Need solution for Part B, the answer I provided was incorrect. The Arrhenius equ

Question

Need solution for Part B, the answer I provided was incorrect.

The Arrhenius equation shows the relationship between the rate constant k and the temperture T in kelvins and is typically written as k = Ae-Ea/RT where R is the gas constant (8.314 J/mol .k) ,A is a constant called the frequency factor ,and Ea is the activation energy for the reaction. However, a more practical form of this equation is In k2/k1 = Ea/R (1/T1 - 1/T2) which is mathmatically equivalent to In k1/k2 = Ea/R (1/T1 - 1/T2) where k1 and k2 are the rate constant for a single reaction at two different absolute tempertures (T1 and T2) The activation energy of a certain is 43.8 KJ/mol. At 30 degrees C, the rate constant is 0.0190s-1 .At what temperature in degrees Celsius would reaction would this reaction go twice as fast? Express your answer with the appropriate units. Given that the intial rate constant is 0.0190s-1 ai an intial temperature of 30 degrees C. what would the rate constant be at a temperature of 180 degrees C for the same reactin described in part A? Express your answer with the appropriate units.

Explanation / Answer

1) since the reaction rate increases two folds

Thre fore k2/k1 = 2 ; where k1 & k2 are the rate constants of the reaction at temperature T1 & T2 respectively

Now, as per Arrhenius equation:-

ln(k2/k1) = (Ea/R)*[(1/T1) - (1/T2)] ; where Ea = activation energy in J/mole ; R = universal gas constant = 8.314 J/(mole*K)

or, ln2 = (43800/8.314)*[(1/303) - (1/T2)]

or, 315.581 = T2

Thus, temperature in 0C = 315.581 - 273 = 42.581 0C = 42.6 0C

2) as per Arrhenius equation:-

ln(k2/k1) = (Ea/R)*[(1/T1) - (1/T2)] ; where Ea = activation energy in J/mole ; R = universal gas constant = 8.314 J/(mole*K)

ln(k2/0.019) = (43800/8.314)*[(1/303) - (1/453)]

or, ln(k2/0.019) = 5.757

or, k2/0.019 = e5.757

or, k2/0.019 = 316.398

or, k2 = 6.013 s-1

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