Enter your answer into the provided box. The energy needed for the following pro
ID: 473514 • Letter: E
Question
Enter your answer into the provided box. The energy needed for the following process is 1.960 times 10^4 kJ/mol: Li(g) rightarrow Li^3+(g) + 3e^- If the first ionization energy of lithium is 5.20 times 10^2 kJ/mol, calculate the second ionization energy of lithium-that is, the energy required for the process: Li^+(g) rightarrow Li^2+(g) + e^- A hydrogen-like ion is an ion containing only one electron. The energies of the electron in a hydrogen-like ion are given by E_n = -(2.18 times 10^-18 J) (Z^2) (1/n^2) where E_n is the energy of the electron in the hydrogen-like ion, n is the principal quantum number, and Z is the atomic number of the element. Second ionization energy of lithium: (Enter your answer in scientific notation)Explanation / Answer
we know sum of IE1 + IE2 + IE3 = total ionization energy
IE1 = 5.2 x 102 KJ/mol
if we calculate IE3 from given data than we can find IE2 also
Li+2 is hydrogen like atom contain only 1 electron in it.
Li+2 ------------> Li+3 + e- = IE3
En = (-2.18 x 10-18 J) (32)(1/12) because electron in 1s orbital so n=1 and Z = 3 foe Li
En = -1.962 x 10-17J per 1 atom
for 1 mole
En = -1.962 x 10-17 x 6.023 x 1023
E2 = -1.134 x 107 J / mol = -1.134 x 104 KJ/mol.
so E3 (third ionization energy ) = 1.134 x 104 KJ/mol
now we know
total energy = IE1 + IE2 + IE3
1.960 x 104 KJ/mol = 5.20 x 102 + IE2 + 1.134 x 104 KJ/mol
IE2 = (1.960 x 104) - (5.20 x 102) - (1.134 x 104)
IE2 = 7.74 x 103 KJ/mol
answer = 7.74 x 103 KJ/mol
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.