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

Medium

One mole of acetone requires less heat to vaporise than $1$ mol of water. Which of the two liquids has higher enthalpy of vaporisation?

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

Vedclass pdf generator app on play store

Solution

(B) One mole of acetone requires less heat to vaporise than $1$ mole of water.
Enthalpy of vaporisation is the amount of energy required to vaporise one mole of a liquid at constant temperature and pressure.
Since water requires more heat for vaporisation,it has a higher enthalpy of vaporisation compared to acetone.
This can be represented as $(\Delta H_{vap})_{water} > (\Delta H_{vap})_{acetone}$.

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

Given $H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}$; $\Delta H = -44 \, Kcal$ and $2Na_{(s)} + 2HCl_{(g)} \rightarrow 2NaCl_{(s)} + H_{2(g)}$; $\Delta H = -152 \, Kcal$,calculate $\Delta H$ for the reaction $Na_{(s)} + 0.5 Cl_{2(g)} \rightarrow NaCl_{(s)}$ in $Kcal$.

Medium

View Solution

Given that $\Delta H_f(H) = 218 \ kJ/mol$,express the $H-H$ bond energy in $kcal/mol$.

DifficultTS EAMCET 2009

View Solution

Assertion $(A)$: Formation of $H_2O_{(\ell)}$ from $H_{2(g)}$ and $O_{2(g)}$ is more exothermic than formation of $H_2O_{(g)}$ from $H_{2(g)}$ and $O_{2(g)}$.
Reason $(R)$: $H_2O_{(\ell)}$ has more attraction force than $H_2O_{(g)}$ and it is more stable than $H_2O_{(g)}$.

Medium

View Solution

The heat of reaction does not depend upon

Medium

View Solution

The enthalpy of combustion of propane,graphite and dihydrogen at $298 \ K$ are: $-2220.0 \ kJ \ mol^{-1}$,$-393.5 \ kJ \ mol^{-1}$ and $-285.8 \ kJ \ mol^{-1}$ respectively. The magnitude of the enthalpy of formation of propane $(C_{3}H_{8})$ is ......... $kJ \ mol^{-1}$. (Nearest integer)

MediumJEE MAIN 2022

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

One mole of acetone requires less heat to vaporise than $1$ mol of water. Which of the two liquids has higher enthalpy of vaporisation?

Similar Questions

Given $H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}$; $\Delta H = -44 \, Kcal$ and $2Na_{(s)} + 2HCl_{(g)} \rightarrow 2NaCl_{(s)} + H_{2(g)}$; $\Delta H = -152 \, Kcal$,calculate $\Delta H$ for the reaction $Na_{(s)} + 0.5 Cl_{2(g)} \rightarrow NaCl_{(s)}$ in $Kcal$.

Given that $\Delta H_f(H) = 218 \ kJ/mol$,express the $H-H$ bond energy in $kcal/mol$.

Assertion $(A)$: Formation of $H_2O_{(\ell)}$ from $H_{2(g)}$ and $O_{2(g)}$ is more exothermic than formation of $H_2O_{(g)}$ from $H_{2(g)}$ and $O_{2(g)}$.Reason $(R)$: $H_2O_{(\ell)}$ has more attraction force than $H_2O_{(g)}$ and it is more stable than $H_2O_{(g)}$.

The heat of reaction does not depend upon

The enthalpy of combustion of propane,graphite and dihydrogen at $298 \ K$ are: $-2220.0 \ kJ \ mol^{-1}$,$-393.5 \ kJ \ mol^{-1}$ and $-285.8 \ kJ \ mol^{-1}$ respectively. The magnitude of the enthalpy of formation of propane $(C_{3}H_{8})$ is ......... $kJ \ mol^{-1}$. (Nearest integer)

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Assertion $(A)$: Formation of $H_2O_{(\ell)}$ from $H_{2(g)}$ and $O_{2(g)}$ is more exothermic than formation of $H_2O_{(g)}$ from $H_{2(g)}$ and $O_{2(g)}$.
Reason $(R)$: $H_2O_{(\ell)}$ has more attraction force than $H_2O_{(g)}$ and it is more stable than $H_2O_{(g)}$.