Use bond energies to estimate the enthalpy of formation of HBr(g).
BE(H-H) = 436 kJ/mol
BE(Br-Br) = 192 kJ/mol
BE(H-Br) = 366 kJ/mol
A) +262 kJ/mol
B) +104 kJ/mol
C) +52 kJ/mol
D) -52 kJ/mol
E) -104 kJ/mol

Question

Use bond energies to estimate the enthalpy of formation of HBr(g).
BE(H-H) = 436 kJ/mol 
BE(Br-Br) = 192 kJ/mol
BE(H-Br) = 366 kJ/mol
A) +262 kJ/mol
B) +104 kJ/mol
C) +52 kJ/mol
D) -52 kJ/mol
E) -104 kJ/mol

Quizplus · Accepted Answer

The enthalpy of formation of HBr can be calculated using the bond energies of H-H, H-Br, and Br-Br as follows:

∆H = ∑BE(bonds broken) - ∑BE(bonds formed)

The bonds broken are one H-H bond and one Br-Br bond, and the bond formed is one H-Br bond. 
∆H = (436 kJ/mol + 192 kJ/mol) - 366 kJ/mol
∆H = 262 kJ/mol

However, since the reaction involves the formation of a bond (H-Br), the enthalpy change is exothermic, meaning that the actual value of ∆H will be negative. Therefore, the correct answer choice is D), -52 kJ/mol, which is the closest negative value to the calculated positive value.

Use Bond Energies to Estimate the Enthalpy of Formation of HBr(g)