212) the rate of a 1st order reaction is 0.08 mol/l/s at 10 mins and 0.03 mol/l/s at 20 mins after initiation. the half life of the reaction is?
a)3/log(8/3)
b) log(3/8)/3
c)3+log(3/8)
d)log(8/3)/3
1) on introducing a catalyst at 500k the rate of 1st order reaction increases by 2 times. the activation energy in the presence of a catalyst is 5 kj/mol. then what will be the Ea in the absence of the catalyst.?
a)9.6 kj/mol
b)4.2 kj/mol
c) 12.5 kj/mol
d) 3.8 kj/mol
-
UP 0 DOWN 0 0 8
8 Answers
21
13571)
lnK=lnA-Ea/RT
lnK1=lnA-E1/RT
lnK2=lnA-E2/RT
ln(K2/K1)=(E1-E2)/RT
E1-E2=RTln2
1From the radioactive decay law,R=R0e-kt
where R0 is the initial decay rate.
now substitute the given values of rates given at the 2 given times.
0.08=R0e-10k ....(1)
0.03=R0e-20k ....(2)
dividing (1) by (2),we get
e10k=8/3
=>k=1/10 ln (8/3) =1/10 *(2.303 log (8/3))
also t1/2=0.693/k =0.693/0.23 log (8/3)
therefore t1/2=3/log (8/3)
1RATE=K[CONC.]
C1C2=83
2.303/K*log[c1c2]=10
now t(half)=ln2/k.........
get A as ans