Q3 / 160 · AP EAMCET 2019 (21 Apr Shift 2) · Chemistry

The electron gain enthalpy $Δ_{eg} H$ of $Cl (g)$ 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

A
$x + 349$
B
$x$
Correct answer
C
$x - 349$
D
$\frac{x - 349}{17}$

Solution

C l_{2} \overset{∆_{e g} H}{\to} 2 C l^{-} + 2 e^{-} As ground state energy is x k J m o l e^{- 1}, \begin{matrix} x & = & ∆ E_{g r o u n d s t a t e} - ∆_{e g} H \\ ∆ E_{g r o u n d s t a t e} & = & x + ∆_{e g} H \\ ∆ E_{g r o u n d s t a t e} & = & x - 349 \end{matrix}

Hence, the correct answer is option: 'C'.

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