Given that bond energy of H-H and Cl- Cl are 430 kJ mol^-1 and 240 kJ mol^-1 respectively and ΔH_f for HCl is -90 kJ mol^-1. Bond and ΔH_f for HCl is -90 kJ mol^-1.Bond enthalpy of HCl is:

290 kJ mol^-1

380 kJ mol^-1

425 kJ mol^-1

425 kJ mol^-1

Solution

380 kJ mol^-1

increment straight H subscript reaction space equals increment subscript straight H minus straight H end subscript space plus increment straight H subscript Cl minus Cl end subscript minus 2 space increment straight H subscript HCl equals negative negative 90 space kJ Or space increment straight H subscript straight H minus Cl end subscript space equals space fraction numerator 430 plus 240 minus left parenthesis negative 90 right parenthesis over denominator 2 end fraction space equals space 760 over 2 space equals space 380 space kJ space mol to the power of negative 1 end exponent

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