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Chemistry Question on Thermodynamics

Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below: 12Cl2(g)−>[12ΔdissH⊖]Cl(g)−>[ΔegH⊖]Cl−(g)−>[ΔHydH⊖]Cl−(aq) {\frac{1}{2} Cl_{2} (g) ->[\frac{1}{2} \Delta_{diss} H^{\ominus}] Cl(g) ->[\Delta_{eg} H^{\ominus}] Cl^{-} (g) ->[\Delta_{Hyd} H^{\ominus}] Cl^{-} (aq)} using the data, ΔdissHCl2⊖=240 kJ mol−1 {\Delta_{diss} H^{\ominus}_{Cl_2} = 240 \, kJ \, mol^{-1}}, ΔegHCl⊖=−349 kJ mol−1 {\Delta_{eg} H^{\ominus}_{Cl} = - 349 \, kJ \, mol^{-1}}, ΔhydHCl−⊖=−381 kJ mol−1 {\Delta_{hyd} H^{\ominus}_{Cl^{-}} = - 381 \, kJ \, mol^{-1}} , well be

A

\ce+152 kJ mol−1\ce{+ 152 \, kJ \, mol^{-1}}

B

\ce−850 kJ mol−1\ce{- 850 \, kJ \, mol^{-1}}

C

\ce−610 kJ mol−1\ce{ - 610 \, kJ \, mol^{-1}}

D

\ce+120 kJ mol−1\ce{+ 120 \, kJ \, mol^{-1}}

Answer

\ce−610 kJ mol−1\ce{ - 610 \, kJ \, mol^{-1}}

Explanation

Solution

Answer (c) −610 kJ mol−1 { - 610 \, kJ \, mol^{-1}}