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Answer 1

Haii there !!

Ques 31

31 log = k2/k1 R ( 1/T1 , -1/T2 )
0.61=Fg/2.300 x 831 = ( 1/300-1/310)
Ea = 1.07.25 Kj/mol
K = AC = -Ea/RT
A= 2.12 x 10^-9
= 9.15 x 10^14 Answer

Ques 32

KT = 0.63/2 x 3600 = 9.62 x 10^-5
log = 9.62x 10-5/3.96 x 10^-5

1/295 - 1/T

0.994 x 2.303 x 8.314/
106 x 10^3 = 0.0032

T = 312.51 Answer

Enjoy !!

Answer 2

Answer:

32

Determine the temperature at which the half-life for the decomposition of N20s is 2 hr. Given at 298 K the rate constant is 3.46 x 10-45-1, and Ea = 106 kJ/mole. (R = 8.314 JK-1 mol-1) 3 Tb temnoturmuotti 1.1 MA 16 Nha Lanted to be