The rate constant of a chemical reaction increased from 0.100 s1 to 3.00 s1 upon

ID: 481490 • Letter: T

Question

The rate constant of a chemical reaction increased from 0.100 s1 to 3.00 s1 upon raising the temperature from 25.0 C to 35.0 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.

Express your answer numerically.

Part C

What is the activation energy of the reaction?

Express your answer numerically in kilojoules per mole.

A) Here T1 = Final temp = 35oc = 35 + 273 = 308 k

T2Initial temp = 25 oc = 25 + 273 = 298 K

Now, [ 1 / T2 - 1 / T1 ]

= [ 1 / 298 - 1 / 308]

= 0.3355 - 0.003246

= 0.0109

B) Here K1 = Initial rate = 0.100 s1

K2 = Final rate = 3.00 s1

Then ln [ 0.100 / 0.300 ]

= ln 0.333

= - 1.099

C ) Activation energy at two different temp is given by,

ln [K1 / K2 ] = Ea / R [ 1 / T2 - 1/T1 ]

Therefore,

Ea = ( [ R x T1 x T2 ] / [ T1 - T2 ] ) x ln [ K1 / K2 ]

Ea = ( [ 8.314 x 308 K x 298 K ] / [ 308 - 298 ]) x - 1.099

= [ 763092.17 / 10 k ] x ( - 1.099 )

= ( 76309.21 ) x ( - 1.099 )

= - 83863.83 J/mol

= - 83863.83 / 1000 [ 1 KiloJoule = 1000 J ]

= - 83.863 KJ / mol

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