± Reaction Rates and Temperature The rate constant of a chemical reaction increa

Question

± Reaction Rates and Temperature

The rate constant of a chemical reaction increased from 0.100 s1 to 2.80 s1 upon raising the temperature from 25.0 C to 41.0 C .

Part A

Calculate the value of (1T21T1) where T1 is the initial temperature and T2 is the final temperature.

Express your answer numerically.

Part B

Calculate the value of ln(k1k2) 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?

Express your answer numerically in kilojoules per mole.

Explanation / Answer

Answer - We are given, T1 = 25+273.15 = 298.15 K , T2 = 41.0+273.15 = 314.15 K , k1 = 0.100 s-1 , k2 = 2.80 s-1 , Ea = ?

Part A) We need to calculate the 1/T2 – 1/T1

= 1/314.15 K - 1/298.15 K

= -0.00017

Part B) ln k1/k2

= ln 0.100 / 2.80

= ln 0.0357

= -3.33

Part C) Activation energy of the reaction- We know the intergraded Arrhenius equation

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

-3.33 = Ea/ 8.314 J/mol.K * (-0.00017)

-3.33 = Ea/ 8.314 J/mol.K * -0.00042

So, Ea = -3.33 * 8.314 J. mol-1.K-1 / -0.00017

             = 162178.6 J/mol

             = 162.2 kJ/mol

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