The electron gain enthalpy `Delta e_(g)H` is -349 kJ `mol^(-1)` . If the ground state energy of Cl(g) is x kJ `mol^(-1)` . The ground state energy (in kJ `mol^(-1)` ) of `Cl^(-)` (g) is

Options

A. x + 349
B. x
C. x - 349
D. `(x - 349)/(17)`

Answer

C

Explanation

`Cl(g) +e^(-) to Cl^(-) (g)`,
Electron gain enthalpy co Cl(g),
`Deltae_(g) H = - 349 kJ mol^(-1)`
Given, ground energy of `Cl(g) = x kJ mol^(-1)`
Let the ground state energy of `Cl^(-) (g) = y kJ mol^(-1)`
Now ,
`Deltae_(g) H = E_(P)-E_(R)`
`:. - 349 kJ mol^(-1) = (y -x) kH//mol`
`rArr y = (x - 349) kJ//mol`