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Hint:
Addition of a suitable catalyst to the reaction mixture can decrease the activation energy of the intermediate and hence it will require less amount of energy per molecule to obtain products. Change in temperature can also alter the rate of the reaction.
Complete step by step solution:
The decomposition reaction of hydrogen peroxide can be given as below.
\[{{H}_{2}}{{O}_{2(l)}}\to {{H}_{2}}{{O}_{(l)}}+\frac{1}{2}{{O}_{2(g)}}\]
Now, we will discuss the way by which the rate of this chemical reaction can be increased.
- The best way to increase the decomposition rate of hydrogen peroxide is to use a catalyst.
- If we use $Mn{{O}_{2}}$ as a catalyst, then it will decrease the activation energy of the reaction. As the activation energy of the reaction gets decreased, it requires comparatively less energy for reactant molecule to transform into the product molecule. So, in this way less energy is required for the decomposition of hydrogen peroxide. Thus, we can say that $Mn{{O}_{2}}$ will increase the rate of decomposition of hydrogen peroxide if we add it as a catalyst.
- Activation energy is the minimum energy required by the reactant to get converted into the product molecule.
- There is an alternative way also available to increase the rate of this reaction by giving high temperatures.
- In most of the chemical reactions, the rate of the reaction increases upon an increase in the reaction temperatures. Increase in the temperature increases the kinetic energy of the molecules, so that number of collisions between them also increases which in turn result in the formation of more product molecules. Thus, we can say that an increase in temperature of the reaction will also increase the rate of decomposition of hydrogen peroxide.
Note: Actually, we cannot predict which catalyst will increase the rate of the reaction by just a theoretical explanation. It also requires practical evidence to confirm that catalyst increases the rate of the reaction. Remember that increasing the temperature of the reaction will increase the rate of the reaction in most of the cases.
Addition of a suitable catalyst to the reaction mixture can decrease the activation energy of the intermediate and hence it will require less amount of energy per molecule to obtain products. Change in temperature can also alter the rate of the reaction.
Complete step by step solution:
The decomposition reaction of hydrogen peroxide can be given as below.
\[{{H}_{2}}{{O}_{2(l)}}\to {{H}_{2}}{{O}_{(l)}}+\frac{1}{2}{{O}_{2(g)}}\]
Now, we will discuss the way by which the rate of this chemical reaction can be increased.
- The best way to increase the decomposition rate of hydrogen peroxide is to use a catalyst.
- If we use $Mn{{O}_{2}}$ as a catalyst, then it will decrease the activation energy of the reaction. As the activation energy of the reaction gets decreased, it requires comparatively less energy for reactant molecule to transform into the product molecule. So, in this way less energy is required for the decomposition of hydrogen peroxide. Thus, we can say that $Mn{{O}_{2}}$ will increase the rate of decomposition of hydrogen peroxide if we add it as a catalyst.
- Activation energy is the minimum energy required by the reactant to get converted into the product molecule.
- There is an alternative way also available to increase the rate of this reaction by giving high temperatures.
- In most of the chemical reactions, the rate of the reaction increases upon an increase in the reaction temperatures. Increase in the temperature increases the kinetic energy of the molecules, so that number of collisions between them also increases which in turn result in the formation of more product molecules. Thus, we can say that an increase in temperature of the reaction will also increase the rate of decomposition of hydrogen peroxide.
Note: Actually, we cannot predict which catalyst will increase the rate of the reaction by just a theoretical explanation. It also requires practical evidence to confirm that catalyst increases the rate of the reaction. Remember that increasing the temperature of the reaction will increase the rate of the reaction in most of the cases.
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