Question

4) A particular irreversible chemical reaction follows the rate law rate = k[A] At a particular temperature the value for the rate constant is k = 4.22 x 10+s! a) In an experiment carried out at that temperature the initial concentration of A in a system is [A]o = 9.42 x 102M. What will the concentration of A be in the system at t = 400.s? [8 points) b) A plot of Ink vs (1/T) for the rate constant for this reaction is given below. Based on this plot, find the value for E., the activation energy of the reaction. [8 points) . 7.5 Ink) .8.0 .85 9.0 9.5 00310 00315 00320 00325 00330 00335 1/T (K)

Answer 1

Date 1 Page RANKA ca) Rate = kra] and Given that :- Ke 4.22 x 10 4 g) LA)= 9:42 X 10-20 t400g from the unit of rate constant (K), it is well predicted, it is First Orden rah, ta 2-303 log [A] * 400 = 20303 Tog Ао A 40 22 x 10-4 400 x 4032x10-4 - logicament 2.3.3 antitog ( 0704329) A 100.07329 101838 A AD 1-1838 A Go 957x102 M = A 9.42 X 102 101838

Pago RANK According to Arhenius eah Ko te talRt with matc thee yo tits as then Ea/RT Comparing yo function q + sloße= - so for finding actruction Energy Co, we have to find slope from graph -865 - (-8.0) m= R 1 slope Yay A - 2 0.00325 - 0.00815 =-0.5 5000 0.0001 =-5000 slobe Now putting this into Eqh D - 5000 = - Ea 8.314 EQ = 3.374 X 500D fa - 41.5.70 KJ