Question

1.The decomposition of N2O5 at 308K has a rate constant of 1.35x10
-4 s
-1
. The Arrhenius constant
for this reaction is 4.79x1013 s
-1
. Calculate the activation energy of this reaction. (R=8.31 JK- 1mol-
2.he same reaction is set up at a different temperature, and the rate constant is found to have a
value of 2.4x10
-3 s
-1
. Given that the Arrhenius constant for this reaction is 6.22x10
13 s
-1
, but all
other conditions remain the same, calculate the temperature at which this reaction occurs

Answer 1

KT -1 $31X30S 2559.45 The decomposition of Nzos bollowus ist andere kinetics Arrhenius egn is given by In x = In A - Es where to rate constant A = Arrahencen ena constant Given, k= 1.35X10451 ER zactivation énergy A = 4.79x 10'34' 8 = $.31Jk'mon and T = 30$X putting all these in the eq? — In (1-35x1024) = 1n (4.79x105) - et Ea 772-303 10g (1-35x134) =(og (4.39x105}]x a-303 => log (1-35X10%) = log (4.79x103) – Ea 4.79x1013 5994.48 log 1:35*104) = log (3-5X10%) = 17 +log(3-55) = 17+0:55=17-55 → Ex = 5894.48 x17.55=103448:124 J.mort n. Ex = 103.44 KJ. mort -EURT K=Ae .... Another terem ct Arrhenius ean X = A, e ki com o te th] ΑΙ K2=Az e -Ea/RT2 K2 A2 A1 = 4-79x10361 → K2= L (43-1)] K2 =2.4x10?5+ A2 = 6.22x1013 T2 = ??? 4.79x1d? : le des T, = 308K 1.35810 2.4 x 10-3 Ea 2303X2557.48 -E/RTi - e K1 =1.35X10-457 6:22x103 765625x103)=(7.7%107) e Cavalta-tos) Ea 7 in (5:625X10-2) = \n(7-77102)+ ene (tz- 108) = lng (5-625x1032) = 109 (7-7X10") + En → -2 +0.75=-1+0.89+ 2-303R (tz - 368) 2:303R ( tz-kos) 308) ► 308 – tz = 1.148831X2-303 = 0.00021 7 0.00324-0.00021 = ta = 0.00303 T2=330.03K -7.14 = Ea 2:303 R ( 21303R (sor - tz)=114 103448. 124 . T2 = O 00303 C

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