Question

The lattice energy of solid KCl is $181{\text{kcalmol}}^{-1}$ and the enthalpy of solution of $KCl$ in $H_{2}O$ is $1.0{\text{kcalmol}}^{-1}$. If the hydration enthalpies of $K^{+}$ and $C{l}^{-}$ ions are in the ratio of 2:1, then the enthalpy of hydration of $K^{+}$ is $-x{\text{kcalmol}}^{-1}$. Find the value of $x$.

Answer 1

$KCl\left(s\right)\to K^{+}\left(g\right)+C{I}^{-}\left(g\right),\Delta H_1=181kcalmo{l}^{-1}$
$KCl\left(s\right)+\left(aq\right)\to K^{+}\left(aq\right)+C{l}^{-}\left(aq\right),\Delta H_2=1.0kcalmo{l}^{-1}$
Let the enthalpy of hydration of`$K^{+}$ be $2\times Cl$
$K^{+}\left(g\right)+\left(aq\right)\to K^{+}\left(aq\right),\Delta H_3=2a$
$C{I}^{-}\left(g\right)+\left(aq\right)\to C{l}^{-}\left(aq\right),\Delta H_4=a$
$\Delta H_3=-\Delta H_1+\Delta H_2-\Delta H_4$
$2a=-181+1-a$
$3a=-180,a=-60$
${\Delta }_{hyd}{H}^{0}of{K}^{+}=2a=-60\times 2=-120$
$-20x=-120$
$x=6$