Question

calculate the bond energy of HBr.bond energy of H2 is 436kj/mol and that of Br2 is 193 kj/mol

Answer 1

your question is incomplete. you didn't mention about enthalpy of formation of HBr. we assume enthalpy of formation is $\Delta H^{\circ}$

bond energy of H2 is 436 Kj/mol

means, H-----H → 2H, $\Delta H_1$ = 436 Kj/mol

bond energy of Br2 is 193 Kj/mol

means, Br-----Br → 2Br, $\Delta H_2$ = 193 Kj/mol

Let bond energy of HBr is $\Delta H_3$.

now, we know, H2 + Br2 → 2HBr

so, enthalpy of formation of HBr = $\Delta H_1+\Delta H_3 - 2(\Delta H_3)$

or, $\Delta H^{\circ}=436+193-2\Delta H_3$

or, $(436+193)-\Delta H^{\circ}=2\Delta H_3$

or, $\Delta H_3=\frac{1}{2}(629-\Delta H^{\circ})$

hence, bond energy of HBr is $\Delta H_3=\frac{1}{2}(629-\Delta H^{\circ})$

Answer 2

Answer:

314.5kj/mol

Explanation:

bond energy of H2 = 436 kj/mol

so bond energy for H =436/2 = 218 kj/mol

bond energy of Br2=193 kj/mol

so bond energy of Br= 193/2 = 96.5 kj/mol

therefore

H-Br = 218 + 96.5 = 314.5 kj/mol.