Question

Question: (a) Use bond-dissociation enthalpies from Table 4-2 (page 203), calculate the heat of reaction for each step in the free-radical bromination of methane.

(b) Calculate the overall heat of reaction.

Short answer

Answer

(a) 190kJ/mol (initiation step)

73kJ/mol (Propagation step I)

-112kJ/mol (Propagation step II)

(b)-39kJ/mol

Step-by-step solution

Free radicals

An atom or group of atoms containing odd or unpaired electronsis known as the free radical. The unpaired electron is represented by a single unpaired dot in the formula. Free radicals are electrically neutral. They are highly reactive species formed by homolytic fission of a covalent bond.

Steps involved in a free radical chain reaction

In a free-radical chain reaction, free radicals are generally created in the initiation steps. A free radical and a reactant is combined to yield a product and another free radical in the propagation steps. Lastly, the number of free radicals generally decreasesin the termination steps.

Bond dissociation enthalpy (BDE)

It may be defined as the amount of enthalpy required to break a bond homolyticallysothat each bonded atom retains one of the bond’s two electrons.

Mathematically,.

$∆H^0=\sum _{}\left(BDEbondsbroken\right)-\sum _{}\left(BDEbondsformed\right)$

Explanation

(a) The mechanism consists of three parts which are the initiation step, propagation step I, and propagation step II.

Mechanism for free-radical bromination of ethane

Therefore, the enthalpy of the reaction for the initiation step is .190kJ/mol .

Therefore, the enthalpy of the reaction for the propagation step I is 73kJ/mol

Therefore, the enthalpy of the reaction for the propagation step II is .-112kJ/mol

b.

Therefore, the enthalpy of the overall reaction is $-39\text{\hspace{0.17em}\hspace{0.17em}kJ/mol}$.