Question

Enthalpy, H

$Cs^+(g) + F(g)$ $\Delta H_1 = 79.4kJ/mol$

$Cs(g) + \frac{1}{2}F_2(g)$ $\Delta H_2 = 375.7kJ/mol$

$Cs(g) + \frac{1}{2}F_2(g)$ $\Delta H_3 = 76.5kJ/mol$

$Cs(s) + \frac{1}{2}F_2(g)$ Overall change $CsF(s)$ $\Delta H_4 = -328.2kJ/mol$

$Cs^+(g) + F^-(g)$ $\Delta H_5 = -756.9kJ/mol$

• A. $\Delta H^0_{formation \space of \space CsF} = -1107kJ/mole$
• B. $\Delta H^0_{Lattice \space energy \space of \space CsF} = -553.5kJ/mole$
• C. $\Delta H^0_{Lattice \space energy \space of \space CsF} = -756.9kJ/mole$
• D. $\Delta H^0_{formation \space of \space CsF} = -553.5kJ/mole$

Answer 1

Answer: (C), (D)

C, D

Answer 2

The Born-Haber cycle is used to calculate the lattice energy or the enthalpy of formation of an ionic compound by applying Hess's Law. The cycle breaks down the formation of an ionic solid from its elements into a series of steps for which the enthalpy changes are known or can be determined.

Let's identify each enthalpy change from the given diagram:

1. Enthalpy of Sublimation of Cesium ($\Delta H_{sub}$):
   This is the energy required to convert solid Cesium to gaseous Cesium.
   $Cs(s) \rightarrow Cs(g)$
   From the diagram, the enthalpy change from $Cs(s) + \frac{1}{2}F_2(g)$ to $Cs(g) + \frac{1}{2}F_2(g)$ is $\Delta H_3$.
   $\Delta H_{sub} = \Delta H_3 = 76.5 \text{ kJ/mol}$

2. Ionization Enthalpy of Cesium ($IE$):
   This is the energy required to remove one electron from a gaseous Cesium atom.
   $Cs(g) \rightarrow Cs^+(g) + e^-$
   From the diagram, the enthalpy change from $Cs(g) + \frac{1}{2}F_2(g)$ to $Cs^+(g) + \frac{1}{2}F_2(g)$ is $\Delta H_2$.
   $IE = \Delta H_2 = 375.7 \text{ kJ/mol}$

3. Bond Dissociation Enthalpy of Fluorine ($\frac{1}{2}BE$):
   This is the energy required to break the bond in half a mole of gaseous F₂ to form gaseous F atoms.
   $\frac{1}{2}F_2(g) \rightarrow F(g)$
   From the diagram, the enthalpy change from $Cs^+(g) + \frac{1}{2}F_2(g)$ to $Cs^+(g) + F(g)$ is $\Delta H_1$.
   $\frac{1}{2}BE = \Delta H_1 = 79.4 \text{ kJ/mol}$

4. Electron Gain Enthalpy of Fluorine ($EGE$):
   This is the energy change when an electron is added to a gaseous Fluorine atom to form a gaseous Fluoride ion.
   $F(g) + e^- \rightarrow F^-(g)$
   From the diagram, the enthalpy change from $Cs^+(g) + F(g)$ to $Cs^+(g) + F^-(g)$ is $\Delta H_4$.
   $EGE = \Delta H_4 = -328.2 \text{ kJ/mol}$

5. Lattice Energy of Cesium Fluoride ($LE$):
   This is the energy released when gaseous Cesium ions and gaseous Fluoride ions combine to form one mole of solid Cesium Fluoride.
   $Cs^+(g) + F^-(g) \rightarrow CsF(s)$
   From the diagram, the enthalpy change from $Cs^+(g) + F^-(g)$ to $CsF(s)$ is $\Delta H_5$.
   $LE = \Delta H_5 = -756.9 \text{ kJ/mol}$

Now let's evaluate the given options:

(C) $\Delta H^0_{Lattice \space energy \space of \space CsF} = -756.9 \text{ kJ/mole}$
As identified above, the lattice energy ($LE$) is directly given as $\Delta H_5 = -756.9 \text{ kJ/mol}$.
Therefore, option (C) is correct.

(D) $\Delta H^0_{formation \space of \space CsF} = -553.5 \text{ kJ/mole}$
The standard enthalpy of formation ($\Delta H^0_{f}$) of CsF is the overall enthalpy change for the reaction:
$Cs(s) + \frac{1}{2}F_2(g) \rightarrow CsF(s)$
According to Hess's Law, the overall enthalpy change is the sum of the enthalpy changes of all the individual steps in the Born-Haber cycle:
$\Delta H^0_{f} = \Delta H_{sub} + IE + \frac{1}{2}BE + EGE + LE$
$\Delta H^0_{f} = \Delta H_3 + \Delta H_2 + \Delta H_1 + \Delta H_4 + \Delta H_5$
Substituting the values:
$\Delta H^0_{f} = 76.5 \text{ kJ/mol} + 375.7 \text{ kJ/mol} + 79.4 \text{ kJ/mol} + (-328.2 \text{ kJ/mol}) + (-756.9 \text{ kJ/mol})$
$\Delta H^0_{f} = 531.6 \text{ kJ/mol} - 328.2 \text{ kJ/mol} - 756.9 \text{ kJ/mol}$
$\Delta H^0_{f} = 203.4 \text{ kJ/mol} - 756.9 \text{ kJ/mol}$
$\Delta H^0_{f} = -553.5 \text{ kJ/mol}$
Therefore, option (D) is correct.

(A) $\Delta H^0_{formation \space of \space CsF} = -1107 \text{ kJ/mole}$
This value is incorrect as calculated above.

(B) $\Delta H^0_{Lattice \space energy \space of \space CsF} = -553.5 \text{ kJ/mole}$
This value is incorrect. -553.5 kJ/mol is the enthalpy of formation, not the lattice energy.

Based on the calculations, both options (C) and (D) are correct.