where R is the gas constant (8.314 J/molK) The activation energy of a certain re

Question

where R is the gas constant (8.314 J/molK)

The activation energy of a certain reaction is 38.2 kJ/mol . At 26  C , the rate constant is 0.0120s1. At what temperature in degrees Celsius would this reaction go twice as fast?

Express your answer with the appropriate units.

Part B

Given that the initial rate constant is 0.0120s1 at an initial temperature of 26  C , what would the rate constant be at a temperature of 200  C for the same reaction described in Part A?

Express your answer with the appropriate units.

Explanation / Answer

Given:

K1=0.0120 s-1

Temp. T1=26? =299 K

Ae=31.6 kJ/mol=31600 J/mol

T2=200 ? = 473 K

From arrhenius equation:

ln(K2/K1­)=Ea/R (1/T1 -1/T2)

ln(K2/0.012)=31600/8.314 *(1/299 -1/473)

K2=1.057 s-1

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.