Solution Found!
Use bond-dissociation enthalpies (Table 4-2, p. 143) to
Chapter 5, Problem 40SP(choose chapter or problem)
Use bond-dissociation enthalpies (Table 4-2, p. 143) to calculate values of \(\Delta H^{\circ}\) for the following reactions.
\(C H_{3}-C H_{3}+I_{2} \rightarrow C H_{3} C H_{2} I+H I\)
\(C H_{3} C H_{2} C l+H I \rightarrow C H_{3} C H_{2} I+H C l\)
\(\left(C H_{3}\right)_{3} C-O H+H C l \rightarrow\left(C H_{3}\right)_{3} C-C l+H_{2} O\)
\(C H_{3} C H_{2} C H_{3}+H_{2} \rightarrow C H_{3} C H_{3}+C H_{4}\)
\(C H_{3} C H_{2} O H+H B r \rightarrow C H_{3} C H_{2}-B r+H_{2} O\)
Equation Transcription:
Text Transcription:
\Delta H^{\circ}
CH_3-CH_3+I_2 \rightarrow CH_3CH_2I+HI
CH_3CH_2Cl +HI \rightarrow CH_3CH_2I+HCl
(CH_3)_3C-OH+HCl \rightarrow (CH_3)_3C-Cl+H_2O
CH_3CH_2CH_3+H_2 \rightarrow CH_3CH_3+CH_4
CH_3CH_2OH+HBr \rightarrow CH_3CH_2-Br+H2O
Questions & Answers
QUESTION:
Use bond-dissociation enthalpies (Table 4-2, p. 143) to calculate values of \(\Delta H^{\circ}\) for the following reactions.
\(C H_{3}-C H_{3}+I_{2} \rightarrow C H_{3} C H_{2} I+H I\)
\(C H_{3} C H_{2} C l+H I \rightarrow C H_{3} C H_{2} I+H C l\)
\(\left(C H_{3}\right)_{3} C-O H+H C l \rightarrow\left(C H_{3}\right)_{3} C-C l+H_{2} O\)
\(C H_{3} C H_{2} C H_{3}+H_{2} \rightarrow C H_{3} C H_{3}+C H_{4}\)
\(C H_{3} C H_{2} O H+H B r \rightarrow C H_{3} C H_{2}-B r+H_{2} O\)
Equation Transcription:
Text Transcription:
\Delta H^{\circ}
CH_3-CH_3+I_2 \rightarrow CH_3CH_2I+HI
CH_3CH_2Cl +HI \rightarrow CH_3CH_2I+HCl
(CH_3)_3C-OH+HCl \rightarrow (CH_3)_3C-Cl+H_2O
CH_3CH_2CH_3+H_2 \rightarrow CH_3CH_3+CH_4
CH_3CH_2OH+HBr \rightarrow CH_3CH_2-Br+H2O
ANSWER:
Solution 40SP:
Here, we are going to calculate the bond enthalpy change for each of the reaction.
Step 1:
The overall enthalpy change for a reaction is the sum of the dissociation enthalpies
of the bonds broken minus the sum of the dissociation enthalpies of the bonds formed.
o = [Bond dissociation energy for bonds broken] -[Bond dissociation energy for bonds formed]