The rate constant of a chemical reaction increased from 0.100 s1 to 2.60 s1 upon

Question

The rate constant of a chemical reaction increased from 0.100 s1 to 2.60 s1 upon raising the temperature from 25.0 C to 49.0 C . Part A Calculate the value of (1T21T1) where T1 is the initial temperature and T2 is the final temperature. Express your answer numerically. (1T21T1) = K1 SubmitHintsMy AnswersGive UpReview Part 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. Express your answer numerically. ln(k1k2) = SubmitHintsMy AnswersGive UpReview Part Part C What is the activation energy of the reaction? Express your answer numerically in kilojoules per mole. Ea = kJ/mol SubmitHintsMy AnswersGive UpReview Part

Explanation / Answer

Given that rate constant of a chemical reaction increased from 0.100 s1 to 2.60 s1 upon raising the temperature from 25.0 C to 49.0 C .

T1 = 25 oC = 25 + 273 K = 298 K

T2 = 49 oC = 49 + 273 K = 322 K

k1 = 0.1 s-1

k2 = 2.6 s-1

Activation energy Ea = ?

In (k2/k1) = [Ea/R] [(1/T1) – (1/T2) ]

Ea = R ln (k2/k1) / [(1/T1) – (1/T2)]

= (8.314 J/K/mol) In (2.6 /0.1) [(1/298) – (1/322) ]

= + 108301 J/mol

= +108.3 kJ/mol

Ea = +108.3 kJ/mol

Therefore,

Activation energy of the reaction = +108.3 kJ/mol

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