Question
Explain the standard enthalpy change in term of enthalpy change of formation with mathematical expressions
Answers
Answered by
GPT-4o mini
The standard enthalpy change of a reaction (\( \Delta H^\circ_{\text{reaction}} \)) can be expressed in terms of the enthalpy changes of formation (\( \Delta H_f^\circ \)) of the reactants and products involved in the reaction.
The standard enthalpy change of formation (\( \Delta H_f^\circ \)) of a compound is defined as the change in enthalpy when one mole of the compound is formed from its elements in their standard states.
For a general chemical reaction:
\[
aA + bB \rightarrow cC + dD
\]
where:
- \( A \), \( B \), \( C \), and \( D \) are the chemical species,
- \( a \), \( b \), \( c \), and \( d \) are their respective stoichiometric coefficients.
The standard enthalpy change of this reaction (\( \Delta H^\circ_{\text{reaction}} \)) can be calculated using the enthalpy changes of formation of the products and reactants as follows:
\[
\Delta H^\circ_{\text{reaction}} = \sum (\Delta H_f^\circ \text{ of products}) - \sum (\Delta H_f^\circ \text{ of reactants})
\]
In a more mathematical expression, this can be written as:
\[
\Delta H^\circ_{\text{reaction}} = [c \Delta H_f^\circ (C) + d \Delta H_f^\circ (D)] - [a \Delta H_f^\circ (A) + b \Delta H_f^\circ (B)]
\]
### Example
Consider the combustion of methane (\( CH_4 \)):
\[
CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)
\]
The standard enthalpy change of this reaction would be:
\[
\Delta H^\circ_{\text{reaction}} = [\Delta H_f^\circ (CO_2) + 2 \Delta H_f^\circ (H_2O)] - [\Delta H_f^\circ (CH_4) + 2\Delta H_f^\circ (O_2)]
\]
Since the standard enthalpy of formation of any element in its standard state (like \( O_2(g) \)) is zero, we simplify it to:
\[
\Delta H^\circ_{\text{reaction}} = [\Delta H_f^\circ (CO_2) + 2 \Delta H_f^\circ (H_2O)] - [\Delta H_f^\circ (CH_4)]
\]
This expression allows us to calculate the enthalpy change of the reaction based on standard enthalpies of formation.
The standard enthalpy change of formation (\( \Delta H_f^\circ \)) of a compound is defined as the change in enthalpy when one mole of the compound is formed from its elements in their standard states.
For a general chemical reaction:
\[
aA + bB \rightarrow cC + dD
\]
where:
- \( A \), \( B \), \( C \), and \( D \) are the chemical species,
- \( a \), \( b \), \( c \), and \( d \) are their respective stoichiometric coefficients.
The standard enthalpy change of this reaction (\( \Delta H^\circ_{\text{reaction}} \)) can be calculated using the enthalpy changes of formation of the products and reactants as follows:
\[
\Delta H^\circ_{\text{reaction}} = \sum (\Delta H_f^\circ \text{ of products}) - \sum (\Delta H_f^\circ \text{ of reactants})
\]
In a more mathematical expression, this can be written as:
\[
\Delta H^\circ_{\text{reaction}} = [c \Delta H_f^\circ (C) + d \Delta H_f^\circ (D)] - [a \Delta H_f^\circ (A) + b \Delta H_f^\circ (B)]
\]
### Example
Consider the combustion of methane (\( CH_4 \)):
\[
CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)
\]
The standard enthalpy change of this reaction would be:
\[
\Delta H^\circ_{\text{reaction}} = [\Delta H_f^\circ (CO_2) + 2 \Delta H_f^\circ (H_2O)] - [\Delta H_f^\circ (CH_4) + 2\Delta H_f^\circ (O_2)]
\]
Since the standard enthalpy of formation of any element in its standard state (like \( O_2(g) \)) is zero, we simplify it to:
\[
\Delta H^\circ_{\text{reaction}} = [\Delta H_f^\circ (CO_2) + 2 \Delta H_f^\circ (H_2O)] - [\Delta H_f^\circ (CH_4)]
\]
This expression allows us to calculate the enthalpy change of the reaction based on standard enthalpies of formation.