- A$\text{HCl > HBr > HI}$
- B$\text{HBr > HI > HCl}$
- ✓$\text{HI > HI > HCl}$
- D$\text{HCl > HI > HBr}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
${H_2}C = CH - CH = C{H_2}\xrightarrow[{0{\,^o}C}]{{HBr}}$ $\begin{array}{*{20}{c}}
{{H_2}C = CH - CH - C{H_3}} \\
{\,\,\,\,\,\,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Br\,\,\,\,\,\,}
\end{array}\xrightarrow{{ + 25{\,^o}C}}$ $\begin{array}{*{20}{c}}
{C{H_2}CH = CHC{H_3}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{Br\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$
These provide an example of $......1......$ control at low temperature and $......2......$ control at higher temperature
$(a)\,B{r_2}(l) \to B{r_2}(g)$
$(b)\,{H_2}O(s) \to {H_2}O(g)$
$(c)\,{N_2}\,\left[ {1\,atm,\,{{100}\,^o}C} \right] \to {N_2}\,\left[ {1\,atm,\,{{150}\,^o}C} \right]$
$(d)\,{N_2}\,(g) + 3{H_2}(g) \to 2N{H_3}(g)$
$(e)\,CaC{O_3}(s) \to CaO(s) + C{O_2}(g)$


Step $I\,\,:$ ${H_2}O(g)\, \to \,H(g)\, + \,OH(g)\,;$ $\Delta H = 498\,kJ\,mol^{-1}$
Step $II\,\,:$ $OH(g)\, \to \,H(g)\, + \,O(g)\,;$ $\Delta H = 428\,kJ\,mol^{-1}$
The bond enthalpy of the $O-H$ bond is....$kJ\,mol^{-1}$
$C _{2} H _{6} \rightarrow C _{2} H _{4}+ H _{2}$
the reaction enthalpy $\Delta_{ r } H =...........{ kJ\, mol ^{-1}}$.
(Round off to the Nearest Integer).
[Given : Bond enthalpies in $kJ$ $mol$ $^{-1}:C-C : 347, C = C : 611 ; C - H : 414, H - H : 436]$