Question

In: Chemistry

Determine the activation energy (kJ) from the following table. Temperature (K) 427 417 &

Determine the activation energy (kJ) from the following table.

Temperature (K) 427 417 409 407 399

k (s^-1) 0.00108 0.000410 0.000208 0.00016 0.0000763

A.25
B.77
C.89
D.134

You have discovered a new enzyme and want to characterize its function. The reaction catalyzed is 20 times more rapid at 25° C than at 4° C. Calculate the activation energy (kJ) for the reaction.

A.24
B.41
C.83
D.100

Expert Solution

logK2/K1 = Ea/2.303R[1/T1 -1/T2]

log0.00108/0.0000763 = Ea/2.303*8.314 [1/399 -1/427]

1.1508 = Ea/19.147 (0.0025-0.00234)

Ea = 1.1508*19.147/(0.0025-0.00234)

= 1.1508*19.147/0.00016

= 137714 J = 137.714KJ

logK2/K1 = Ea/2.303R[1/T1 -1/T2]

T2 = 25C = 25+273 = 298K

T1 = 4C = 4+273 = 277K

log20 = Ea/2.303*8.314 [1/277-1/298]

1.3010 = Ea/19.147 (0.0036-0.00335)

Ea = 1.3010*19.147/(0.0036-0.00335)

= 24.91/0.00025 = 99640J = 99.640KJ = 100KJ

queen_honey_blossom answered 2 years ago

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