(Given ${\Delta _{fus}}H = 6\, kJ\, mol^{-1}$ at $0\,^oC$,
$C_p(H_2O, l) =75.3\, J\, mol^{-1} \, K^{-1}$ ,
$C_p(H_2O, s) = 36.8\, J\, mol^{-1} \, K^{ -1}$ )
$(a)$ Energy change of $1\,mol$, $H_2O\,(l)$, at $5\,^oC$
$\to \,1\,mol$, $H_2O\,(l)$ , $0\,^oC$
$(b)$ Energy change of $1\,mol$, $H_2O\,(l)$, at $0\,^oC$
$\to \,1\,mol$, $H_2O\,(s)$ (ice) , $0\,^oC$
$(c)$ Energy change of $1\,mol$, Ice $(s)$, at $0\,^oC$
$\to \,1\,mol$, Ice $(s)$ , $-5\,^oC$
Total $\Delta H$
$ = \,{C_P}\,[{H_2}O\,(l)]\,\,\Delta T\,\, + \,\Delta H$ freezing $ + \,\,{C_P}\,[{H_2}O\,(s)]\,\,\Delta T$
$ = \,(75.3\,\,J\,\,mo{l^{ - 1}}\,{K^{ - 1}})\,( - 5)\,K\, + ( - \,6\, \times \,{10^3}\,\,J\,mo{l^{ - 1}}\,{K^{ - 1}})$ $+ \,(36.8\,\,J\,mo{l^{ - 1}}\,{K^{ - 1}})\,( - 5)\,K$
$\Delta H\,\, = \,\, - \,6.56\,\,kJ\,mo{l^{ - 1}}$ (exothermic process)
So, $\Delta H\,\, = \,\,6.56\,\,kJ\,mo{l^{ - 1}}$
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$C_{(graphite)} + CO_{2(g)} \rightarrow 2CO _{(g)}$
are $170\, kJ$ and $170\, J K^{-1},$ respectively. This reaction will be spontaneous at ............ $\mathrm{K}$
$Image$
The observed pattern of electrophilic substitution can be explained by
$(A)$ the steric effect of the halogen
$(B)$ the steric effect of the tert-butyl group
$(C)$ the electronic effect of the phenolic group
$(D)$ the electronic effect of the tert-butyl group
The order of basicity is