Correct Answer - a.Ea=76.623kJmol-1
b.k=1.042×10-2s-1
a.
k298K=
2.303
t1
log(
100
100-10
)=
2.303
t1
log
10
9
=
2.303
t1
×(0.0458)
=
0.1055
t1
∴t1=
0.1055
k28K
...(i)
b.
k308K=
2.303
t2
log(
100
100-25
)=
2.303
t2
log(
4
3
)
=
2.303
t2
×(0.125)
=
0.2879
t2
ort2=
0.2879
k308K
....(ii)
Since t1=t2
hence Eqs.(i)
and (ii)
are equal.
∴
0.1055
9
k298K0=
0.2879
k308K
or
k308K
k298K
=2.7289...(iii)
Using Arrhenius equation,
log
k308K
k298K
=
Ea
2.303R
(
T2-T1
T1T2
)
From Eq.(iii)
,
∴log2.7289=
Ea
2.303×8.314
×
10
298×308
orEa=76.623kJmol-1
c.
Calculation of k
at 318K:
logk=logA-
Ea
2.303RT
=log(4×1010)s-1
-
76.623×1000Jmol-1
2.303×8.314JK-1mol-1×318K
=10.6021-12.5843
=-1.9822
or k=Antilog(-1.9822)=Antilogbar(1).0178
=1.042×10-1s-1
