Respuesta :
Answer : The value of second ionization energy of Ca is 1010 kJ.
Explanation :
The formation of calcium oxide is,
[tex]Ca(s)+\frac{1}{2}O_2(g)\overset{\Delta H_f}\rightarrow CaO(s)[/tex]
[tex]\Delta H_f^o[/tex] = enthalpy of formation of calcium oxide = -635 kJ
The steps involved in the born-Haber cycle for the formation of [tex]CaO[/tex]:
(1) Conversion of solid calcium into gaseous calcium atoms.
[tex]Ca(s)\overset{\Delta H_s}\rightarrow Ca(g)[/tex]
[tex]\Delta H_s[/tex] = sublimation energy of calcium = 193 kJ
(2) Conversion of gaseous calcium atoms into gaseous calcium ions.
[tex]Ca(g)\overset{\Delta H_I_1}\rightarrow Ca^{+1}(g)[/tex]
[tex]\Delta H_I_1[/tex] = first ionization energy of calcium = 590 kJ
(3) Conversion of gaseous calcium ion into gaseous calcium ions.
[tex]Ca^{+1}(g)\overset{\Delta H_I_2}\rightarrow Ca^{+2}(g)[/tex]
[tex]\Delta H_I_2[/tex] = second ionization energy of calcium = ?
(4) Conversion of molecular gaseous oxygen into gaseous oxygen atoms.
[tex]O_2(g)\overset{\Delta H_D}\rightarrow OI(g)[/tex]
[tex]\frac{1}{2}O_2(g)\overset{\Delta H_D}\rightarrow O(g)[/tex]
[tex]\Delta H_D[/tex] = dissociation energy of oxygen = [tex]\frac{498}{2}=249kJ[/tex]
(5) Conversion of gaseous oxygen atoms into gaseous oxygen ions.
[tex]O(g)\overset{\Delta H_E_1}\rightarrow O^-(g)[/tex]
[tex]\Delta H_E_1[/tex] = first electron affinity energy of oxygen = -141 kJ
(6) Conversion of gaseous oxygen ion into gaseous oxygen ions.
[tex]O^-(g)\overset{\Delta H_E_2}\rightarrow O^{2-}(g)[/tex]
[tex]\Delta H_E_2[/tex] = second electron affinity energy of oxygen = 878 kJ
(7) Conversion of gaseous cations and gaseous anion into solid calcium oxide.
[tex]Ca^{2+}(g)+O^{2-}(g)\overset{\Delta H_L}\rightarrow CaO(s)[/tex]
[tex]\Delta H_L[/tex] = lattice energy of calcium oxide = -3414 kJ
To calculate the overall energy from the born-Haber cycle, the equation used will be:
[tex]\Delta H_f^o=\Delta H_s+\Delta H_I_1+\Delta H_I_2+\Delta H_D+\Delta H_E_1+\Delta H_E_2+\Delta H_L[/tex]
Now put all the given values in this equation, we get:
[tex]-635kJ=193kJ+590kJ+\Delta H_I_2+249kJ+(-141kJ)+878kJ+(-3414kJ)[/tex]
[tex]\Delta H_I_2=1010kJ[/tex]
Therefore, the value of second ionization energy of Ca is 1010 kJ.
The second ionization energy of Calcium is 1010 kJ.
Given here,
The enthalpy of formation of calcium oxide [tex]\rm \bold { \Delta Hf(CaO)}[/tex] = -635 kJ
The first ionization energy of calcium IE (1)= 590 kJ
The first electron affinity energy of oxygen EA = -141 kJ
The dissociation energy of oxygen DE = 249 kJ
The second electron affinity of oxygen EA = 878 kJ
The lattice energy of calcium oxide U = -3414 kJ
To calculate the second ionization energy of Calcium,
Born-Haber cycle, the equation
[tex]\rm \bold{ \Delta H(f) = SE+ IE(1)+ IE(2) + DE + EA(1) +EA( 2) + LE}[/tex]
Put the value,
-635 kJ = 590 kJ + EI(1) + (-141 kJ) + 249 kJ + 878 kJ + ( -3414 kJ)
Solve it for IE(2)
IE(2) = 1010 kJ
Hence, we can conclude the second ionization energy of Calcium is 1010 kJ.
To know more about Born-Haber cycle, refer to the link:
https://brainly.com/question/6545392?referrer=searchResults
