(A) The enthalpy change of a reaction is calculated as $\Delta H = \Sigma(\text{Bond Energies of reactants}) - \Sigma(\text{Bond Energies of products}

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(A) The enthalpy change of a reaction is calculated as $\Delta H = \Sigma(\text{Bond Energies of reactants}) - \Sigma(\text{Bond Energies of products})$.Energy required to break reactant bonds: $(2 \times 436 \ kJ/mol) + (1 \times 496 \ kJ/mol) = 872 + 496 = 1368 \ kJ$.Energy released during the formation of product bonds ($2$ moles of $H_2O$ contain $4$ $O-H$ bonds): $4 \times 464 \ kJ/mol = 1856 \ kJ$.$\Delta H = 1368 \ kJ - 1856 \ kJ = -488 \ kJ/mol$.

Class 11 Chemistry Thermodynamics Heat of reaction, Bond energy and Hess law

Medium

Given that the bond energy of $H-H$ bonds is $436 \ kJ/mol$,$O-H$ bonds is $464 \ kJ/mol$,and $O=O$ bonds is $496 \ kJ/mol$,what is the approximate heat of reaction for $2H_2 + O_2 \longrightarrow 2H_2O$ in $kJ/mol$?

A
$-488$
B
$-440$
C
$440$
D
$488$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Chemistry Questions

Chapter

Thermodynamics

Subtopic

Heat of reaction, Bond energy and Hess law

The heat of formation of $H_2O$ is $-286 \, kJ/mol$ and $H_2O_2$ is $-188 \, kJ/mol$. The enthalpy change for the reaction $2H_2O_2 \to 2H_2O + O_2$ is......$kJ$.

Medium

View Solution

For the reaction,$H_{2(g)} + I_{2(g)} \to 2HI_{(g)}$,$\Delta H = -12.40 \ kcal$. According to this,the heat of formation of $HI$ will be......$kcal$.

Medium

View Solution

The enthalpy of neutralization of $HCN$ by $NaOH$ is $-12.13 \, kJ \, mol^{-1}$. The enthalpy of ionisation of $HCN$ will be $... \, kJ \, mol^{-1}$.

Difficult

View Solution

Heat of combustion of two isomers $x$ and $y$ are $17 \ kJ/mol$ and $12 \ kJ/mol$ respectively. From this information,it may be concluded that:

Medium

View Solution

In the following reaction,how much enthalpy change in $kJ$ is associated with the formation of $5.67 \ mol$ of $HCl$ gas? $H_{2(g)} + Cl_{2(g)} \to 2HCl_{(g)} \; ; \Delta H = -184.6 \ kJ$

Difficult

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Given that the bond energy of $H-H$ bonds is $436 \ kJ/mol$,$O-H$ bonds is $464 \ kJ/mol$,and $O=O$ bonds is $496 \ kJ/mol$,what is the approximate heat of reaction for $2H_2 + O_2 \longrightarrow 2H_2O$ in $kJ/mol$?

Similar Questions

The heat of formation of $H_2O$ is $-286 \, kJ/mol$ and $H_2O_2$ is $-188 \, kJ/mol$. The enthalpy change for the reaction $2H_2O_2 \to 2H_2O + O_2$ is......$kJ$.

For the reaction,$H_{2(g)} + I_{2(g)} \to 2HI_{(g)}$,$\Delta H = -12.40 \ kcal$. According to this,the heat of formation of $HI$ will be......$kcal$.

The enthalpy of neutralization of $HCN$ by $NaOH$ is $-12.13 \, kJ \, mol^{-1}$. The enthalpy of ionisation of $HCN$ will be $... \, kJ \, mol^{-1}$.

Heat of combustion of two isomers $x$ and $y$ are $17 \ kJ/mol$ and $12 \ kJ/mol$ respectively. From this information,it may be concluded that:

In the following reaction,how much enthalpy change in $kJ$ is associated with the formation of $5.67 \ mol$ of $HCl$ gas? $H_{2(g)} + Cl_{2(g)} \to 2HCl_{(g)} \; ; \Delta H = -184.6 \ kJ$

Vedclass Test Series

Exam Paper Generator

Online Exam Module