Question

Use bond energies to estimate the enthalpy of formation of HBr(g).

H₂(g) + Br₂(g) 2HBr(g)

BE(H—H) = 436 kJ/mol
BE(Br—Br) = 192 kJ/mol
BE(H—Br) = 366 kJ/mol
a. +262 kJ
b. –52 kJ
c. –104 kJ
d. +104 kJ
e. +52 kJ

Answer 1

To estimate the enthalpy of formation of HBr(g), we need to calculate the energy change involved in breaking the bonds in the reactants and forming the bonds in the products. The enthalpy of formation of HBr(g) is approximately +332 kJ/mol.

Step-by-step explanation:

To estimate the enthalpy of formation of HBr(g), we need to calculate the energy change involved in breaking the bonds in the reactants and forming the bonds in the products. First, we calculate the energy required to break the bonds in H₂(g) and Br₂(g) using the bond dissociation energies provided. Next, we calculate the energy released when 2HBr(g) bonds are formed. Finally, we sum up the energy changes for both the reactants and the products to estimate the enthalpy of formation of HBr(g).

The energy required to break the bonds in H₂(g) is 2 x 436 kJ/mol = 872 kJ/mol. The energy required to break the bond in Br₂(g) is 192 kJ/mol. Therefore, the total energy required to break the bonds in the reactants is 872 kJ/mol + 192 kJ/mol = 1,064 kJ/mol.

The energy released when 2HBr(g) bonds are formed is 2 x 366 kJ/mol = 732 kJ/mol. Therefore, the total energy released in forming the bonds in the products is 732 kJ/mol.

Finally, we subtract the energy released in forming the bonds in the products from the energy required to break the bonds in the reactants to estimate the enthalpy of formation of HBr(g). 1,064 kJ/mol - 732 kJ/mol = 332 kJ/mol. Therefore, the enthalpy of formation of HBr(g) is approximately +332 kJ/mol.