KINETICS OF THE HYDROGEN-BROMIDE REACTION

October 28, 2022
by Bhoomika Sheladiya

Kinetics of hydrogen-bromide reaction as,

\[\displaystyle {{H}_{{2(g)}}}+B{{r}_{{2(g)}}}\underset{{}}{\overset{{h\nu }}{\longrightarrow}}2HB{{r}_{{(g)}}}\]

The mechanism of hydrogen- bromine reaction same as the thermal reaction. The difference is that the first step in the photochemical reaction is the decomposition of Br2 into Br atom by the absorption of a quantum of energy hν,

\[\displaystyle B{{r}_{2}}+h\nu \underset{{}}{\overset{{{{k}_{1}}}}{\longrightarrow}}2Br\]
\[\displaystyle Br+{{H}_{2}}\underset{{}}{\overset{{{{k}_{2}}}}{\longrightarrow}}HBr+H\]
\[\displaystyle H+B{{r}_{2}}\underset{{}}{\overset{{{{k}_{3}}}}{\longrightarrow}}HBr+Br\]
\[\displaystyle H+HBr\underset{{}}{\overset{{{{k}_{4}}}}{\longrightarrow}}{{H}_{2}}+Br\]
\[\displaystyle 2Br\underset{{}}{\overset{{{{k}_{5}}}}{\longrightarrow}}B{{r}_{2}}\]

The rate formation of HBr is,

\[\displaystyle r=\frac{{d\left[ {HBr} \right]}}{{dt}}\]
\[\displaystyle ={{k}_{2}}\left[ {Br} \right]\left[ {{{H}_{2}}} \right]+{{k}_{3}}\left[ H \right]\left[ {B{{r}_{2}}} \right]-{{k}_{4}}\left[ H \right]\left[ {HBr} \right]---(1)\]

At steady-state approximation to [Br], we get,

\[\displaystyle \frac{{d\left[ {Br} \right]}}{{dt}}={{k}_{1}}{{I}_{a}}-{{k}_{2}}\left[ {Br} \right]\left[ {{{H}_{2}}} \right]+{{k}_{3}}\left[ H \right]\left[ {B{{r}_{2}}} \right]+{{k}_{4}}\left[ H \right]\left[ {HBr} \right]-{{k}_{5}}{{\left[ {Br} \right]}^{2}}=0---(2)\]

Applying steady-state approximation to [H], we get,

\[\displaystyle \frac{{d\left[ H \right]}}{{dt}}={{k}_{2}}\left[ {Br} \right]\left[ {{{H}_{2}}} \right]-{{k}_{3}}\left[ H \right]\left[ {B{{r}_{2}}} \right]-{{k}_{4}}\left[ H \right]\left[ {HBr} \right]=0----(3)\]

Now, adding equation (2) to equation (3),

\[\displaystyle {{k}_{1}}{{I}_{a}}-{{k}_{5}}{{\left[ {Br} \right]}^{2}}=0\]
\[\displaystyle \left[ {Br} \right]={{\left( {\frac{{{{k}_{1}}{{I}_{a}}}}{{{{k}_{5}}}}} \right)}^{{\frac{1}{2}}}}----(4)\]

From equation(3),

\[\displaystyle \left[ H \right]=\frac{{{{k}_{2}}\left[ {Br} \right]\left[ {{{H}_{2}}} \right]}}{{{{k}_{3}}\left[ {B{{r}_{2}}} \right]+{{k}_{4}}\left[ {HBr} \right]}}----(5)\]

Substituting equation (4) into equation (5) and we get,

\[\displaystyle \left[ H \right]=\frac{{{{k}_{2}}{{{\left( {\frac{{{{k}_{1}}{{I}_{a}}}}{{{{k}_{5}}}}} \right)}}^{{\frac{1}{2}}}}\left[ {{{H}_{2}}} \right]}}{{{{k}_{3}}\left[ {B{{r}_{2}}} \right]+{{k}_{4}}\left[ {HBr} \right]}}----(6)\]

Substituting equation (4) and equation (6) into equation (5) and we get,

\[\displaystyle r=\frac{{d\left[ {HBr} \right]}}{{dt}}=\frac{{2{{k}_{2}}{{{\left( {\frac{{{{k}_{1}}}}{{{{k}_{5}}}}} \right)}}^{{\frac{1}{2}}}}{{I}_{a}}^{{\frac{1}{2}}}\left[ {{{H}_{2}}} \right]}}{{1+\frac{{{{k}_{4}}\left[ {HBr} \right]}}{{{{k}_{3}}\left[ {B{{r}_{2}}} \right]}}}}----(7)\]

The rate law given by equation (7) agrees with the observed rate law, that the square root of the Intensity Ia of the absorbed radiation.

The quantum yield for this reaction is 0.01.

Comparison of Hydrogen-bromine and Hydrogen-Chlorine reaction

The quantum yield of Hydrogen-chlorine is high, whereas the Hydrogen-bromine is low, although both are chain reactions.

Facts explained that reaction (ii) in the mechanism for the former reaction is exothermic and takes place spontaneously. Reaction (ii) In the mechanism, the endothermic reaction occurs extremely slowly at ordinary temperature.

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About the author

Bhoomika Sheladiya

BSc. (CHEMISTRY) 2014- Gujarat University
MSc. (PHYSICAL CHEMISTRY) 2016 - School of Science, Gujarat University

Junior Research Fellow (JRF)- 2019
AD_HOC Assistant Professor-(July 2016 to November 2021)

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