit is ${\text{kJ}}.{\text{mo}}{{\text{l}}^{ - 1}}$. Formation reactions and their enthalpy change values may be manipulated to determine the balanced equation and the enthalpy change value of chemical reactions.
It can be calculated by subtracting the enthalpy change of reactants, $\Delta {{\text{H}}_{\text{r}}}$ from the enthalpy change of products, $\Delta {{\text{H}}_{\text{p}}}$.
So when hydrocarbons are involved in the reaction which gives carbon dioxide and water molecules, it is called combustion. Consider the reaction of formation of ${\text{C}}{{\text{O}}_{2\left( {\text{g}} \right)}}$.
${{\text{C}}_{\left( {\text{s}} \right)}}{\text{ + }}{{\text{O}}_{2\left( {\text{g}} \right)}}{\text{C}}{{\text{O}}_{2\left( {\text{g}} \right)}}$
This is also a combustion reaction since carbon dioxide is formed and the carbon is burnt in the presence of oxygen. Thus the enthalpy of both will be the same.
i.e. the enthalpy of formation of ${\text{C}}{{\text{O}}_{2\left( {\text{g}} \right)}}$ is same as that of the enthalpy of combustion of ${{\text{C}}_{\left( {\text{s}} \right)}}$ which is equal to $ - 94.4{\text{Kcal}}$.
Hence the correct option is B.
Note:
For complex organic compounds, it is difficult to directly measure its formation. Instead, the compound is combusted and the products analyzed to determine the enthalpy change of formation, for the original compound. Also the standard heat of combustion is the enthalpy change for the combustion of one mole of compound.
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