
A hydrogenation reaction is carried out at $500{\text{K}}$. If the same reaction is carried out in presence of a catalyst at the same rate, the temperature required is $400{\text{K}}$. Calculate the activation energy of the reaction if the catalyst lowers the activation barrier by $20{\text{ kJ mo}}{{\text{l}}^{ - 1}}$.
Answer
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Hint: To answer this question you must recall the Arrhenius equation. The Arrhenius equation gives a relation between the rate of the reaction and its activation energy. We shall substitute the values to form 2 equations and compare them.
Formula used:
$k = A{e^{ - \dfrac{{{E_a}}}{{RT}}}}$
Where, $k$ is the rate with which the reaction proceeds
$A$ is the Arrhenius constant
$R$ is the gas constant
$T$ is the temperature at which the reaction proceeds
And ${E_a}$ is the activation energy or the activation barrier of the reaction
Complete step by step answer:
For the reaction carried out at $500{\text{K}}$, the Arrhenius equation can be written as
${k_{500}} = A{e^{ - \dfrac{{{E_{500}}}}{{R \times 500}}}}$
For the reaction carried out at $400{\text{K}}$, the Arrhenius equation can be written as
${k_{400}} = A{e^{ - \dfrac{{{E_{400}}}}{{R \times 400}}}}$
It is given in the question that both the reactions proceed at the same rate. Thus we can equate the two equations and we get
$A{e^{ - \dfrac{{{E_{400}}}}{{R \times 400}}}} = A{e^{ - \dfrac{{{E_{500}}}}{{R \times 500}}}}$
Simplifying:
\[ \Rightarrow \dfrac{{{E_{400}}}}{{400}} = \dfrac{{{E_{500}}}}{{500}}\]
Or, ${E_{500}} = \dfrac{{{E_{400}}}}{4} \times 5$
In the question, it is given that the activation energy is reduced by $20{\text{kJ mo}}{{\text{l}}^{ - 1}}$ for the second reaction. Thus, we can write, ${E_{500}} = {E_{400}} + 20$
Substituting this value of the activation energy in the previous equation, we get,
${E_{400}} + 20 = {E_{400}} \times 1.25$
Therefore,${E_{400}} = \dfrac{{20}}{{0.25}} = 80{\text{ kJ mo}}{{\text{l}}^{ - 1}}$
And, ${E_{500}} = 100{\text{ kJ mo}}{{\text{l}}^{ - 1}}$
Note:
Activation energy is the energy that we need to provide to compounds in order for a chemical reaction to take place. The activation energy $\left( {{E_a}} \right)$ is commonly measured in joules per mole. Activation energy can be considered as the magnitude of the energy barrier separating the initial and final thermodynamic states, namely the reactants and products. For a chemical reaction to occur at a good rate, the temperature of the system should be so high that the number of molecules with energy levels greater than or equal to the activation energy is appreciable.The activation energy of a reaction can be reduced by addition of suitable catalysts that make the reaction more feasible and faster.
Formula used:
$k = A{e^{ - \dfrac{{{E_a}}}{{RT}}}}$
Where, $k$ is the rate with which the reaction proceeds
$A$ is the Arrhenius constant
$R$ is the gas constant
$T$ is the temperature at which the reaction proceeds
And ${E_a}$ is the activation energy or the activation barrier of the reaction
Complete step by step answer:
For the reaction carried out at $500{\text{K}}$, the Arrhenius equation can be written as
${k_{500}} = A{e^{ - \dfrac{{{E_{500}}}}{{R \times 500}}}}$
For the reaction carried out at $400{\text{K}}$, the Arrhenius equation can be written as
${k_{400}} = A{e^{ - \dfrac{{{E_{400}}}}{{R \times 400}}}}$
It is given in the question that both the reactions proceed at the same rate. Thus we can equate the two equations and we get
$A{e^{ - \dfrac{{{E_{400}}}}{{R \times 400}}}} = A{e^{ - \dfrac{{{E_{500}}}}{{R \times 500}}}}$
Simplifying:
\[ \Rightarrow \dfrac{{{E_{400}}}}{{400}} = \dfrac{{{E_{500}}}}{{500}}\]
Or, ${E_{500}} = \dfrac{{{E_{400}}}}{4} \times 5$
In the question, it is given that the activation energy is reduced by $20{\text{kJ mo}}{{\text{l}}^{ - 1}}$ for the second reaction. Thus, we can write, ${E_{500}} = {E_{400}} + 20$
Substituting this value of the activation energy in the previous equation, we get,
${E_{400}} + 20 = {E_{400}} \times 1.25$
Therefore,${E_{400}} = \dfrac{{20}}{{0.25}} = 80{\text{ kJ mo}}{{\text{l}}^{ - 1}}$
And, ${E_{500}} = 100{\text{ kJ mo}}{{\text{l}}^{ - 1}}$
Note:
Activation energy is the energy that we need to provide to compounds in order for a chemical reaction to take place. The activation energy $\left( {{E_a}} \right)$ is commonly measured in joules per mole. Activation energy can be considered as the magnitude of the energy barrier separating the initial and final thermodynamic states, namely the reactants and products. For a chemical reaction to occur at a good rate, the temperature of the system should be so high that the number of molecules with energy levels greater than or equal to the activation energy is appreciable.The activation energy of a reaction can be reduced by addition of suitable catalysts that make the reaction more feasible and faster.
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