Use the data given below to construct a born-haber cycle to determine the second ionization energy of ca. δ h°(kj) ca(s)→ca(g) 193 ca(g)→ca (g) e− 590 2o(g)→o2(g) - 498 o(g) e−→o−(g) - 141 o−(g) e−→o2−(g) 878 ca(s) 12o2(g)→cao(s) - 635 ca2 (g) o2−(g)→cao(s) - 34142o(g) > o2(g) - 498o(g) + e- > o-(g) - 141o-(g) + e- > o2-(g) 878ca(s) + 1/2 o2(g) > cao(s) - 635

Answer :  The value of second ionization energy of Ca is 1010 kJ.

Explanation :  

The formation of calcium oxide is,

= enthalpy of formation of calcium oxide = -635 kJ

The steps involved in the born-Haber cycle for the formation of :

(1) Conversion of solid calcium into gaseous calcium atoms.

= sublimation energy of calcium = 193 kJ

(2) Conversion of gaseous calcium atoms into gaseous calcium ions.

= first ionization energy of calcium = 590 kJ

(3) Conversion of gaseous calcium ion into gaseous calcium ions.

= second ionization energy of calcium = ?

(4) Conversion of molecular gaseous oxygen into gaseous oxygen atoms.

= dissociation energy of oxygen =

(5) Conversion of gaseous oxygen atoms into gaseous oxygen ions.

= first electron affinity energy of oxygen = -141 kJ

(6) Conversion of gaseous oxygen ion into gaseous oxygen ions.

= second electron affinity energy of oxygen = 878 kJ

(7) Conversion of gaseous cations and gaseous anion into solid calcium oxide.

= lattice energy of calcium oxide = -3414 kJ

To calculate the overall energy from the born-Haber cycle, the equation used will be:

Now put all the given values in this equation, we get:

Therefore, the value of second ionization energy of Ca is 1010 kJ.

a:

by the combination of an unlimited number of elements