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Chapter 4: Q 41P (page 234)

Use bond-dissociation enthalpies (Table 4-2, p.203) to calculate values of螖Hfor the following reactions.

(a)

(b)

(c)

(d)

(e)

Short Answer

Expert verified

(a) 螖H = -36kJ/mol

(b) 螖H = -17kJ/mol

(c) 螖H = -19kJ/mol

(d) 螖H= -54kJ/mol

(e) 螖H = -41kJ/mol

Step by step solution

01

Bond dissociation enthalpy (BDE)

It may be defined as the amount of enthalpy required to break a bond homolytically in such a way that each bonded atom retains one of the bond鈥檚 two electrons.

Mathematically, 螖Ho = 鈭� (BDE of bonds broken) - 鈭� (BDE of bonds formed)

02

Calculation of  ΔH

(a)

Hence, 螖H value for the reaction is 螖H = -36kJ/mol.

(b)

Hence, 螖H value for the reaction is 螖H = -17kJ/mol.

(c)

Hence, 螖H value for the reaction is 螖H = -19kJ/mol.

(d)

Hence, 螖Hvalue for the reaction is 螖H= -54kJ/mol.

(e)

Hence, 螖H value for the reaction is 螖H = -41kJ/mol.

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