Question

Given that
`\Delta H^o_f` [Br(g)] = 111.9 kJ ⋅ mol⁻¹
`\Delta H^o_f` [Br(g)] = 111.9 kJ⋅mol⁻¹
`\Delta H^o_f` [C(g)] = 716.7 kJ ⋅ mol⁻¹
`\Delta H^o_f` [C(g)] = 716.7 kJ⋅mol⁻¹
`\Delta H^o_f` [CBr₄(g)] = 29.4 kJ ⋅ mol⁻¹
`\Delta H^o_f` [CBr₄(g)] = 29.4 kJ⋅mol⁻¹ Calculate the average molar bond enthalpy of the carbon–bromine bond in a CBr₄ molecule.

Answer 1

Answer : The average molar bond enthalpy of the carbon–bromine bond in a CBr₄ molecule is -283.72 kJ

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as
`\Delta H^o`

The equation used to calculate enthalpy change is of a reaction is:

`\Delta H^o_(rxn)=\sum [n* \Delta H^o_f(product)]-\sum [n* \Delta H^o_f(reactant)]`

The equilibrium reaction follows:

`C(g)+4Br(g)\rightleftharpoons CBr_4(g)`

The equation for the enthalpy change of the above reaction is:

`\Delta H^o_(rxn)=[(n_((CBr_4))* \Delta H^o_f_((CBr_4)))]-[(n_((Br))* \Delta H^o_f_((Br)))+(n_((C))* \Delta H^o_f_((C)))]`

We are given:

`\Delta H^o_f_((CBr_4(g)))=29.4kJ/mol\\\Delta H^o_f_((C(g)))=716.7kJ/mol\\\Delta H^o_f_((Br(s)))=111.9kJ/mol`

Putting values in above equation, we get:

`\Delta H^o_(rxn)=[(1* 29.4)]-[(4* 111.9)+(1* 716.7)]=-1134.9kJ/mol`

Now we have to calculate the average molar bond enthalpy of the carbon–bromine bond in a CBr₄ molecule.

The enthalpy change of reaction = E(bonds broken) - E(bonds formed)

The enthalpy change of reaction = - E(bonds formed)

`\Delta H=-[4* B.E_(C-Br)]`

Now put all the given values in the above expression, we get:

`-1134.9=-[4* B.E_(C-Br)]`

`B.E_(C-Br)=-283.72kJ`

Therefore, the average molar bond enthalpy of the carbon–bromine bond in a CBr₄ molecule is -283.72 kJ