What is meant by the term average bond enthalpy? Why is there a difference in the bond enthalpy of the $O-H$ bond in ethanol $(C_{2}H_{5}OH)$ and water $(H_{2}O)$?
(N/A) The similar bonds in a molecule do not possess the same bond enthalpies. For example,in an $H_{2}O$ $(H-O-H)$ molecule,after the breaking of the first $O-H$ bond,the second $O-H$ bond undergoes some change because of the changed chemical environment.
Therefore,in polyatomic molecules,the term mean or average bond enthalpy is used. It is obtained by dividing the total bond dissociation enthalpy by the number of bonds broken.
For example:
$H_{2}O_{(g)} \rightarrow H_{(g)} + OH_{(g)}$; $\Delta_{a} H_{1}^{0} = 502 \ kJ \ mol^{-1}$
$OH_{(g)} \rightarrow H_{(g)} + O_{(g)}$; $\Delta_{a} H_{2}^{0} = 427 \ kJ \ mol^{-1}$
Average $O-H$ bond enthalpy $= \frac{502 + 427}{2} = 464.5 \ kJ \ mol^{-1}$
The bond enthalpies of the $O-H$ bond in $C_{2}H_{5}OH$ and $H_{2}O$ are different because of the different chemical (electronic) environment around the oxygen atom.
Therefore,in polyatomic molecules,the term mean or average bond enthalpy is used. It is obtained by dividing the total bond dissociation enthalpy by the number of bonds broken.
For example:
$H_{2}O_{(g)} \rightarrow H_{(g)} + OH_{(g)}$; $\Delta_{a} H_{1}^{0} = 502 \ kJ \ mol^{-1}$
$OH_{(g)} \rightarrow H_{(g)} + O_{(g)}$; $\Delta_{a} H_{2}^{0} = 427 \ kJ \ mol^{-1}$
Average $O-H$ bond enthalpy $= \frac{502 + 427}{2} = 464.5 \ kJ \ mol^{-1}$
The bond enthalpies of the $O-H$ bond in $C_{2}H_{5}OH$ and $H_{2}O$ are different because of the different chemical (electronic) environment around the oxygen atom.
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Similar Questions
Given:
$C + 2S \to CS_2 ; \Delta H_f^o = +117.0 \, kJ \, mol^{-1} \dots (1)$
$C + O_2 \to CO_2 ; \Delta H_f^o = -393 \, kJ \, mol^{-1} \dots (2)$
$S + O_2 \to SO_2 ; \Delta H_f^o = -297 \, kJ \, mol^{-1} \dots (3)$
The heat of reaction for $CS_2 + 3O_2 \to CO_2 + 2SO_2$ is:
.....$kJ \, mol^{-1}$
$C + 2S \to CS_2 ; \Delta H_f^o = +117.0 \, kJ \, mol^{-1} \dots (1)$
$C + O_2 \to CO_2 ; \Delta H_f^o = -393 \, kJ \, mol^{-1} \dots (2)$
$S + O_2 \to SO_2 ; \Delta H_f^o = -297 \, kJ \, mol^{-1} \dots (3)$
The heat of reaction for $CS_2 + 3O_2 \to CO_2 + 2SO_2$ is:
.....$kJ \, mol^{-1}$
Medium
View SolutionIf enthalpies of formation of $C_2H_{4(g)}$,$CO_{2(g)}$ and $H_2O_{(l)}$ at $25 \ ^\circ C$ and $1 \ atm$ pressure are $52$,$-394$ and $-286 \ kJ \ mol^{-1}$ respectively,the enthalpy of combustion of $C_2H_{4(g)}$ will be.....$kJ \ mol^{-1}$.
MediumAIIMS 1998
View SolutionOn the basis of the following equations,the heat of dimerisation of $NO_2$ will be:
$(i) \ N_2 + 2O_2 \to 2NO_2, \Delta H = 67.9 \ kJ$
$(ii) \ N_2 + 2O_2 \to N_2O_4, \Delta H = 9.3 \ kJ$
$(i) \ N_2 + 2O_2 \to 2NO_2, \Delta H = 67.9 \ kJ$
$(ii) \ N_2 + 2O_2 \to N_2O_4, \Delta H = 9.3 \ kJ$
Medium
View SolutionThe atomization enthalpies of $NH_{3(g)}$ and $N_2H_{4(g)}$ are $+150 \ kJ \ mol^{-1}$ and $+310 \ kJ \ mol^{-1}$ respectively. The $\Delta H(N-N)$ bond enthalpy in $kJ \ mol^{-1}$ is:
Medium
View SolutionCalculate $\Delta H$ for the reaction: $H_{2(g)} + O_{2(g)} \rightarrow H_2O_{2(g)}$ given the bond energies: $BE_{H-H} = 436 \ kJ/mol$,$BE_{O=O} = 499 \ kJ/mol$,$BE_{O-O} = 142 \ kJ/mol$,and $BE_{O-H} = 460 \ kJ/mol$. (in $kJ$)
Easy
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