(D) The process involves three steps:$1. \text{Dissociation of } \frac{1}{2} Cl_{2(g)}$ $\rightarrow Cl_{(g)}: \Delta H_1 = \frac{1}{2} \times 240 = 1

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(D) The process involves three steps:$1. \text{Dissociation of } \frac{1}{2} Cl_{2(g)}$ $\rightarrow Cl_{(g)}: \Delta H_1 = \frac{1}{2} \times 240 = 120 \, kJ \, mol^{-1}$$2. \text{Electron gain of } Cl_{(g)} \rightarrow Cl^{-}_{(g)}: \Delta H_2 = -350 \, kJ \, mol^{-1}$$3. \text{Hydration of } Cl^{-}_{(g)} \rightarrow Cl^{-}_{(aq)}: \Delta H_3 = -380 \, kJ \, mol^{-1}$Total enthalpy change $\Delta H^{\circ} = \Delta H_1 + \Delta H_2 + \Delta H_3$$\Delta H^{\circ} = 120 + (-350) + (-380) = -610 \, kJ \, mol^{-1}$The magnitude is $610$.

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

Difficult

The enthalpy change for the conversion of $\frac{1}{2} Cl_{2(g)}$ to $Cl^{-}_{(aq)}$ is $......$ $kJ \, mol^{-1}$ (Nearest integer).
Given:
$\Delta_{dis}H^{\circ}_{Cl_{2(g)}} = 240 \, kJ \, mol^{-1}$
$\Delta_{eg}H^{\circ}_{Cl_{(g)}} = -350 \, kJ \, mol^{-1}$
$\Delta_{hyd}H^{\circ}_{Cl^{-(g)}} = -380 \, kJ \, mol^{-1}$

JEE MAIN 2023 All JEE MAIN

A
$600$
B
$620$
C
$630$
D
$610$

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Thermodynamics

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Heat of reaction, Bond energy and Hess law

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Given $C + O_2 \rightarrow CO_2$ : $\Delta H = -x \ kJ$ and $2CO + O_2 \rightarrow 2CO_2$ : $\Delta H^\circ = -y \ kJ$,find the enthalpy of formation of carbon monoxide.

Difficult

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The standard enthalpy of formation of $NH_3$ is $-46.0 \ kJ \ mol^{-1}.$ If the enthalpy of formation of $H_2$ from its atoms is $-436 \ kJ \ mol^{-1}$ and that of $N_2$ is $-712 \ kJ \ mol^{-1},$ the average bond enthalpy of $N-H$ bond in $NH_3$ is ................ $kJ \ mol^{-1}$

MediumJEE MAIN 2014

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The bond enthalpies of $H_2$,$X_2$,and $HX$ are in the ratio of $2 : 1 : 2$. If the standard enthalpy of formation of $HX$ is $-50 \ kJ \ mol^{-1}$,the bond enthalpy of $H_2$ is .... $kJ \ mol^{-1}$.

Difficult

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$C_{(diamond)} + O_2 \to CO_2; \Delta H = -395.3 \ kJ/mole$
$C_{(graphite)} + O_2 \to CO_2; \Delta H = -393.4 \ kJ/mole$
$C_{(graphite)} \to C_{(diamond)}; \Delta H = ?$

Medium

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Consider the following data:
$\Delta_{f}H^{\Theta}(CH_{4}, g) = -X \ kJ \ mol^{-1}$
Enthalpy of sublimation of graphite = $Y \ kJ \ mol^{-1}$
Dissociation enthalpy of $H_{2} = Z \ kJ \ mol^{-1}$
The bond enthalpy of $C-H$ bond is given by:

DifficultJEE MAIN 2026

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Given $C + O_2 \rightarrow CO_2$ : $\Delta H = -x \ kJ$ and $2CO + O_2 \rightarrow 2CO_2$ : $\Delta H^\circ = -y \ kJ$,find the enthalpy of formation of carbon monoxide.

The standard enthalpy of formation of $NH_3$ is $-46.0 \ kJ \ mol^{-1}.$ If the enthalpy of formation of $H_2$ from its atoms is $-436 \ kJ \ mol^{-1}$ and that of $N_2$ is $-712 \ kJ \ mol^{-1},$ the average bond enthalpy of $N-H$ bond in $NH_3$ is ................ $kJ \ mol^{-1}$

The bond enthalpies of $H_2$,$X_2$,and $HX$ are in the ratio of $2 : 1 : 2$. If the standard enthalpy of formation of $HX$ is $-50 \ kJ \ mol^{-1}$,the bond enthalpy of $H_2$ is .... $kJ \ mol^{-1}$.

$C_{(diamond)} + O_2 \to CO_2; \Delta H = -395.3 \ kJ/mole$$C_{(graphite)} + O_2 \to CO_2; \Delta H = -393.4 \ kJ/mole$$C_{(graphite)} \to C_{(diamond)}; \Delta H = ?$

Consider the following data:$\Delta_{f}H^{\Theta}(CH_{4}, g) = -X \ kJ \ mol^{-1}$Enthalpy of sublimation of graphite = $Y \ kJ \ mol^{-1}$Dissociation enthalpy of $H_{2} = Z \ kJ \ mol^{-1}$The bond enthalpy of $C-H$ bond is given by:

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$C_{(diamond)} + O_2 \to CO_2; \Delta H = -395.3 \ kJ/mole$
$C_{(graphite)} + O_2 \to CO_2; \Delta H = -393.4 \ kJ/mole$
$C_{(graphite)} \to C_{(diamond)}; \Delta H = ?$

Consider the following data:
$\Delta_{f}H^{\Theta}(CH_{4}, g) = -X \ kJ \ mol^{-1}$
Enthalpy of sublimation of graphite = $Y \ kJ \ mol^{-1}$
Dissociation enthalpy of $H_{2} = Z \ kJ \ mol^{-1}$
The bond enthalpy of $C-H$ bond is given by: