Answered

Consider the reaction of glucose with oxygen:

[tex]\[ C_6H_{12}O_6(s) + 6O_2(g) \rightarrow 6CO_2(g) + 6H_2O(l) \][/tex]

The enthalpy of formation \(\left(\Delta H_f\right)\) for \( C_6H_{12}O_6(s) \) is \(-1,273.02 \, \text{kJ/mol}\), \(\Delta H_f\) for \( CO_2(g) \) is \(-393.5 \, \text{kJ/mol}\), and \(\Delta H_f\) for \( H_2O(l) \) is \(-285.83 \, \text{kJ/mol}\).

What is [tex]\(\Delta H_f\)[/tex] for [tex]\( O_2(g) \)[/tex]?



Answer :

GinnyAnswer
To determine the enthalpy of formation for \( O_2(g) \) and the enthalpy change of the reaction, we can follow a detailed, step-by-step process:

1. Identify the Given Data:
- Enthalpy of formation (\(\Delta H_f\)) for glucose (\( C_6H_{12}O_6(s) \)): \(-1273.02 \ \text{kJ/mol}\)
- Enthalpy of formation (\(\Delta H_f\)) for carbon dioxide (\( CO_2(g) \)): \(-393.5 \ \text{kJ/mol}\)
- Enthalpy of formation (\(\Delta H_f\)) for water (\( H_2O(l) \)): \(-285.83 \ \text{kJ/mol}\)

2. Write the Balanced Chemical Equation:
[tex]\[ C_6H_{12}O_6(s) + 6O_2(g) \rightarrow 6CO_2(g) + 6H_2O(l) \][/tex]

3. Calculate the Enthalpy of the Reaction (\(\Delta H_{reaction}\)):

The enthalpy change for the reaction (\(\Delta H_{reaction}\)) can be calculated using the formula:
[tex]\[ \Delta H_{reaction} = \left[ \sum \Delta H_f (\text{products}) \right] - \left[ \sum \Delta H_f (\text{reactants}) \right] \][/tex]

For this reaction, the products are \( 6CO_2(g) \) and \( 6H_2O(l) \), and the reactants are \( C_6H_{12}O_6(s) \) and \( 6O_2(g) \).

4. Determine the Enthalpy of Formation of Products:
[tex]\[ 6 \Delta H_f (CO_2) + 6 \Delta H_f (H_2O) = 6 \times (-393.5 \ \text{kJ/mol}) + 6 \times (-285.83 \ \text{kJ/mol}) = -2361.0 \ \text{kJ} + (-1714.98 \ \text{kJ}) = -4075.98 \ \text{kJ} \][/tex]

5. Determine the Enthalpy of Formation of Reactants:
[tex]\[ \Delta H_f (C_6H_{12}O_6) + 6 \Delta H_f (O_2) = -1273.02 \ \text{kJ} + 6 \times 0 \ \text{kJ} = -1273.02 \ \text{kJ} \][/tex]

Note: The enthalpy of formation (\(\Delta H_f\)) of \( O_2(g) \) in its standard state is zero (\(0 \ \text{kJ/mol}\)).

6. Calculate the Enthalpy Change for the Reaction:
[tex]\[ \Delta H_{reaction} = \big( -4075.98 \ \text{kJ} \big) - \big( -1273.02 \ \text{kJ} \big) = -4075.98 \ \text{kJ} + 1273.02 \ \text{kJ} = -2802.96 \ \text{kJ} \][/tex]

7. Conclusion:
- The enthalpy change for the reaction (\(\Delta H_{reaction}\)) is \(-2802.96 \ \text{kJ}\).
- The enthalpy of formation (\(\Delta H_f\)) for \( O_2(g) \) is \( 0 \ \text{kJ/mol} \) (since it is an element in its standard state).

Therefore, the enthalpy of formation for [tex]\( O_2(g) \)[/tex] is [tex]\( 0 \ \text{kJ/mol} \)[/tex].