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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 : 0

Explanation / 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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