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How much energy in kJ is released when 7.1 g of Chlorine are completely converted to Cl$^{-}$ ions in the gaseous state?

Note: The electron gain enthalpy of Cl (g) is -3.7 eV, and 1 eV = 96.49 kJ mole$^{-1}$

$\begin{array}{1 1} 71.4\;kJ \\ 35.7\;kJ \\ 757.5\;kJ \\ 96.49\;kJ \end{array}$

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Answer:71.4 kJ
Cl (g) + e$^{-} \rightarrow$ Cl$^{-}$ (g) + 3.7 e.v
Energy released when 1 mole (35.5g) of Chlorine changes completeley into Cl$^{-}$ (g) = 3.7 e.v
We know that 1 e.v = 94.49 kJ mole$^{-1}$
$\Rightarrow$ Cl (g) + e$^{-} \rightarrow$ Cl$^{-}$ (g) + 3.7 $\times$ 96.49 kJ mol$^{-1}$
$\Rightarrow$ Cl (g) + e$^{-} \rightarrow$ Cl$^{-}$ (g) + 357.013 kJ
$\Rightarrow$ Energy released when 7.1 g of Chlorine completely changes $ = \large\frac{7.1}{35.5}$ $\times 357.013 = 71.4$ kJ
answered Jul 24, 2014 by balaji.thirumalai

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