Thermodynamics

Thermodynamics

Question

Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below

1 half Cl subscript 2 space left parenthesis straight g right parenthesis space rightwards arrow with 1 half space increment subscript diss straight H to the power of minus on top space Cl space left parenthesis straight g right parenthesis space rightwards arrow with increment subscript eg straight H to the power of minus on top space Cl to the power of minus space left parenthesis straight g right parenthesis space rightwards arrow with increment subscript hyd straight H to the power of minus on top Cl to the power of minus space left parenthesis aq right parenthesis
The energy involved in conversion of 1 half Cl subscript 2 space left parenthesis straight g right parenthesis space to space Cl to the power of minus space left parenthesis straight g right parenthesis

left parenthesis using space the space data comma space increment subscript diss straight H subscript cl subscript 2 end subscript superscript minus space equals space 240 space kJ space mol to the power of minus comma space increment subscript eg space straight H subscript cl superscript minus space equals space minus space 349 space kJ space mol to the power of minus comma space
increment subscript eg space straight H subscript cl superscript minus space equals space minus 381 space kJ space mol to the power of minus right parenthesis

  • 152 kJ mol-

  • -610 kJ mol-

  • -850  kJ mol-

  • +120  kJ mol-

Answer

B.

-610 kJ mol-

1 half space Cl subscript 2 space left parenthesis straight g right parenthesis space rightwards arrow with space on top space Cl subscript aq superscript minus
space increment space straight H space equals space 1 half space increment straight H subscript diss space of space Cl subscript 2 space plus space increment subscript eq space Cl space plus space increment subscript hyd Cl to the power of minus
equals space plus space 240 over 2 space minus 349 minus 381
space equals space minus space 610 space kJ space mol to the power of negative 1 end exponent

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