The rate of formation of \({\bf{B}}{{\bf{r}}_{\bf{2}}}\) is \({\bf{6}}.{\bf{0}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{6}}}}{\rm{ }}{\bf{mol}}/{\bf{L}}/{\bf{s}}\) in a reaction described by the following net ionic equation:

\(5B{r^ - } + BrO_3^ - + \,6{H^ + }3B{r_2} + 3{H_2}O{\rm{ }}\)

Write the equations that relate the rates of consumption of the reactants and the rates of formation of the products.

The rate of reaction is the change in the quantity of a reactant or product per unit time.

Rate expression for a reaction is the mathematical depiction of the change in species concentration over time.

\(\begin{align}{\rm{rate of reaction = }} - \frac{{{\rm{change in concentration of reactant}}}}{{{\rm{time interval}}}}\\{\rm{rate of reaction = }} - \frac{{\Delta \left( {{\rm{concentration of reactant}}} \right)}}{{\Delta {\rm{t}}}}\end{align}\)

Where,

Brackets indicate molar concentrations,

Symbol delta (Δ) indicates ‘change in’.

The rate of the reaction will be equal to the rate of consumption of the reactants divided by its stoichiometric coefficients.

Similarly, the rate of the reaction will be equal to the rate of formation of the products divided by its stoichiometric coefficients.

Therefore, \({\rm{rate of }}formation/consumption\;{\rm{ = }} \pm \frac{1}{{stoichiometriccoefficient}} \times \frac{{\Delta \left( {reactant\;or\;product} \right)}}{{\Delta {\rm{t}}}}\)

From the given equation,

\(5B{r^ - } + BrO_3^ - + \,6{H^ + }3B{r_2} + 3{H_2}O{\rm{ }}\)

We know that, 5 bromide ions react with bromate ions and 6 hydrogen ions to form 3 bromine molecules and 3 molecules of water. Hence, for the above given reaction,

Rate of reactants:

Rate of consumption of bromide ion \(\left( {B{r^ - }} \right) = - \frac{1}{5}\frac{{\Delta \left( {B{r^ - }} \right)}}{{\Delta t}}\)

Rate of consumption of bromate ions \(\left( {{\bf{BrO}}_3^ - } \right) = - \frac{{\Delta \left( {{\bf{BrO}}_3^ - } \right)}}{{\Delta t}}\)

Rate of consumption of hydrogen ions \(\left( {{H^ + }} \right) = - \frac{1}{6}\frac{{\Delta \left( {{H^ + }} \right)}}{{\Delta t}}\)

Rate of products:

Rate of formation of bromine molecules \(\left( {B{r_2}} \right) = + \frac{1}{3}\frac{{\Delta \left( {B{r_2}} \right)}}{{\Delta t}}\)

Rate of formation of water molecules\(\left( {{H_2}O} \right) = + \frac{1}{3}\frac{{\Delta \left( {{H_2}O} \right)}}{{\Delta t}}\)

Therefore, equations relating the rates of consumption of the reactants and the rates of formation of the products would be:

\( - \frac{1}{5}\frac{{\Delta \left( {B{r^ - }} \right)}}{{\Delta t}} = - \frac{{\Delta \left( {{\bf{BrO}}_3^ - } \right)}}{{\Delta t}} = - \frac{1}{6}\frac{{\Delta \left( {{H^ + }} \right)}}{{\Delta t}} = + \frac{1}{3}\frac{{\Delta \left( {B{r_2}} \right)}}{{\Delta t}} = + \frac{1}{3}\frac{{\Delta \left( {{H_2}O} \right)}}{{\Delta t}}\)

Activation energy is the energy difference between the initial reagents and the transition state. (In case of transition state the energy is maximum on the reaction coordinate diagram).

In the reaction (a), the reagents are at 35 kJ and the transition state is at 45 kJ, so the activation energy can be calculated as:

\({{\rm{E}}_{\rm{a}}}{\rm{ = 45 kJ - 35 kJ = 10 kJ}}\)

Thus, the activation energy \(\left( {{E_a}} \right)\) of the reaction (a) a is 10 kJ and reaction (b) is 10 kJ.