Question

The alternative reaction pathways for radical chain reactions of $B{r}_2$ with ethane may be represented as follows.
1} $C{H}_{3}C{H}_3+B{r}_2\to C{H}_{3}C{H}_{2}Br+HBr$
2) $C{H}_{3}C{H}_3+B{r}_2\to 2C{H}_{3}Br$

Which statement is correct?

• A. Reaction $\left(1\right)$ is more likely because it is more endothermic
• B. Reaction $\left(2\right)$ is more likely because it is more endothermic
• C. Reaction $\left(1\right)$ is more likely because it is more exothermic
• D. Reaction $\left(2\right)$ is more likely because it is more exothermic

Answer 1

The correct option is C Reaction $\left(1\right)$ is more likely because it is more exothermic

Halogenation is supposed to proceed via free radical chain mechanism.

$C{H}_{3}C{H}_3+B{r}_2\to C{H}_{3}C{H}_{2}Br+HBr$

$C{H}_{3}C{H}_3+B{r}_2\to 2C{H}_{3}Br$

In the first reaction, more energy released than in the second reaction. So the first reaction is more likely as enthalpy is more negative in the first reaction than the second reaction.

More the negative enthalpy, more negative free energy, greater the spontaneity.