A. The activation energy of a certain reaction is 47.8 k J / m o l . At 28 ? C ,

Question

A. The activation energy of a certain reaction is 47.8kJ/mol . At 28 ?C , the rate constant is 0.0190s?1. At what temperature in degrees Celsius would this reaction go twice as fast?
B. Given that the initial rate constant is 0.0190s?1 at an initial temperature of 28 ?C , what would the rate constant be at a temperature of 180 ?C for the same reaction described in Part A?

A. The activation energy of a certain reaction is 47.8kJ/mol . At 28 ?C , the rate constant is 0.0190s?1. At what temperature in degrees Celsius would this reaction go twice as fast?
B. Given that the initial rate constant is 0.0190s?1 at an initial temperature of 28 ?C , what would the rate constant be at a temperature of 180 ?C for the same reaction described in Part A?

Explanation / Answer

Ea = 47800 J/mol

ln (K2 / K1) = (Ea / RT) x (1/T1 - 1/T2)

K2 / K1 = 2 , for reaction to go twice as fast.

=> ln 2 = (47800 / 8.314) x (1/301 - 1/T2)

=> T2 = 312.33 K = 39.33 degree C

B) ln (K2 / 0.019) = (47800 / 8.314) x (1/301 - 1/453)

=> K2 = 11.54 s-1

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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