- A$200\ km$
- B$150\ km$
- C$100\ km$
- ✓$25\ km$
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${H_2}C = CH - CH = C{H_2}\xrightarrow[{0{\,^o}C}]{{HBr}}$ $\begin{array}{*{20}{c}}
{{H_2}C = CH - CH - C{H_3}} \\
{\,\,\,\,\,\,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Br\,\,\,\,\,\,}
\end{array}\xrightarrow{{ + 25{\,^o}C}}$ $\begin{array}{*{20}{c}}
{C{H_2}CH = CHC{H_3}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{Br\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$
These provide an example of $......1......$ control at low temperature and $......2......$ control at higher temperature
$2{H_2}{O_2}(l) \rightleftharpoons {H_2}O(l) + {O_2}(g)$
$(R = 83\, JK^{-1}\, mol^{-1})$
$A + B$ $\rightleftharpoons$ $C + D$ + heat
has reached equilibrium. The reaction may be made to proceed forward by