Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below (1/2)(C l)2(g) overset(1/2) Δ d i s sH ° arrow Cl(g) oversetΔ E A ( )H ° arrow (C l)-(g) oversetΔ h y d ( )H ° arrow (C l)-(.aq.) The energy involved in the conversion of (1/2)Cl2(.g.) to (C l)-(.aq.) (Using the data) Δ d i s sH ° C l2 =240 kJ mol- 1 Δ E AH ° C l =-349 kJ mol- 1 Δ h y d H ° Cl=-381 kJ mol- 1 will be

Question

Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below (1/2)(C l)2(g) overset(1/2) Δ d i s sH ° arrow Cl(g) oversetΔ E A ( )H ° arrow (C l)-(g) oversetΔ h y d ( )H ° arrow (C l)-(.aq.) The energy involved in the conversion of (1/2)Cl2(.g.) to (C l)-(.aq.) (Using the data) Δ d i s sH ° C l2 =240 kJ mol- 1 Δ E AH ° C l =-349 kJ mol- 1 Δ h y d H ° Cl=-381 kJ mol- 1 will be (A) +152 kJ mol- 1 (B) -610 kJ mol- 1 (C) -850 kJ mol- 1 (D) +120 kJ mol- 1

Tardigrade Admin · Accepted Answer

(1/2)Cl2(.g.) arrow Cl-(.aq.) Δ H=(1/2)Δ Hd i s s(C l2)+Δ HE A Cl+Δ Hh y d(.Cl-.) =(240/2)-349-381 =-610 kJ mol- 1

$+ 152 k J m o l^{- 1}$

$- 610 k J m o l^{- 1}$

$- 850 k J m o l^{- 1}$

$+ 120 k J m o l^{- 1}$

$\frac{1}{2} C l_{2} (g) \to C l^{-} (a q)$
$Δ H = \frac{1}{2} Δ H_{d i ss} (C l_{2}) + Δ H_{E A} Cl + Δ H_{h y d} (C l^{-})$
$= \frac{240}{2} - 349 - 381$
$= - 610 k J m o l^{- 1}$

+152kJmol−1

−610kJmol−1

−850kJmol−1

+120kJmol−1

21​Cl2​(g)→Cl−(aq) ΔH=21​ΔHdiss​(Cl2​)+ΔHEA​Cl+ΔHhyd​(Cl−) =2240​−349−381 =−610kJmol−1

$+ 152 k J m o l^{- 1}$

$- 610 k J m o l^{- 1}$

$- 850 k J m o l^{- 1}$

$+ 120 k J m o l^{- 1}$

$\frac{1}{2} C l_{2} (g) \to C l^{-} (a q)$
$Δ H = \frac{1}{2} Δ H_{d i ss} (C l_{2}) + Δ H_{E A} Cl + Δ H_{h y d} (C l^{-})$
$= \frac{240}{2} - 349 - 381$
$= - 610 k J m o l^{- 1}$