Question

          calculate the average molar bond enthalpy of the carbon-bromine bond in a CBr4 molecule....

    
    
calculate the average molar bond enthalpy of the carbon-bromine bond in a CBr4 molecule.

given thay :

deltaH= Br(g)= 111.9 kj
dH= C(g)= 716.7 jk
dH= CBr4(g)= 29.4
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Answer #1
Concepts and reason

• The thermodynamic quantity “enthalpy” is state function that depends on the initial and final states of the system.

• The enthalpy is measured when a system is at the constant pressure; the heat released or absorbed is equal to the change in enthalpy.

• The bond enthalpy is also known as the bond energy, and it is defined as the energy required for breaking the chemical bonds in a molecule.

• The average molar bond enthalpy can be calculated by dividing the bond enthalpy with total number of bonds present in a molecule.

Fundamentals

The arbitrary reaction is given below:

aA +bB—>cC + dD

The formulae to calculate the enthalpy of reaction is given below:

AH.cation = [c(AH° ) +(AH”)]-[a(AH° ) + b(AH”)]
Where,
AH reaction is Enthalpy of reaction,
AH, is Standard change in ent

Consider the formation of molecule from its elements.

C+4Br—
>CBr,

Given that,

AH[Br(g)]=111.9kJ
AH[C(g)]=716.7kJ
AH[CBr, (g)]= 29.4kJ

ΔΗ. =mΔΗ (product)-nΔΗ (reactant)
-[ixΔΗ (CBT)]-[ixΔΗ (C) + 4xΔΗ (Br)]ki
=[29.4]-[(716.7)-4x(111.9)]kJ/mole
=-1134.9kJ

The number of carbon-bromine bonds in molecule is four.

Therefore, the average bond enthalpy is given below:

AH, = BE( bonds broken) - BE(product bonds formed)
-1134.9kJ = -4BE(C-Br)
BE(C-Br)1134.9kJ
= 283.725kJ

Therefore, the average molar bond enthalpy is 283.725kJ.

Ans:

The average molar bond enthalpy of the carbon-bromine bond in a molecule is 283.725kJ.

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