Problem 84 - Chemistry Central Science

Question

Employing the bond enthalpy values listed in Table \(8.4\), estimate the molar enthalpy change occurring upon (a) polymerization of ethylene, (b) formation of nylon 6,6 , (c) formation of polyethylene terephthalate (PET).

Step-by-Step Solution

Answer

The molar enthalpy changes for the polymerization processes are: (a) Polymerization of ethylene: \(\Delta H = 82\:kJ/mol\), (b) Formation of nylon 6,6: \(\Delta H = -141\:kJ/mol\), and (c) Formation of PET: \(\Delta H = -22\:kJ/mol\).

Step 1: (a) Polymerization of ethylene

Ethylene, or C2H4, has the following structure: \[CH_2=CH_2\] The double bond between the two carbon atoms will break during polymerization, and a new single bond will form between each carbon atom and another ethylene molecule. To find the molar enthalpy change, we need to subtract the bond enthalpy of the double bond from the bond enthalpy of two single bonds. Using the bond enthalpy values from Table 8.4: C=C bond enthalpy: \(610\:kJ/mol\) C-C bond enthalpy: \(346\:kJ/mol\) The molar enthalpy change for polymerization of ethylene is: \(\Delta H_{polymerization} = 2 \times 346 - 610 = 82\: kJ/mol\)

Step 2: (b) Formation of nylon 6,6

Nylon 6,6 is formed via condensation polymerization of adipic acid (\(HOOC(CH_2)_4COOH\)) and hexamethylenediamine (\(H_2N(CH_2)_6NH_2\)). During the reaction, an amide bond is formed between the acid and amine groups, and a water molecule is released. Looking at the enthalpy changes, we have: C-O bond enthalpy in the acid: \(358\:kJ/mol\) O-H bond enthalpy in the acid: \(467\:kJ/mol\) N-H bond enthalpy in the amine: \(391\:kJ/mol\) Breaking one C-O and one O-H bond, forming one N-H bond, and a new C-N bond: C-N bond enthalpy: \(293\:kJ/mol\) \(\Delta H_{formation} = -[1 \times (358 + 467) - 1 \times (391 + 293)] = -141\:kJ/mol\)

Step 3: (c) Formation of polyethylene terephthalate (PET)

PET is formed from the reaction between terephthalic acid (\((COOH)_2(C_6H_4\)) and ethylene glycol (\((HOCH_2CH_2OH)_{n}\)). In this reaction, an ester bond is formed between the acid and alcohol groups, and a water molecule is released. Using the bond enthalpy values from Table 8.4: C-O bond enthalpy in acid: \(358\:kJ/mol\) O-H bond enthalpy in the alcohol: \(467\:kJ/mol\) O=C bond enthalpy in the acid: \(745\:kJ/mol\) C-O bond in the ester: \(358\:kJ/mol\) \(\Delta H_{formation} = -[1 \times (358 + 467) - 1 \times (745 + 358)] = -22\:kJ/mol\) In summary: (a) Polymerization of ethylene: \(\Delta H = 82\:kJ/mol\) (b) Formation of nylon 6,6: \(\Delta H = -141\:kJ/mol\) (c) Formation of PET: \(\Delta H = -22\:kJ/mol\)