At standard conditions,if the change in the enthalpy for the following reaction is $-109 \; kJ \ mol^{-1}$:
$H_{2(g)} + Br_{2(g)} \rightarrow 2HBr_{(g)}$
Given that bond energy of $H_2$ and $Br_2$ is $435 \; kJ \ mol^{-1}$ and $192 \; kJ \ mol^{-1}$,respectively,what is the bond energy (in $kJ \ mol^{-1}$) of $HBr$?
$H_{2(g)} + Br_{2(g)} \rightarrow 2HBr_{(g)}$
Given that bond energy of $H_2$ and $Br_2$ is $435 \; kJ \ mol^{-1}$ and $192 \; kJ \ mol^{-1}$,respectively,what is the bond energy (in $kJ \ mol^{-1}$) of $HBr$?
- A$259$
- B$368$
- C$736$
- D$518$
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If the enthalpy of atomisation for $Br_{2(l)}$ is $x \ kJ/mol$ and the bond enthalpy for $Br_{2(g)}$ is $y \ kJ/mol$,what is the relation between them?
DifficultJEE MAIN 2020
View SolutionThe standard enthalpies of formation of $Al_{2}O_{3}$ and $CaO$ are $-1675 \ kJ \ mol^{-1}$ and $-635 \ kJ \ mol^{-1}$ respectively.
For the reaction $3 CaO + 2 Al \rightarrow 3 Ca + Al_{2}O_{3}$,the standard reaction enthalpy $\Delta_{r}H^{\circ}$ is .......... $kJ$.
(Round off to the Nearest Integer).
For the reaction $3 CaO + 2 Al \rightarrow 3 Ca + Al_{2}O_{3}$,the standard reaction enthalpy $\Delta_{r}H^{\circ}$ is .......... $kJ$.
(Round off to the Nearest Integer).
Calculate the enthalpy change for the following reaction:
$H_2C=CH_{2(g)} + H_{2(g)} \longrightarrow H_3C-CH_{3(g)}$
[The bond energies of $C-H, C-C, C=C$ and $H-H$ are $414, 347, 615$ and $435 \ kJ/mol$ respectively.] (in $kJ$)
$H_2C=CH_{2(g)} + H_{2(g)} \longrightarrow H_3C-CH_{3(g)}$
[The bond energies of $C-H, C-C, C=C$ and $H-H$ are $414, 347, 615$ and $435 \ kJ/mol$ respectively.] (in $kJ$)
The $\Delta_f H^{\circ}$ of $AO_{(s)}$,$BO_{2(g)}$ and $ABO_{3(s)}$ is $-635$,$x$ and $-1210 \ kJ \ mol^{-1}$ respectively.
$ABO_{3(s)} \rightarrow AO_{(s)} + BO_{2(g)} ; \Delta_r H^{\circ} = 175 \ kJ \ mol^{-1}$.
What is the value of $x$ (in $kJ \ mol^{-1}$) ?
$ABO_{3(s)} \rightarrow AO_{(s)} + BO_{2(g)} ; \Delta_r H^{\circ} = 175 \ kJ \ mol^{-1}$.
What is the value of $x$ (in $kJ \ mol^{-1}$) ?
Enthalpy change for the reaction,$\frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \to HCl(g)$,is called:
Medium
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