Question

If the bond energies of $H-H,Br-Br$ and $H-Br$ are 433, 192 and 364 $kJmo{l}^{-1}$ respectively, $\Delta H$ for the reaction $H_{2\left(g\right)}+B{r}_{2\left(g\right)}\to 2HB{r}_{\left(g\right)}$ is:

• A. -261 kJ
• B. +103 kJ
• C. +261 kJ
• D. -103 kJ

Answer 1

The correct option is D -103 kJ

The given reaction is

$H_2\left(g\right)+B{r}_2\left(g\right)\to 2HBr\left(g\right)$

In this reaction, one $H-H$ and one $Br-Br$ bond is broken and two new $H-Br$ bond is formed. So the enthalpy of the reaction would be,

$\Delta H=\left[\right(433+192)-(2\times 364\left)\right]kJmo{l}^{-1}$ (during bond cleavage energy is required and during bond formation energy is released)

$\Delta H=-103kJmo{l}^{-1}$

Correct option is D.