Question

# The rate constant of a chemical reaction increased from 0.100 s−1 to 3.10 s−1 upon raising...

The rate constant of a chemical reaction increased from 0.100 s−1 to 3.10 s−1 upon raising the temperature from 25.0 ∘C to 51.0 ∘C .

Part A

Calculate the value of (1/T2−1/T1) where T1 is the initial temperature and T2 is the final temperature.

Part B

Calculate the value of ln(k1/k2) where k1 and k2 correspond to the rate constants at the initial and the final temperatures as defined in part A.

Part C

What is the activation energy of the reaction?

Answer – Given, rate constant, k1 = 0.100 s-1 , rate constant, k2 = 3.10 s-1

T1 =25+273.15 = 298.15 K , T2 = 51 +273.15 = 324.15 K

Part A) In this one we need to calculate (1/T2-1/T1)

= 1/ 324.15 K – 1/ 298.15 K

= - 0.00027 K

Part B) Calculate the value of ln(k1/k2)

= ln 0.100 s-1 / 3.10 s-1

= -3.43

Part C) The activation energy of the reaction- We know formula

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

-3.43 = Ea/ 8.314J.mol-1.K-1 * (-0.00027 K)

Ea = -3.43 *8.314 / -0.00027

= 1.06*105 J/mol

= 1.06*102 kJ/mol

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