Chemistry 121A first order chemical reaction is given as below:A→PAbove reaction is occurring at 500 K. If the same reaction is to be carried out in presence of a catalyst, temperature of 400 K is required, to maintain the same rate of reaction. If catalyst lowers the activation energy by 30 kJ, then find out value of (t25).(t is value of activation energy of the original pathway (uncatalyzed reaction)) (inkJmole)
Question
Chemistry 121A first order chemical reaction is given as below:A→PAbove reaction is occurring at 500 K. If the same reaction is to be carried out in presence of a catalyst, temperature of 400 K is required, to maintain the same rate of reaction. If catalyst lowers the activation energy by 30 kJ, then find out value of (t25).(t is value of activation energy of the original pathway (uncatalyzed reaction)) (inkJmole)
Solution
To solve this problem, we need to use the Arrhenius equation which relates the rate constant of a reaction with the temperature and activation energy. The equation is:
k = Ae^(-Ea/RT)
where: k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature.
Given that the rate of the reaction remains the same with and without the catalyst, we can set up the following equation:
k1 = k2
Ae^(-Ea/RT1) = Ae^(-Ea'/RT2)
where: Ea is the activation energy of the uncatalyzed reaction, Ea' is the activation energy of the catalyzed reaction, T1 is the temperature of the uncatalyzed reaction, and T2 is the temperature of the catalyzed reaction.
We know that Ea' = Ea - 30 kJ, T1 = 500 K, and T2 = 400 K. Substituting these values into the equation, we get:
e^(-Ea/500) = e^(-(Ea-30)/400)
Solving this equation for Ea, we get:
Ea = 30 kJ * (500/100) = 150 kJ/mole
So, the activation energy of the original pathway (uncatalyzed reaction) is 150 kJ/mole.
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