$H_{2(g)} + Cl_{2(g)} \to 2HCl_{(g)}, \Delta H = -44 \ kcal$
$2Na_{(s)} + 2HCl_{(g)} \to 2NaCl_{(s)} + H_{2(g)}, \Delta H = -152 \ kcal$
For the reaction $Na_{(s)} + \frac{1}{2}Cl_{2(g)} \to NaCl_{(s)}, \Delta H = \dots \ kcal$
$2Na_{(s)} + 2HCl_{(g)} \to 2NaCl_{(s)} + H_{2(g)}, \Delta H = -152 \ kcal$
For the reaction $Na_{(s)} + \frac{1}{2}Cl_{2(g)} \to NaCl_{(s)}, \Delta H = \dots \ kcal$
- A$-108$
- B$-98$
- C$-196$
- D$54$
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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 heat of combustion of solid benzoic acid at constant volume is $-321.30 \ kJ$ at $27^{\circ} C$. The heat of combustion at constant pressure is $(-321.30 - x R) \ kJ$. The value of $x$ is . . . . . . .
The enthalpy of neutralization of any strong acid and strong base is constant because......
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View SolutionAverage $C-H$ bond energy is $416 \ kJ \ mol^{-1}$. Which of the following is correct?
Given the following data:
Reaction Energy Change (in $kJ$) $Li_{(s)} \to Li_{(g)}$ $161$ $Li_{(g)} \to Li^{+}_{(g)}$ $520$ $\frac{1}{2} F_{2(g)} \to F_{(g)}$ $77$ $F_{(g)} + e^- \to F^{-}_{(g)}$ (Electron gain enthalpy) $Li^{+}_{(g)} + F^{-}_{(g)} \to LiF_{(s)}$ $-1047$ $Li_{(s)} + \frac{1}{2} F_{2(g)} \to LiF_{(s)}$ $-617$
Based on the data provided,the value of electron gain enthalpy of fluorine would be $kJ\ mol^{-1}$.
| Reaction | Energy Change (in $kJ$) |
| $Li_{(s)} \to Li_{(g)}$ | $161$ |
| $Li_{(g)} \to Li^{+}_{(g)}$ | $520$ |
| $\frac{1}{2} F_{2(g)} \to F_{(g)}$ | $77$ |
| $F_{(g)} + e^- \to F^{-}_{(g)}$ | (Electron gain enthalpy) |
| $Li^{+}_{(g)} + F^{-}_{(g)} \to LiF_{(s)}$ | $-1047$ |
| $Li_{(s)} + \frac{1}{2} F_{2(g)} \to LiF_{(s)}$ | $-617$ |
Based on the data provided,the value of electron gain enthalpy of fluorine would be $kJ\ mol^{-1}$.
DifficultJEE MAIN 2013
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