Répondre :
To calculate the heat of dissociation (\( \Delta H^\circ \)) of gaseous water at 68°C using the provided data, we can use the equation:
\[
\Delta H_{\text{dissociation}}^\circ(T_2) = \Delta H_{\text{dissociation}}^\circ(T_1) + \int_{T_1}^{T_2} C_p \, dT
\]
Where:
- \( \Delta H_{\text{dissociation}}^\circ(T_2) \) is the heat of dissociation at the final temperature (68°C).
- \( \Delta H_{\text{dissociation}}^\circ(T_1) \) is the heat of dissociation at the initial temperature (18°C).
- \( C_p \) is the molar heat capacity of gaseous water at constant pressure.
Given:
- \( \Delta H_{\text{dissociation}}^\circ(291\, \text{K}) = 241750\, \text{J} \)
- \( C_{p(\text{H}_2\text{O})} = 33.56\, \text{J} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \)
- \( C_{p(\text{H}_2)} = 28.83\, \text{J} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \)
- \( C_{p(\text{O}_2)} = 29.12\, \text{J} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \)
We need to integrate the heat capacity (\( C_p \)) of water with respect to temperature (\( T \)) from 18°C to 68°C and add it to the initial heat of dissociation.
Let's calculate:
\[
\Delta H_{\text{dissociation}}^\circ(68^\circ\text{C}) = \Delta H_{\text{dissociation}}^\circ(291\, \text{K}) + \int_{291\, \text{K}}^{T_2} C_{p(\text{H}_2\text{O})} \, dT
\]
First, let's convert 68°C to Kelvin:
\[
T_2 = 68^\circ\text{C} + 273.15 = 341.15\, \text{K}
\]
Now, let's perform the integration:
\[
\Delta H_{\text{dissociation}}^\circ(68^\circ\text{C}) = 241750\, \text{J} + \int_{291\, \text{K}}^{341.15\, \text{K}} 33.56\, \text{J} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \, dT
\]
\[
\Delta H_{\text{dissociation}}^\circ(68^\circ\text{C}) = 241750\, \text{J} + 33.56\, \text{J} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \cdot (341.15\, \text{K} - 291\, \text{K})
\]
\[
\Delta H_{\text{dissociation}}^\circ(68^\circ\text{C}) = 241750\, \text{J} + 33.56\, \text{J} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \cdot 50.15\, \text{K}
\]
\[
\Delta H_{\text{dissociation}}^\circ(68^\circ\text{C}) \approx 241750\, \text{J} + 1685.524\, \text{J/mol}
\]
\[
\Delta H_{\text{dissociation}}^\circ(68^\circ\text{C}) \approx 243435.524\, \text{J/mol}
\]
So, the heat of dissociation of gaseous water at 68°C is approximately \( 243435.524\, \text{J/mol} \).
Merci d'avoir visité notre site Web dédié à Physique/Chimie. Nous espérons que les informations partagées vous ont été utiles. N'hésitez pas à nous contacter si vous avez des questions ou besoin d'assistance. À bientôt, et pensez à ajouter ce site à vos favoris !