Use the table of bond dissociation energies to calculate and compare the enthalp

Question

Use the table of bond dissociation energies to calculate and compare the enthalpy of monobromination and monochlorination for the given alkyl position.

Explanation / Answer

C (s) + 2 H2 (g) --> CH4 (g) (bond dissociation energy of H-H and C-H bonds are 436 KJ/mol and 415 KJ/mol, respectively. dHf of C (g) = 715 KJ/mol) My solution: CH4 - 2*H2 - C = 4*C-H - 2*H-H - C = 4*(415) - (715) - 2*2*(436) = C (s) + 2 H2 (g) --> CH4 (g) (bond dissociation energy of H-H and C-H bonds are 436 KJ/mol and 415 KJ/mol, respectively. dHf of C (g) = 715 KJ/mol) My solution: CH4 - 2*H2 - C = 4*C-H - 2*H-H - C = 4*(415) - (715) - 2*2*(436) = Am I doing it right? Almost... note that they give you the bond DISSOCIATION energy. The 4 CH bongs are being formed, not dissociated, so that would get the negative sign in front of it. Products (negative) while reactants (positive). The bonds in the reactants are the ones that are dissociating. -4*(415) + [(715) + 2*(436)] By the way... in the H2 bond... there's only one bond between the two H's... but we have two of those types of molecules... so it's only 2*436.

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.