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How much energy in kJ is released when 3.55g of Chlorine are completely converted to Cl$^{-}$ ions in the gaseous state?

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

$\begin{array}{1 1} 35.701\;kJ \\ 33.735\;kJ \\ 357.05\;kJ \\ 35.5\;kJ \end{array}$

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Answer: 35.7013 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 3.55 g of Chlorine completely changes $ = \large\frac{3.55}{35.5}$ $\times 357.013 = 35.7013$ kJ
answered Jul 24, 2014 by balaji.thirumalai

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