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If the rate constant for a reaction triples when the temperature rises from 300. K to 310. K, what is the activation energy of the reaction

User ShawnOrr
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Final answer:

The activation energy of a reaction can be calculated using the Arrhenius equation. In this case, if the rate constant triples when the temperature rises from 300 K to 310 K, the activation energy can be determined using the equation ln(2) = (-Ea/R) * (1/T2 - 1/T1), where Ea is the activation energy. Solving this equation gives the value of Ea.

Step-by-step explanation:

The activation energy of a reaction can be calculated using the Arrhenius equation. The Arrhenius equation is given as:

k = Ae^(-Ea/RT)

where k is the rate constant, A is the pre-exponential factor or frequency factor, Ea is the activation energy, R is the ideal gas constant, and T is the temperature in Kelvin. In this case, we know that the rate constant triples when the temperature rises from 300 K to 310 K. Since we are given a doubling of the reaction rate with a 10°C increase in temperature, we can set up the following equation:

ln(2) = (-Ea/R) * (1/T2 - 1/T1)

Substituting the known values, we get:

ln(2) = (-Ea/8.314 J/Kmol) * (1/(310 K) - 1/(300 K))

Simplifying the equation, we can solve for Ea:

Ea = -8.314 J/Kmol * ln(2) / ((1/(310 K) - 1/(300 K)))

User Kumbhani Bhavesh
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