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.
Explore More
Similar Questions
Hess's law is applicable for the determination of heat of
For the reaction $H_{2(g)} + I_{2(g)} \rightarrow 2HI_{(g)}; \Delta H = 12.40 \, Kcal$,the heat of formation of $HI$ is ....... $Kcal$.
Medium
View SolutionThe net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus the amount of energy required to form all the bonds in the product molecules. What will be the enthalpy change for the following reaction: $H_{2(g)} + Br_{2(g)} \to 2HBr_{(g)}$? Given that the bond energy of $H_2$,$Br_2$,and $HBr$ is $435 \ kJ \ mol^{-1}$,$192 \ kJ \ mol^{-1}$,and $368 \ kJ \ mol^{-1}$ respectively.
Medium
View SolutionCalculate the enthalpy of formation of ethylene $(C_2H_4)$ from the following data:
$I. C_{(graphite)} + O_{2(g)} \rightarrow CO_{2(g)}; \Delta H = -393.5 \ kJ$
$II. H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)}; \Delta H = -286.2 \ kJ$
$III. C_2H_{4(g)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 2H_2O_{(l)}; \Delta H = -1410.8 \ kJ$ (in $kJ$)
$I. C_{(graphite)} + O_{2(g)} \rightarrow CO_{2(g)}; \Delta H = -393.5 \ kJ$
$II. H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)}; \Delta H = -286.2 \ kJ$
$III. C_2H_{4(g)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 2H_2O_{(l)}; \Delta H = -1410.8 \ kJ$ (in $kJ$)
The standard enthalpies of formation for $CH_{4(g)}$,$CO_{2(g)}$,and $H_2O_{(l)}$ are $-75$,$-393.5$,and $-286 \ kJ \ mol^{-1}$ respectively. The enthalpy change for the reaction $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)}$ is ......... $kJ$.
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo