(B) The enthalpy of reaction is given by the formula:$\Delta H_{r}^{\circ} = \sum (B.E.)_{\text{reactants}} - \sum (B.E.)_{\text{products}}$Substituti

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(B) The enthalpy of reaction is given by the formula:$\Delta H_{r}^{\circ} = \sum (B.E.)_{\text{reactants}} - \sum (B.E.)_{\text{products}}$Substituting the given values:$-185 = [B.E._{H-H} + B.E._{Cl-Cl}] - [2 \times B.E._{H-Cl}]$$-185 = [435 + 244] - 2x$$-185 = 679 - 2x$$2x = 679 + 185$$2x = 864$$x = 432 \ kJ \ mol^{-1}$Therefore,the bond enthalpy of the $H-Cl$ bond is $432 \ kJ \ mol^{-1}$.

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

Medium

Calculate the bond enthalpy of the $H-Cl$ bond from the following reaction:
$H_{2(g)} + Cl_{2(g)} \rightarrow 2 HCl_{(g)}$,$\Delta_{r} H^{\circ} = -185 \ kJ \ mol^{-1}$
(Given bond enthalpies of $H-H$ and $Cl-Cl$ bonds are $435.0 \ kJ \ mol^{-1}$ and $244 \ kJ \ mol^{-1}$ respectively.)

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A
$340 \ kJ \ mol^{-1}$
B
$432 \ kJ \ mol^{-1}$
C
$370 \ kJ \ mol^{-1}$
D
$864 \ kJ \ mol^{-1}$

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Thermodynamics

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Heat of reaction, Bond energy and Hess law

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Medium

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Calculate the bond enthalpy of the $H-Cl$ bond from the following reaction:$H_{2(g)} + Cl_{2(g)} \rightarrow 2 HCl_{(g)}$,$\Delta_{r} H^{\circ} = -185 \ kJ \ mol^{-1}$(Given bond enthalpies of $H-H$ and $Cl-Cl$ bonds are $435.0 \ kJ \ mol^{-1}$ and $244 \ kJ \ mol^{-1}$ respectively.)

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The enthalpies of formation of $Al_{2}O_{3}$ and $Cr_{2}O_{3}$ are $-1596 \ kJ$ and $-1134 \ kJ$ respectively. $\Delta H$ for the reaction $2Al + Cr_{2}O_{3} \to 2Cr + Al_{2}O_{3}$ is.......$kJ$

The bond dissociation energies of $X_2$,$Y_2$ and $XY$ are in the ratio of $1 : 0.5 : 1$. $\Delta H$ for the formation of $XY$ is $-200 \ kJ \ mol^{-1}$. The bond dissociation energy of $X_2$ will be $...... \ kJ \ mol^{-1}$.

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$AB$,$A_2$,and $B_2$ are diatomic molecules. Enthalpies of dissociation of $AB$,$A_2$,and $B_2$ are in the ratio of $1:1:0.5$. Enthalpy of formation of $AB$,$\Delta_f H = -100 \ kJ \ mol^{-1}$. Find the dissociation enthalpy of $A_2$?Reaction : $\frac{1}{2} A_2 + \frac{1}{2} B_2 \to AB$

If the bond energies of $H-H$,$Br-Br$ and $H-Br$ are $433$,$192$ and $364 \ kJ \ mol^{-1}$ respectively,then $\Delta H^{\circ}$ for the reaction: $H_{2(g)} + Br_{2(g)} \rightarrow 2HBr_{(g)}$ is:

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Calculate the bond enthalpy of the $H-Cl$ bond from the following reaction:
$H_{2(g)} + Cl_{2(g)} \rightarrow 2 HCl_{(g)}$,$\Delta_{r} H^{\circ} = -185 \ kJ \ mol^{-1}$
(Given bond enthalpies of $H-H$ and $Cl-Cl$ bonds are $435.0 \ kJ \ mol^{-1}$ and $244 \ kJ \ mol^{-1}$ respectively.)

$AB$,$A_2$,and $B_2$ are diatomic molecules. Enthalpies of dissociation of $AB$,$A_2$,and $B_2$ are in the ratio of $1:1:0.5$. Enthalpy of formation of $AB$,$\Delta_f H = -100 \ kJ \ mol^{-1}$. Find the dissociation enthalpy of $A_2$?
Reaction : $\frac{1}{2} A_2 + \frac{1}{2} B_2 \to AB$