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Secure https//session.masteringchemistry.com/myct/itemView?offset next&assignmentProblemID-95612127; 1. Chemical Kinetics The Arrhenius Equation 11 of 25 Constants1 Periadic Table Part A The Arrhenius Equation is typically written as Tha activation anargy of a cartain reaction is 36 5 k.I/mol At 20 C, tha rata constant is 0.0130s Express your answer numerically in degrees Celsius View Available Hints) At what temperature would this raaction go twics as fast However, the tollowing more practical form of this equation also exists: where ki and k are the rate constants for a single reaction at two different absolute temperatures and T31, Subrnit Part B Given that the inilial rate constant is 0.0130s at an iniial teriperalure o20 C whst would the rale constant be at a temperature of 00 C Express your answer numerically in inverse seconds. View Available Hint(s) Subrnt Provide Feedback Next ED e 3:40 PM 432010

Explanation / Answer

Answer:

Part A: Given Arrhenius equation

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

Given k1=0.0130 s^-1, T1=20°C=20+273=293 K,

Ea=36.5 kJ/mol=36.5x10^3 J/mol, k2=twice that of k, k2=2x0.0130s^-1.

T2=?

ln[(2x0.0130s^-1)/(0.0130 s^-1)=(36.5x10^3 J/mol/8.314 J/mol.K)[(1/293K)-(1/T2)]

(1/293 K)-(1/T2)=1.5788x10^-4

1/T2=3.1978x10^-3K^-1

T2=312.713 K=312.713-273=39.71°C~40°C

Part B:

Given k1=0.0130s^-1 at T1=20°C=20+273=293 K,

k2=? at T2=100°C=373 K

ln(k2/0.0130 s^-1)=(36.5x10^3 J/mol/8.314 J/mol.K)[(1/298)-(1/373)]

ln(k2/0.0130 s^-1)=2.9622

k2/0.0130 s^-1=e^2.9622

k2=0.251 s^-1~0.25 s^-1.

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