Calculate the enthalpy of reaction for the following reaction using the supplied

Question

Calculate the enthalpy of reaction for the following reaction using the supplied bond dissociation enthalpies Would you expect that the free energy of the reaction would be the enthalpy of the reactions? C6H5CH2 -H + Br - Br Rightarrow C5H3CH2-Br + H-Br Bond dissociation enthalpies: H-Br Rightarrow H+ +Br* Delta H Degree = 366kJ/mol C6H5CH2-H Rightarrow C6H5CH2-H Rightarrow C6H5CH2* + H * Delta H Degrees = 375kj/mol Br-Br Rightarrow Br +Br* Delta H Degree = 193kJ/mol C6H5CH2-H Rightarrow C6H5CH2-Br Rightarrow C6H5CH2* + Br * Delta H Degrees = 266kj/mol

Explanation / Answer

We know formula for bond enthalpy

Delta H = n sum delta H (bond broken) – n sum (bond formed)

In the given reaction one C6H5CH2-H bond is broken and one Br-Br bond is broken.

One C6H5CH2-H and one H-Br bond is formed.

Lets use the value to get find out delta H

Delta H

= [1 mol x bond enthalpy C6H5CH2-H + 1 mol x bond enthalpy Br-Br ]

- [1 mol x bond enthalpy C6H5CH2-Br + 1 mol x bond enthalpy of HBr]

= (375 kJ + 193 kJ ) – (266 kJ + 366 kJ )

= -64 kJ

Since delta S of the reaction does not change so its delta H value will be the same with free energy.

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