The rate constant of the first order reaction is 4.20*10^-4s^-1 at 350.C degrees
ID: 985214 • Letter: T
Question
The rate constant of the first order reaction is 4.20*10^-4s^-1 at 350.C degrees. If the activation energy is 103kj/mol calculate the temperature at which its rate constant is 8.95*10^-4s^-1. Need answer in Celsius degrees please
connect. CHM 112 Spring 2016: CHM 112-M6 CHEMISTRY Quiz Chap.13 Question 2 (of 5) 2. Question Enter your answer in the provided box. Assistance The rate constant of a first-order reaction is 4.20 x 10 sat 350. °C. If the activation energy is 103 kl/mol, ealculate the temperature at which its rate constant is8.95 x 10 4s Show Me Practice This Question Question Help Report a Problem esc : 0Explanation / Answer
This problem cab be solved by using Arrhenius equation.
Arrhenius equation k = A e-Ea/RT
where k = rate of reaction
A = collision frequency
Ea = activation energy
R= universal gas constant = 8.314 J/K/mol
T = temperature
Arrhenius equation can be written as
In (k2/k1) = Ea/R (1/T1 - 1/T2) -- Eq (1)
Given that Ea = 103 kJ/mol = 103000 J/mol
Initial rate of reaction k1 = 4.2 x 10-4
Final rate of reaction k2 = 8.95 x 10-4
Initial temperature T1 = 350oC = 350 + 273 K = 623 K
Final temperature T2 = ?
Substitute all these values in eq (1),
In (k2/k1) = Ea/R (1/T1 - 1/T2) -- Eq (1)
In ( 8.95 x 10-4 / 4.2 x 10-4) = [103000/8.314] [ (1/623) - (1/T2)]
[(1/623) - (1/T2)] = [In ( 8.95 x 10-4 / 4.2 x 10-4)] x [8.314 / 103000 ]
(1/623) - [In ( 8.95 x 10-4 / 4.2 x 10-4)] x [8.314 / 103000 ] = 1/T2
On solving ,
T2 = 647.6 K
= 647.6 -273 oC
= 374.6 oC
T2 = 374.6 oC
Therefore,
required temperature = 374.6 oC
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