Chemistry Chemical Kinetics

Kt = 2.303 log? (A? /A? )
K * 570 = 2.303 log? (100/32)
K = [2.303 / 570] [log?10² - log?2? ]
K = [2.303 / 570] [2 - 5 * 0.301] = [2.303 / 570] * 0.495 = 0.002 = 2.0 * 10? ³ s? ¹
Ans = 2

Industries manily select an optimal temperature which is high enough to increase the reaction rate (molecules colliding with each other at higher fequencies) and at the same time is balanced to avoid any side effect. This effectively yields high quality production without any wastage or leading to harmful results while ensuring safety of the consumers too.

For A
0.693/300 = (2.303/t) log (A? /A? )

For B
0.693/180 = (2.303/t) log (B? /B? )

Given A? = B? & A? = 4B?
Substituting & solving we get t = 900 s

To understand the phenomenon behind this, have a close look at the formula of the first order reaction.

t1/2 = 0.693/k

Here, we can see that the half life is only dependent on k (rate constant) since there is no [A]' in the formula. Hence, we can easily conclude that the half life of a first order reaction is independent of the initial concentration [A]'.

log (k? /k? ) = (Ea / 2.303R) [1/T? - 1/T? ]
log (3.555) = (Ea / (2.303R) [1/303 - 1/313]
1.268 * 8.314 * 303 * 313 = 10Ea
So, Ea = 100 kJ

k = Ae? Ea/RT
ln (K? /K? ) = (Ea/R) [1/T? - 1/T? ]
ln (5) = (Ea/8.314) [1/300 - 1/315]
1.6094 = (Ea/8.314) [15 / (300 * 315)]
Ea = 84297 J

first order reaction
K = (2.303/t) log (a? / (a? - x)
K = (2.303/90) log (a? / 0.25a? ) = 0.0154
t = 60% = (2.303/K) log (a? / a? ) . (2)
= (2.303 / 0.0154) x (1 - 0.602) = 59.51 mins ≈ 60

SN2 is a bimolecular reaction. This is because as per the rate law,
Rate = k [substrate] [nucleophile]

Here, the rate determining step is dependent on both substrate and nucleophile, unlike SN1 who only involves substrate. Therefore, SN2 is proved to be bimolecular.