6 g of Mg ( s ) atoms (atomic mass =24.0 amu ) is converted into Mg ++( g ). Then find the energy required in kJ mol -1 for this process, if the first and second ionisation energies of Mg are 740 and 1440 kJ mol -1, respectively, and enthalpy of sublimation is 100 kg mol -1.

Question

6 g of Mg ( s ) atoms (atomic mass =24.0 amu ) is converted into Mg ++( g ). Then find the energy required in kJ mol -1 for this process, if the first and second ionisation energies of Mg are 740 and 1440 kJ mol -1, respectively, and enthalpy of sublimation is 100 kg mol -1. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Total energy required to convert (6/24)=(1/4) mol of Mg ( s ) into Mg ++ E=(1/4)(100+740+1440) =(2280/4)=570 kJ / mol

Q. 6g of Mg(s) atoms (atomic mass =24.0amu ) is converted into Mg++(g). Then find the energy required in kJmol−1 for this process, if the first and second ionisation energies of Mg are 740 and 1440kJmol−1, respectively, and enthalpy of sublimation is 100kgmol−1.

Total energy required to convert 246​=41​ mol of Mg(s) into Mg++ E=41​(100+740+1440) =42280​=570kJ/mol

Total energy required to convert
$\frac{6}{24} = \frac{1}{4}$ mol of $M g (s)$ into $M g^{++}$
$E = \frac{1}{4} (100 + 740 + 1440)$
$= \frac{2280}{4} = 570 k J / m o l$