?attempt 502190 DEL Bài giáng tham khao Thóng ke NguroI dùng Vietnamese (vi- Quó

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?attempt 502190 DEL Bài giáng tham khao Thóng ke NguroI dùng Vietnamese (vi- Quóc Té » Khoa Ky thuät Hóa hoc » Hóa ý 2 (CH2031) Nguyån Quang Long (OT HK171) Ch dé 9 > Group quiz-171 a. The decomposition of A to produce B can be wrtten as A--B When the intail concentration of A l5 0012 M,the rate s 0.0018 M/min and when the initial concentration of A is 0.024 M.the rate is 00036M/min. - Write the rate law for the reaction. - Ir the activation energy for the reaction is 258 kJ/mol and the rate constant at 60 K is 8.1 x 10n-3 (1/ 9) what willbe the rate constant at 90 K? b. Suppose a reaction has a rate constant of 0.240 energy for this reaction? 103 (1/ s) at 0 (oC) and 2.65 x 10-3 (1/ sat 24 (0C), What is the activation

Explanation / Answer

a)

i) Rate = k[A]

rate is doubling when concentration is doubling.So, this reaction is first order reaction

ii) Arrhenius equation is

k1 = Ae^-Ea/RT1

k2 = Ae^-Ea/RT2

k1/k2 = e^-Ea/RT1/ e^-Ea/RT2

ln(k1/k2) = -Ea/RT1 + Ea/RT2

ln(k1/k2)= - Ea/R( 1/T1 - 1/T2)

log(k1/k2) = - Ea/2.303R(1/T1 - 1/T2)

= -268000(J/mol)/2.303×8.314(J/Kmol)(1/660K - 1/690K)

= -13997K × 6.6×10^-5K^-1

=- 0.92

8.1×10^-3s^-1 /k2 = 1×10^-0.92

k2 = 0.0081s^-1/0.120

= 6.75×10^-2s^-1

b) log(k1/k2) = -(Ea/2.303R)( 1/T1 - 1/T2)

log(0.240×10^-3s^-1/2.65×10^-3s^-1) = - (Ea/2.303 × 8.314(J/Kmol) ( 1/273.15K - 1/297.15K)

log (9.06×10^-2)= -(Ea/ 19.15(J/mol))(0.000296)

Ea = -(log(9.06×10^-2))×19.15(J/mol)/0.000296)

= 67.47kJ/mol

  

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