Question

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$\large\bold\red{Q.}$ A hydrogenation reaction is carried out at 500 K. If the same Reaction is carried out in the presence of a catalyst at the same rate with same frequency factor, the temperature required is 400 K. What is the activation energy of the Reaction, if the catalyst lowers the activation energy barrier by 16 KJ/mol ??​

Answer 1

$\Large\underline{\underline{\sf \blue{Given}:}}$

• Hydrogenation Reaction is carried out at $\sf{(T_1)=500\:K}$

• Same Reaction is carried out in presence of catalyst $\sf{(T_)=400\:K}$

• Rate of both reaction is same

•°• $\sf{R_1=R_2=R}$

• Catalyst lower activation energy by $\sf{E_{a1}-E_{a2}=16\:KJ/mol}$

$\Large\underline{\underline{\sf \blue{To\:Find}:}}$

• Activation Energy of the Reaction = ?

$\Large\underline{\underline{\sf \blue{Solution}:}}$

Arrhenius Equation :

$\large{\boxed{\sf K=Ae^{-E_a/RT} }}$

Here ,

K = Rate Constant

A = Pre-exponential factor

T = Temperature in kelvin

$\sf{E_a}$ = Activation Energy

R = Rate

In logorithmic Expression :

$\large{\boxed{\sf logK=\frac{E_a}{2.303RT}}}$

In absence of Catalyst :

$\implies{\sf logK_1=\dfrac{E_{a1}}{2.303RT_1} }$

In presence of Catalyst :

$\implies{\sf logK_2=\dfrac{E_{a2}}{2.303RT_2} }$

When ,

$\sf{logK_1=logK_2}$

$\implies{\sf \dfrac{E_{a1}}{2.303RT_1}=\dfrac{E_{a2}}{2.303RT_2} }$

$\implies{\sf \dfrac{E_{a1}}{T_1}=\dfrac{E_{a2}}{T_2} }$

$\sf{E_{a2}=E_{a1}-16}$ Given

$\implies{\sf \dfrac{E_{a1}}{500}=\dfrac{E_{a1}-16}{400} }$

$\implies{\sf \dfrac{E_{a1}}{E_{a1}-16}=\dfrac{500}{400} }$

$\implies{\sf 4E_{a1}=5(E_{a1}-16) }$

$\implies{\sf 4E_{a1}=5E_{a1}-80}$

$\implies{\sf 80=5E_{a1}-4E_{a1}}$

${\boxed{\sf E_{a1}=80\:KJ/mol}}$

$\implies{\sf E_{a2}=E_{a1}-16 }$

$\implies{\sf E_{a2}=80-16}$

${\boxed{\sf E_{a2}=64\:KJ/mol} }$

$\Large\underline{\underline{\sf \blue{Answer}:}}$

•°• Activation Energy of the Reaction is 80 KJ/mol