Question

10 of 15 Questions < O Assignment Score: 1210/1500 Resources Hint Check Answer 6 Question 1 of Attempts 100/100 Correct Question 10 of 15 > 0/100 7 Question 6 of Attempts CH3 H3C-c. + CH3 CH3 H3C-C-Br Step 2 Br-Br → + Br. In Progress CH3 100/100 8 Question 1 of 0 Attempts Correct Overall reaction CH3 H3C-C-H CH3 + Br-Br - CH3 H3C-Ć-Br CH₃ + H-Br 9 Question 3 of Attempts 90/100 Correct A + B, calculate the enthalpy of each step and the enthalpy of Using the table of bond dissociation energies for A-B — the overall reaction. 0/100 10 Question 0 of 0 Attempts Bond broken AH, kJ/mol H-H 436 11 Question 3 of Attempts 80/100 Correct Br-Br 194 100/100 12 Question 1 of 0 Attempts H-Br (CH2)2C-H (CH2)2C-Br 366 400. 292 Correct 90/100 13 Question 2 of 0 Attempts Correct Step 1: AH = kJ/mol Step 2: AH = kJ/mol 50/100 14 Question 2 of 0 Attempts In Progress Overall reaction: AH = kJ/mol 15 Question 1 of - Attempts 100/100 Correct

Answer 1

Answer:-

As we know that in the first step , the dissociation of (CH₃)₃C-H bond takes place and the formation of tertiary butyl free radical ((CH₃)₃C^.) occurs.

Step:- 1

(CH₃)₃C-H $\rightarrow$   (CH₃)₃C^. + H^.

Since we know that

enthalpy of the dissociation of (CH₃)₃C-H bond ($\Delta$H) = 400 kJ /mol

therefore

enthalpy of reaction of Step 1 ($\Delta$H¹_rxn) = H_products - H_reactants

enthalpy of reaction of Step 1 ($\Delta$H¹_rxn) = 0 - enthalpy of the dissociation of (CH₃)₃C-H bond ($\Delta$H)

enthalpy of reaction of Step 1 ($\Delta$H¹_rxn) = 0 - 400 kJ /mol

enthalpy of reaction of Step 1 ($\Delta$H¹_rxn) = - 400 kJ /mol (i.e the the answer)

In Step 2 , the dissociation of Br-Br bond takes place which forms bromine free radicals (Br ^.) and one bromine free radical (Br ^.) reacts with tertiary butyl free radical ((CH₃)₃C^.) to form the tertiary butyl bromide ((CH₃)₃C-Br) and other bromine free radical (Br ^.).

Step:- 2

      (CH₃)₃C^. + Br-Br   $\rightarrow$  (CH₃)₃C-Br + Br ^.

Since we know that the enthalpy of the bond dissociation is that energy which is required to break the chemical bond and it is also equal to the enthalpy of the bond formation i.e the energy which is required to form the chemical bond.

So

enthalpy of the dissociation of Br-Br bond ($\Delta$H) = 194 kJ /mol

enthalpy of the dissociation of (CH₃)₃C-Br bond ($\Delta$H) = 292 kJ /mol

therefore

enthalpy of reaction of Step 2 ($\Delta$H²_rxn) = H_products - H_reactants

enthalpy of reaction of Step 2 ($\Delta$H²_rxn) = 292 kJ /mol - 194 kJ /mol

enthalpy of reaction of Step 2 ($\Delta$H²_rxn) = 98 kJ /mol (i.e the the answer)

As we know that overall reaction is as follows:-

      (CH₃)₃C-H + Br-Br   $\rightarrow$  (CH₃)₃C-Br + HBr

So

enthalpy of the dissociation of Br-Br bond ($\Delta$H) = 194 kJ /mol

enthalpy of the dissociation of (CH₃)₃C-Br bond ($\Delta$H) = 292 kJ /mol

enthalpy of the dissociation of H-Br bond ($\Delta$H) = 366 kJ /mol

enthalpy of the dissociation of (CH₃)₃C-H bond ($\Delta$H) = 400 kJ /mol

therefore

enthalpy of overall reaction ($\Delta$H_rxn) = H_products - H_reactants

enthalpy of overall reaction ($\Delta$H_rxn) = [enthalpy of the dissociation of (CH₃)₃C-Br bond ($\Delta$H) + enthalpy of the dissociation of H-Br bond ($\Delta$H)] - [enthalpy of the dissociation of (CH₃)₃C-H bond ($\Delta$H) + enthalpy of the dissociation of Br-Br bond ($\Delta$H) ]

enthalpy of overall reaction ($\Delta$H_rxn) = (292 kJ /mol + 194 kJ /mol) - (400 kJ /mol + 366 kJ /mol )

enthalpy of overall reaction ($\Delta$H_rxn) = (486 kJ /mol) - (766 kJ /mol )

enthalpy of overall reaction ($\Delta$H_rxn) = 486 kJ /mol - 766 kJ /mol

enthalpy of overall reaction ($\Delta$H_rxn) = - 280 kJ /mol (i.e the the answer)