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:  

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

The equilibrium reaction follows:

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

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

$\Delta H^o_{rxn}=[(n_{(CBr_4)}\times \Delta H^o_f_{(CBr_4)})]-[(n_{(Br)}\times \Delta H^o_f_{(Br)})+(n_{(C)}\times \Delta H^o_f_{(C)})]$

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:

$\Delta H^o_{rxn}=[(1\times 29.4)]-[(4\times 111.9)+(1\times 716.7)]=-1134.9kJ/mol$

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)

$\Delta H=-[4\times B.E_{C-Br}]$

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

$-1134.9=-[4\times B.E_{C-Br}]$

$B.E_{C-Br}=-283.72kJ$

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