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You are given Avogadro's no . Of X atoms...

You are given Avogadro's no . Of `X` atoms . If half of the atoms of `X` transfer one electron to the other half of ` X` atoms ` 409 kJ` must be added . If these ` X^-` ions are subsequently converted to ` X^+`, and additional ` 733k J` must be added . Calculate ` IP` and ` EA ` of X in `eV` Uses `(1 eV = 1. 602 xx 10^(-19) J ` and `N_a = 6. 023 xx 10^(23))` .

Text Solution

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Give ` X rarr X^+ + e, Delta H = IP_1 a eV`
` X =e rarr X^- , Delta H = EA_1 =- b eV`
If `N/ 12` atoms of X lose electrons which are atlen up by remaining ` Z/2` of X to give` X^-` then
`a xx N/2 - b xx N/2 = (409 xx 10^3 )/( 1.602 xx 10^(19))eV`.
or ` a-b= (409 xx 10^3 xx 2)/(1.602 xx10^(-19) xx 6.023 xx 10^(23))`
`:. a -b= 8. 4 77`
Now , `N//2 ` of ` X^-` lose two electrons to giev ` X^+`
` X^- rarr X= e , Delta H =+ EA_1 = +b`
`X rarr X^+ + e, Delta = + IP_1 = +a`
`:. a xx N/2 + b xx N/2 = ( 733 xx10^3)/(1.602 xx 10^(-19)) eV`
or `a +b = (733 xx 10^3 xx20)/( 1.602 xx 10^(19) xx 6. 023 xx10^(23))`
`a +b = 15. 194 `
` :. a = 11 . 7835 eV` and ` b=3, 358 eV`.
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