Question

Using KF as an example, write equations that refer to $\mathbf{∆}{\mathbf{H}}_{\mathbf{soln}}$and $\mathbf{∆}{\mathbf{H}}_{\mathbf{hyd}}$Lattice energy was defined in chapter 13 as $\mathbf{∆}\mathbf{H}$ the reaction. K⁺(g)+F^-(g) →KF(s). Show how would you utilize Hess’s law to calculate ${\mathbf{\Delta H}}_{\mathbf{sol}}$from ${\mathbf{\Delta H}}_{\mathbf{hyd}}$and ${\mathbf{\Delta H}}_{\mathbf{LE}}$for KF, where ${\mathbf{\Delta H}}_{\mathbf{LE}}$= lattice energy,${\mathbf{\Delta H}}_{\mathbf{sol}}$for KF, as for other soluble ionic compounds, is a relatively small number. How can this be since${\mathbf{\Delta H}}_{\mathbf{hyd}}$and${\mathbf{\Delta H}}_{\mathbf{LE}}$are relatively large negative numbers?

Short answer

Equation that refers to${\mathrm{\Delta H}}_{\mathrm{sol}}\mathrm{KF}\left(\mathrm{S}\right)\to \mathrm{K}\left(\mathrm{aq}\right)+{\mathrm{F}}^{-}\left(\mathrm{aq}\right)$ and${\mathrm{\Delta H}}_{\mathrm{hyd}}=\mathrm{K}\left(\mathrm{aq}\right)+{\mathrm{F}}^{-}\left(\mathrm{aq}\right)\to \mathrm{K}\left(\mathrm{g}\right)+{\mathrm{F}}^{-}\left(\mathrm{g}\right)$

Step-by-step solution

Step 1- Hess’s law

Its states that the total enthalpy change for a multi-step reaction is equal to the sum of the enthalpy changes for each individual stage.

Step 2- Calculating enthalpies for individual step

${\mathrm{\Delta H}}_1$is the enthalpy of the given reaction $\mathrm{KF}\left(\mathrm{S}\right)\to \mathrm{K}\left(\mathrm{g}\right)+{\mathrm{F}}^{-}\left(\mathrm{g}\right)$…. (I)

${\mathrm{\Delta H}}_1={\mathrm{\Delta H}}_{\mathrm{L}.\mathrm{E}}$

${\mathrm{\Delta H}}_2$is the enthalpy of the given reaction $\mathrm{K}\left(\mathrm{g}\right)+{\mathrm{F}}^{-}\left(\mathrm{g}\right)\to \mathrm{K}\left(\mathrm{aq}\right)+{\mathrm{F}}^{-}\left(\mathrm{aq}\right)$…(ii)

${\mathrm{\Delta H}}_2={\mathrm{\Delta H}}_{\mathrm{H}.\mathrm{E}}$

${\mathrm{\Delta H}}_2$is the enthalpy of hydration i.e amount of energy released when an ion is dissolved in a large amount of water.

Step 3- Calculating enthalpy of solution.

It is determined by the enthalpy of lattice energy and the degree of hydration. Depending on whether the reaction is exothermic or endothermic, it might be positive or negative.

${\mathrm{\Delta H}}_{\mathrm{sol}}={\mathrm{\Delta H}}_{\mathrm{H}.\mathrm{E}}+{\mathrm{\Delta H}}_{\mathrm{L}.\mathrm{E}}$

The overall reaction is formed by adding equations (i) and (ii)

$\mathrm{KF}\left(\mathrm{S}\right)\to \mathrm{K}\left(\mathrm{aq}\right)+{\mathrm{F}}^{-}\left(\mathrm{aq}\right)$

So the enthalpy of the solution is $∆{\mathrm{H}}_{\mathrm{sol}}$

$$\begin{gathered} ={\mathrm{\Delta H}}_1+{\mathrm{\Delta H}}_2 \\ ={\mathrm{\Delta H}}_{\mathrm{H}.\mathrm{E}}+{\mathrm{\Delta H}}_{\mathrm{L}.\mathrm{E}} \end{gathered}$$

${\mathrm{\Delta H}}_1$is a positive value because it is the reverse of lattice energy and ${\mathrm{\Delta H}}_2$is a large negative number. So the exothermic and endothermic processes are getting canceled out and ${\mathrm{\Delta H}}_{\mathrm{sol}}$is close to zero.