The Born-Haber cycle for KCl is evaluated with the following data:
$$\Delta_f H^{\ominus}$$ for KCl = $$-436.7$$ kJ mol$$^{-1}$$; $$\Delta_{sub} H^{\ominus}$$ for K = 89.2 kJ mol$$^{-1}$$;
$$\Delta_{ionization} H^{\ominus}$$ for K = 419.0 kJ mol$$^{-1}$$; $$\Delta_{electron gain} H^{\ominus}$$ for Cl$$_{(g)}$$ = $$-348.6$$ kJ mol$$^{-1}$$
$$\Delta_{bond} H^{\ominus}$$ for Cl$$_2$$ = 243.0 kJ mol$$^{-1}$$
The magnitude of lattice enthalpy of KCl in kJ mol$$^{-1}$$ is _________ (Nearest integer)

We begin with the Born-Haber cycle, which relates the standard enthalpy of formation of an ionic crystal to several simpler enthalpy changes. For potassium chloride, the cycle can be written symbolically as

$$\Delta_fH^\theta(\text{KCl}_{(s)}) \;=\; \Delta_{sub}H^\theta(\text{K}_{(s)}) \;+\; \Delta_{ion}H^\theta(\text{K}_{(g)}) \;+\; \frac12\,\Delta_{bond}H^\theta(\text{Cl}_2) \;+\; \Delta_{eg}H^\theta(\text{Cl}_{(g)}) \;+\; \Delta_{latt}H^\theta(\text{KCl})$$

where

• $$\Delta_{sub}H^\theta$$ is the enthalpy of sublimation of K.
• $$\Delta_{ion}H^\theta$$ is the first-ionization enthalpy of K.
• $$\frac12\,\Delta_{bond}H^\theta$$ is half the bond dissociation enthalpy of Cl2, because only one chlorine atom is required.
• $$\Delta_{eg}H^\theta$$ is the electron-gain (electron affinity) enthalpy of Cl(g). It is exothermic and therefore given as a negative value.
• $$\Delta_{latt}H^\theta$$ is the lattice enthalpy of KCl, the quantity we need. By convention, the formation of the lattice from gaseous ions releases energy, so this term will be negative; its magnitude is asked for in the question.

We now insert all the numerical data supplied:

$$\Delta_fH^\theta = -436.7\ \text{kJ mol}^{-1}$$

$$\Delta_{sub}H^\theta = 89.2\ \text{kJ mol}^{-1}$$

$$\Delta_{ion}H^\theta = 419.0\ \text{kJ mol}^{-1}$$

$$\Delta_{eg}H^\theta = -348.6\ \text{kJ mol}^{-1}$$

$$\Delta_{bond}H^\theta(\text{Cl}_2) = 243.0\ \text{kJ mol}^{-1}$$ so that $$\frac12\,\Delta_{bond}H^\theta = \frac{243.0}{2} = 121.5\ \text{kJ mol}^{-1}$$

Substituting every value into the Born-Haber equation, we have

$$-436.7 \;=\; 89.2 \;+\; 419.0 \;+\; 121.5 \;+\; (-348.6) \;+\; \Delta_{latt}H^\theta$$

Now we gather the known numerical terms step by step.

First add the sublimation and ionization enthalpies:

$$89.2 + 419.0 = 508.2$$

Next add the half bond enthalpy for chlorine:

$$508.2 + 121.5 = 629.7$$

Now include the electron-gain enthalpy (remember, it is negative):

$$629.7 + (-348.6) = 629.7 - 348.6 = 281.1$$

Thus the equation becomes

$$-436.7 = 281.1 + \Delta_{latt}H^\theta$$

We isolate the lattice enthalpy term by subtracting 281.1 from both sides:

$$\Delta_{latt}H^\theta = -436.7 - 281.1 = -717.8\ \text{kJ mol}^{-1}$$

The negative sign shows that energy is released when the crystal lattice forms. The problem asks for the magnitude (absolute value) of this lattice enthalpy, so we quote

$$|\Delta_{latt}H^\theta| \approx 718\ \text{kJ mol}^{-1}$$

So, the answer is $$718$$.