The correct relationship between heat of fusion $(\Delta H_{fus})$,heat of vaporization $(\Delta H_{vap})$,and heat of sublimation $(\Delta H_{sub})$ is:
- A$\Delta H_{fus} = \Delta H_{vap} + \Delta H_{sub}$
- B$\Delta H_{vap} = \Delta H_{fus} + \Delta H_{sub}$
- C$\Delta H_{sub} = \Delta H_{vap} + \Delta H_{fus}$
- D$\Delta H_{sub} = \Delta H_{vap} - \Delta H_{fus}$
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Similar Questions
The standard heat of formation $\left(\Delta_{f} H_{298}^{0}\right)$ of ethane in $kJ/mol$,if the heat of combustion of ethane,hydrogen and graphite are $-1560$,$-286$,and $-393.5 \; kJ/mol,$ respectively is ........... $kJ/mol$.
The value of enthalpy of neutralization is highest for the reaction between:
Easy
View SolutionGiven 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
View SolutionWhich substance has a standard molar enthalpy of formation equal to zero at $298 \ K$?
Medium
View SolutionThe reaction $CH_{4(g)} + Cl_{2(g)} \to CH_3Cl_{(g)} + HCl_{(g)}$ has $\Delta H = -25 \, kcal$. Given bond energies $BE(C-H) = 84 \, kcal$,$BE(H-Cl) = 103 \, kcal$,$BE(C-Cl) = x$,and $BE(Cl-Cl) = y$. If $\frac{x}{y} = \frac{9}{5}$,then find the value of $y$. (in $, kcal$)
Medium
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