At standard conditions, if the change in the enthalpy for the following reaction is –109 kJ mol^–1. H2(g)+Br2(g)⟶2HBr(g)

Question

At standard conditions, if the change in the enthalpy for the following reaction is –109 kJ mol^–1.

$\require{mhchem}\ce{H_{2(g)} + Br_{2(g)} -> 2HBr_{(g)}}$

Given that bond energy of $\ce{H2}$ and $\ce{Br2}$ is 435 kJ mol^–1 and 192 kJ mol^–1, respectively, what is the bond energy (in kJ mol^–1) of $\ce{HBr}$?

(1) 259 

(2) 368

(3) 736 

(4) 518

Comments

Thermodynamics, enthalpy change of a reaction

Answer 1

Ans: (2) 368

Sol. $\ce{H2(g) + Br2(g) -> 2HBr(g)};\ \Delta H=-109$

$\Delta H=\Sigma (BE)_R-\Sigma(BE)_P$

$=(BE_{H–H})+(BE_{Br–Br})-2(BE_{H–Br})$

$-109=(435)+(192)-2(BE_{H–Br})$

$BE_{H–Br}=368\ kJmol^{-1}$

Comments

That is a pretty interesting answer