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Thermodynamics

Question
CBSEENCH11006218

Estimate the enthalpy change for the reaction:

straight H subscript 2 left parenthesis straight g right parenthesis space plus space Br subscript 2 left parenthesis straight g right parenthesis space rightwards arrow space space 2 HBr left parenthesis straight g right parenthesis

Given that bond enthalpy of H – H, Br – Br and H – Br bonds are 435 kJ mol–1, 192 kJ mol–1 and 368 kJ mol–1 respectively.

Solution

We know that
          increment subscript straight r straight H to the power of 0 space equals space sum from blank to blank of Bond space enthalpies subscript left parenthesis Reactants right parenthesis end subscript space minus space sum from blank to blank of Bond space enthalpies subscript left parenthesis Products right parenthesis end subscript
space space space space space space space space equals space open square brackets straight H to the power of 0 subscript straight H minus straight H end subscript space plus space straight H subscript Br space minus Br end subscript superscript 0 close square brackets space minus space open square brackets 2 space cross times space straight H subscript straight H space minus space Br end subscript superscript 0 close square brackets
               equals space open square brackets 435 space kJ space mol to the power of negative 1 end exponent space plus space 192 space kJ space mol to the power of negative 1 end exponent close square brackets space minus space open square brackets 2 cross times 368 space kJ space mol to the power of negative 1 end exponent close square brackets
equals space 627 space kJ space mol to the power of negative 1 end exponent space minus space 736 space kJ space mol to the power of negative 1 end exponent
space equals negative 109 space kJ space mol to the power of negative 1 end exponent