Given $:$
$\Delta H^{\ominus}_{sub}[C(graphite)] = 710 \ kJ \ mol^{-1}$
$\Delta H^{\ominus}_{C-H} = 414 \ kJ \ mol^{-1}$
$\Delta H^{\ominus}_{H-H} = 436 \ kJ \ mol^{-1}$
$\Delta H^{\ominus}_{C=C} = 611 \ kJ \ mol^{-1}$
The $\Delta H^{\ominus}_{f}$ for $CH_2=CH_2$ is $............ \ kJ \ mol^{-1}$ $(nearest \ integer \ value)$
$\Delta H^{\ominus}_{sub}[C(graphite)] = 710 \ kJ \ mol^{-1}$
$\Delta H^{\ominus}_{C-H} = 414 \ kJ \ mol^{-1}$
$\Delta H^{\ominus}_{H-H} = 436 \ kJ \ mol^{-1}$
$\Delta H^{\ominus}_{C=C} = 611 \ kJ \ mol^{-1}$
The $\Delta H^{\ominus}_{f}$ for $CH_2=CH_2$ is $............ \ kJ \ mol^{-1}$ $(nearest \ integer \ value)$
- A$15$
- B$5$
- C$35$
- D$25$
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Find the value of enthalpy of formation of $PCl_5(s)$ given the following thermochemical equations:
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View SolutionThe 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 .... $kJ \ mol^{-1}$.
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