bohr-haber cycle for BaO crystal
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14 The Born-Haber Cycle
The formation of an ionic solid MX from its elements may be formulated by two different path interrelating different thermochemical quantities.
Born-Haber cycle for the formation of an ionic solid MX
Figure. Born-Haber cycle for the formation of an ionic solid MX.
The various terms are:
S = Sublimation energy of the metal
M_{(s)} \rightarrow M_{(g)}
Since energy is required for the process (endothermic), its value is reported as a positive quantity.
D = Dissociation energy of X2 gas into atoms
X_{2(g)} \rightarrow 2X_{(g)}
This process is also endothermic and its value is reported as a positive quantity.
I = Ionization energy, i.e., energy required for the removal of an electron from an isolated gaseous atom M. The process is endothermic and is reported as a positive quantity.
M_{(g)} \rightarrow M_{(g)}^+ + e^-
E = Electron affinity, i.e., amount of energy released when a neutral gaseous atom gains an electron. The process is exothermic and its value is given as a negative quantity.
X_{(g)} +e \rightarrow X_{(g)}^-
U0 = Lattice energy; it is the amount of energy released when one mole of gaseous positive and negative ions condense to form an ionic solid.
M_{(g)}^+ + X_{(g)}^- \rightarrow MX{(s)}
The process is exothermic and its value is always negative.
ΔHf = Heat of formation; it is defined as the amount of energy released when one mole of a substance is formed from its elements.
M_{(s)} + \frac12 X_{2(g)} \rightarrow MX_{(s)}
This is a single step reaction. Energy is released in the process and its value is reported as a negative quantity.
Since the two processes are independent of each other, the two can be equated, i.e., heat of formation of MX must be equal to the summation of all other terms.
\begin{array}{c} \Delta H_f = S+ \frac12 D + I + E + U_0 \\ U_0= \Delta H_f - S - \frac12 D - I -E \end{array}
The lattice energy of a solid can be calculated with the help of other thermochemical quantities which can be measured experimentally. The quantitative value of each term (along with their signs) are substituted in previous equation.
As an example we may calculate the lattice energy of NaCl crystal from the following data by the use of Born-Haber cycle.
\begin{array} {r} Sublimation \ energy \ (S)=\ 26 \ kcal/mole \\ Dissociation \ energy \ for \ Cl_2 (D)=\ 54 \ kcal/mole \\ Ionization \ energy \ for \ Na_{(g)} \ (I)=117 \ kcal/mole \\ Electron \ affinity \ for \ Cl_{(g)} \ (E)= \ 84 \ kcal/mole \\ Heat \ of \ formation \ of NaCl (\Delta H_f)=-99 \ kcal/mole \end{array}
Now substituting the values in expression
\begin{array}{l}U_0=(-99)-(+26)-\frac12(+54)-(+117)-(-84) \\ U_0=-185 \ kcal/mole \end{array}
In another follow-up example we may calculate the heat of formation(ΔHf) of KF from its elements from the followin data by the use of Born-Haber cycle.
\begin{array}{r}Dissociation \ energy \ of \ potassium (S)=21 \ kcal/mole \\ Dissociation \ energy \ of F_2 \ (D)=38 \ kcal/mole \\ Ionization \ energy \ of K_{(g)} \ (D)=99 \ kcal/mole \\ Electron \ affinity \ for F_{(g)} \ (E)=-80 \ kcal/mole\\ Lattice \ energy \ of \ KF\ (U_0)=-193 \ kcal/mole \end{array}
Now substituting the values in expression
\begin{array} {l} \Delta H_f=S+\frac12D+I+E+U_0 \\ \Delta H_f=21+\frac12(38)+99+(-80)+(193)=-134 \ kcal/mole \end{array}
This can be represented as:
K_{(s)}+\frac12F_{2(g)} \rightarrow KF_{(s)}+134 \ kcal/mole.
The lattice energies of ionic solids are generally greater than the average dissociation energies of diatomic molecules (which are held by covalent bonds) suggesting thereby that ionic solids are held together by very strong attractive forces. The Born-Habre cycle is useful in explaining the stability of BaO despite the fact that very large amount of energy is required in producing bipositive Ba+ ions and binegative O2- ions. Both these processes are highly endothermic. The heat of formation of this compound is negative because of the fact that very large amount of energy is released in the form of lattice energy (U0) when doubly charged Ba2+ ions and O2- ions condense to give BaO.
Role of Lattice Enargy
Stability of ionic solids: Greater the lattice energy, greater is the stability of an ionic solid. Because more energy will be needed to pull apart the +ive and -ive ions.
Melting points: Salts with high lattice energies require a greater input of thermal energy to breakdown the crystal lattice. Consequently, salts have high melting points.
Solubilities of ionic solids: The magnitude of lattice energy of a solid may give us an idea about the solubility of that substance in different solvents. Ionic solids in general are insoluble in non-polar (covalent) solvents such as carbontetrachloride, where as these are soluble in polar solvent, the strong forces of attraction between its ions (lattice energy) must be overcome. This energy can be overcome by the ion-solvent interactions. The solvation of ions is referred to in terms of solvation energy which is always negative, i.e., energy is released in the process.
The formation of an ionic solid MX from its elements may be formulated by two different path interrelating different thermochemical quantities.
Born-Haber cycle for the formation of an ionic solid MX
Figure. Born-Haber cycle for the formation of an ionic solid MX.
The various terms are:
S = Sublimation energy of the metal
M_{(s)} \rightarrow M_{(g)}
Since energy is required for the process (endothermic), its value is reported as a positive quantity.
D = Dissociation energy of X2 gas into atoms
X_{2(g)} \rightarrow 2X_{(g)}
This process is also endothermic and its value is reported as a positive quantity.
I = Ionization energy, i.e., energy required for the removal of an electron from an isolated gaseous atom M. The process is endothermic and is reported as a positive quantity.
M_{(g)} \rightarrow M_{(g)}^+ + e^-
E = Electron affinity, i.e., amount of energy released when a neutral gaseous atom gains an electron. The process is exothermic and its value is given as a negative quantity.
X_{(g)} +e \rightarrow X_{(g)}^-
U0 = Lattice energy; it is the amount of energy released when one mole of gaseous positive and negative ions condense to form an ionic solid.
M_{(g)}^+ + X_{(g)}^- \rightarrow MX{(s)}
The process is exothermic and its value is always negative.
ΔHf = Heat of formation; it is defined as the amount of energy released when one mole of a substance is formed from its elements.
M_{(s)} + \frac12 X_{2(g)} \rightarrow MX_{(s)}
This is a single step reaction. Energy is released in the process and its value is reported as a negative quantity.
Since the two processes are independent of each other, the two can be equated, i.e., heat of formation of MX must be equal to the summation of all other terms.
\begin{array}{c} \Delta H_f = S+ \frac12 D + I + E + U_0 \\ U_0= \Delta H_f - S - \frac12 D - I -E \end{array}
The lattice energy of a solid can be calculated with the help of other thermochemical quantities which can be measured experimentally. The quantitative value of each term (along with their signs) are substituted in previous equation.
As an example we may calculate the lattice energy of NaCl crystal from the following data by the use of Born-Haber cycle.
\begin{array} {r} Sublimation \ energy \ (S)=\ 26 \ kcal/mole \\ Dissociation \ energy \ for \ Cl_2 (D)=\ 54 \ kcal/mole \\ Ionization \ energy \ for \ Na_{(g)} \ (I)=117 \ kcal/mole \\ Electron \ affinity \ for \ Cl_{(g)} \ (E)= \ 84 \ kcal/mole \\ Heat \ of \ formation \ of NaCl (\Delta H_f)=-99 \ kcal/mole \end{array}
Now substituting the values in expression
\begin{array}{l}U_0=(-99)-(+26)-\frac12(+54)-(+117)-(-84) \\ U_0=-185 \ kcal/mole \end{array}
In another follow-up example we may calculate the heat of formation(ΔHf) of KF from its elements from the followin data by the use of Born-Haber cycle.
\begin{array}{r}Dissociation \ energy \ of \ potassium (S)=21 \ kcal/mole \\ Dissociation \ energy \ of F_2 \ (D)=38 \ kcal/mole \\ Ionization \ energy \ of K_{(g)} \ (D)=99 \ kcal/mole \\ Electron \ affinity \ for F_{(g)} \ (E)=-80 \ kcal/mole\\ Lattice \ energy \ of \ KF\ (U_0)=-193 \ kcal/mole \end{array}
Now substituting the values in expression
\begin{array} {l} \Delta H_f=S+\frac12D+I+E+U_0 \\ \Delta H_f=21+\frac12(38)+99+(-80)+(193)=-134 \ kcal/mole \end{array}
This can be represented as:
K_{(s)}+\frac12F_{2(g)} \rightarrow KF_{(s)}+134 \ kcal/mole.
The lattice energies of ionic solids are generally greater than the average dissociation energies of diatomic molecules (which are held by covalent bonds) suggesting thereby that ionic solids are held together by very strong attractive forces. The Born-Habre cycle is useful in explaining the stability of BaO despite the fact that very large amount of energy is required in producing bipositive Ba+ ions and binegative O2- ions. Both these processes are highly endothermic. The heat of formation of this compound is negative because of the fact that very large amount of energy is released in the form of lattice energy (U0) when doubly charged Ba2+ ions and O2- ions condense to give BaO.
Role of Lattice Enargy
Stability of ionic solids: Greater the lattice energy, greater is the stability of an ionic solid. Because more energy will be needed to pull apart the +ive and -ive ions.
Melting points: Salts with high lattice energies require a greater input of thermal energy to breakdown the crystal lattice. Consequently, salts have high melting points.
Solubilities of ionic solids: The magnitude of lattice energy of a solid may give us an idea about the solubility of that substance in different solvents. Ionic solids in general are insoluble in non-polar (covalent) solvents such as carbontetrachloride, where as these are soluble in polar solvent, the strong forces of attraction between its ions (lattice energy) must be overcome. This energy can be overcome by the ion-solvent interactions. The solvation of ions is referred to in terms of solvation energy which is always negative, i.e., energy is released in the process.
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