Respuesta :
(4xC-Cl bond + 2xF-F bond) - (4xC-F bond + 2xCl-Cl bond)
ΔH°rxn=(4x331 + 2x155) - (4x439 + 2x243) = -608kJ/mol-1
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ΔH°rxn=(4x331 + 2x155) - (4x439 + 2x243) = -608kJ/mol-1
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Molar bond enthalpy is defined as the enthalpy of reaction calculated by the subtraction of standard enthalpies of reactants and enthalpies of the products. The estimate value of [tex]\Delta \text{H}_{\text {rxn}}[/tex] for the given equation is -608 kJ/[tex]\text {mol}^{-1}[/tex].
Given equation,
[tex]\begin{aligned}\text{CCl}_{4(\text{g})}+2\text{F}_{2(\text{g})}\rightarrow\text{CF}_{4(\text{g})}+2\text{Cl}_2\end{aligned}[/tex]
- Enthalpy for 4 (C-Cl bonds) = 4 x 331 = 1324 joules.
- Enthalpy for 2 (F - F bonds) = 2 x 155 = 310 joules.
- Enthalpy for 4 (C - F bonds) = 4 x 439 = 1756 joules.
- Enthalpy for 2 (Cl - Cl bonds) = 2 x 243 = 486 joules.
Now, molar bond enthalpy:
[tex][(4\times\text{C}-\text{Cl bond})+(2\times\text{F}-\text{F bond})][/tex][tex]-[(4\times\text{C}-\text{F bond})+(2\times\text{Cl}-\text{Cl bond})]\end{aligned}[/tex]
[tex]\begin{aligned}\Delta\text{H}\degree\text{rxn}&=(1324+310)-(1756+486)\\&=-608\text{kJ/mol}^{-1}\end{aligned}[/tex]
Therefore, the molar bond enthalpy of the reaction is -608 kJ/[tex]\text {mol}^{-1}[/tex].
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https://brainly.com/question/13656766?referrer=searchResults
