Given that [br(g)] = 111.9 kj ⋅ mol⁻¹ [br(g)] = 111.9 kj⋅mol⁻¹ [c(g)] = 716.7 kj ⋅ mol⁻¹ [c(g)] = 716.7 kj⋅mol⁻¹ [cbr₄(g)] = 29.4 kj ⋅ mol⁻¹ [cbr₄(g)] = 29.4 kj⋅mol⁻¹ calculate the average molar bond enthalpy of the carbon–bromine bond in a cbr₄ molecule.

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

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

The equilibrium reaction follows:

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

We are given:

Putting values in above equation, we get:

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)

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

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

heat or electricity in the atmosphere- c.